number theory
Exponential equation
IMO
Hi

Determine all pairs $(x, y)$ of integers such that \[1+2^{x}+2^{2x+1}= y^{2}.\]

$x < 0$: LHS integer iff $x =-1$, but then $LHS = 2 \neq y^{2}$.
$(x,y) = (0,2)$ is a solution.
for $x = 1,2$ no solution.
so assume $x > 2$. LHS is odd, so writing $y = 2n+1$ gives us $2^{x-2}(1+2^{x+1}) = n(n+1)$. $n,n+1$ are coprime, and so are $2^{x-2}, 1+2^{x+1}$. so $n = 2^{x-2}, n+1=1+2^{x+1}$ or vice versa, but both lead to a contradiction

But $( x,y)=(4,+-23)$ is also a solution

as i said about problem 1 i can't believe these are imo problems.i will continue from berti's post since i'm too lazy to write the steps again :P .it is clear now that $4n/2^{x}$ or $4(n+1)/2^{x}$ are integers.
case 1:$4n=a2^{x}\Rightarrow a^{2}*2^{x}+4a=8*2^{x}+4$.it is clear now that $a=1$ or $a=2$.in both cases we get a contradiction.
case 2:$4(n+1)=a*2^{x}$again it is easy to see that $a=1$or $a=2$or $a=3$.the only good one is $a=3 \Rightarrow x=4$ :P

This one wasnt so hard and I have solved it but my solution is reallz long but correct  :lol:

this is really a very straightforward problem..

Say that $x \geq 3$ (the other cases are easily solved).

$y^{2}\equiv 1$ is an odd quadratic residue modulo $2^{x}$. I posted a problem some time ago about such odd quadratic residues. ZetaX proved that all residues $\equiv 1 \left( \bmod ~ 8 \right)$ are quadratic residues modulo $2^{x}$ and that they correspond to exactly four numbers. If I remember correctly, you needed to see in which conditions $x_{1}^{2}\equiv x_{2}^{2}$ takes place.

Next, $y \equiv 1,-1,2^{x-1}-1 \, \, \textrm{or}\, \, 2^{x-1}+1 \left( \bmod ~ 2^{x}\right)$, so $y \equiv \pm 1 \left( \bmod ~ 2^{x-1}\right)$. Take $y = k \cdot 2^{x-1}\pm 1$, so that the given relation rewrites as $\left( k^{2}-8 \right) 2^{x-2}= 1 \mp k$. Now, we have $k^{2}-8 \geq 1-k$ or $k^{2}-8 \geq 1+k$, except for a small set of numbers, which can be manually checked.

yes again easy number theory problem  :)  but this year there are less pefect scorers i think going to be
for now I know just three of them also I am not sure please note this :

Iurei Boreico
Alexandr Magazinov
Chinese guy -

Another easy one!

My solution:

It is easy to show that $x \geq 0$

Now it is obvious for x=0.

Now y is odd. Say: $y=2p+1$ where $p \geq 0$

Let’s check for x=1 and 2, and opine that $x\geq 3$

Say: $z=2^{x-2}$

Then, $z+8z^{2}=p(p+1)$
implies, $z(1+8z) = p(p+1)$

it is clear that, $p \geq z$

And one of p or p+1 is even, so: either: $p=kz$ or $p+1=kz$

if $p=kz$ then $z=\frac{1-k}{k^{2}-8}$ and it can be easily shown that, maintaining all the restrictions there is no value for k for which z can be postive.

So, $p+1=kz$

for it:
$z=\frac{1+k}{k^{2}-8}$

this time, for z to be postive: $k \leq 3$ checking the values k=3 leads z an integer. And it leads to the soution: $(2,23), (2,-23)$

i think this was an ugly problem...the short list must have had ugly number theory problems.

Last year's number theory problem was very easy too, but in my opinion, nicer.

Hi all.
My sol. was similar offical solution; it became different at step $1+m = 2^{x-2}(m^{2}-8)$
(I assumed $y \geq  0$ since $(x,y); (x,-y)$ both are solutions.)
Let $2^{x-2}= a$ (then $a > 0$; it's easy to check theres aren't solutions if x<0 so a is integer.)
Then $m = ak-1$ where k - positive integer. So $ak = a(a^{2}k^{2}-2ak-7)$;
$k(a^{2}k-2a-1)=7$ It means k|7, so k = 1 or 7. If k=7, a isn't integer. If k=1 a = -2 (but $a > 0$) or 4.
So k=1, a=4; that gives x=4, y =23 and the final answer is:
${(0;2);(0;-2);(4;23);(4;-23)}$

The answer is $(0, \pm 2), (4, \pm 23)$.  These obviously work.

[hide=Mega Fakesolve, please ignore]Notice that we can rewrite the original equation as $$(2+1)(2^x+1) = y^2+2 \implies 3(2^x+1) = y^2+2.$$Now obviously, $y = 3k+1$ for some integer $k$.  Then after expanding, $$2^x = 3k^2+2k.$$If $x = 0$, we have the solution $y = \pm 2$.  Otherwise, we can again substitute $k = 2n$ to get $$2^x = 12n^2+4n \implies 2^{x-2} = 3n^2+n = n(3n+1).$$ Consequently, $n$ must be a power of $2$.  However, $3n+1$ is odd when $n$ is even, so $$n = 1 \implies 2^{x-2} = 4 \implies x = 4 \implies y = \pm 23.$$We have exhausted all cases, and so this is the answer.$\square$[/hide]

$x$ cannot be less than $-1$ because if $x=-a$ where $a>1$ is an integer, then $2^x+2^{2x+1} = \frac{1}{2^a} + \frac{1}{2^{2a-1}} = \frac{2^{a-1}+1}{2^{2a-1}}$ which cannot be simplified because the numerator is odd and the denominator is even.

$1+2^x+2^{2x+1}\equiv 1 + (-1)^x + (-1)^{2x+1}\equiv (-1)^x\pmod 3$, but since perfect squares must be $0$ or $1\pmod 3$, $x$ must be even so we let $x=2k$. Then, $y$ can be written as either $3a+1$ or $3a-1$.

If $y=3a+1$, $2^{2k}+2^{4k+1}=9a^2+6a$. If $a$ is odd then $k=0$, giving $1+2=9a^2+6a$ and $a=-1$ satisfies it so the pairs $(0, -2)$ and $(0, 2)$ work.
If $k>0$, then $9a^2+6a$ must be even, so $a$ must be even, so we can write $a=2a_1$ and $y=3\cdot 2^na_n+1$. We get $2^{2k-2}+2^{4k-2}=9a_1^2+3a_1$. We don't have an integer solution for $a_1$ when $k=1$, so if $k>1$, the LHS must be even, so $a_1$ must be even and we let $a_1=2a_2$ which gives us $2^{2k-3} + 2^{4k-2}=18a_2^2 + 3a_2$, but we can't get the LHS to be odd, so we write $a_2=2a_3$. That gives us $2^{2k-4}+2^{4k-3} = 36a_3^2 + 3a_3$ and when the LHS is odd, we get $1+32=36a_3^2+3a_3$ and $a_3=-1$ is the integer solution, so the pairs $(4, 23)$ and $(4,-23)$ work.
The equation $2^{2k-2}+2^{4k-2}=9a_1^2+3a_1$ is of the form $2^{2k-n-1}+2^{4k-n}=9\cdot 2^{n-1}a_n^2 + 3a_n$. When $a_n=2a_{n+1}$, we get $2^{2k-n-1}+2^{4k-n}=9\cdot 2^{n+1}a_{n+1}^2 + 6a_{n+1}$ which simplifies to $2^{2k-n-2}+2^{4k-n-1}=9\cdot 2^{n}a_{n+1}^2 + 3a_{n+1}$ so by induction, this holds for all integers $n\geq 1$.

Clearly, to check for solutions, we only need to consider odd $n$s, since for the exponent of $2^{2k-n-1}$ to be $0$, $k=\frac{n+1}{2}$ must be an integer.
Plugging in $k=\frac{n+1}{2}$, we get $1 + 2^{n+2}=9\cdot 2^{n-1}a_n^2 + 3a_n$.
When $n=5$, we get $1+128=144a_5^2 + 3a_5$ and each time $n$ increases by 2, $2^{n+2}$ and the coefficient of $a_5^2$ are both multiplied by $4$. The RHS would be much too large if $|a_n|>1$, and if $a_n=0$, the RHS would be too small, meaning $a_5=\pm 1$, but that is impossible since $144a_n^2 - 2^{n+2}>4$ for $n>3$, so there are no more solutions when $y=3a+1$.

Now for the case $y=3a-1$ we get $2^{2k}+2^{4k+1}=9a^2-6a$. Notice that this case is identical to the previous ones, except our solutions we got for $(k, a_n)$ can be replaced with $(k, -a_n)$, and the value of $y^2$ will be the same, because $y=3\cdot 2^n \cdot -a_n - 1 = -(3\cdot 2^na_n+1)$, so we are done looking for solutions in all our cases and the only solutions are $\boxed{(0, 2), (0, -2), (4, 23), (4, -23)}$.

If $x$ is negative, then it is obvious $1<1+2^x+2^{2x+1}<4$, so it can't be a square. Checking $x=0,1,2$, we get $(x,y)=(0,2),(0,-2)$ as a solution pair. Now assume $x>2$ and $y>0$, as the other is symmetric. Then note that $y$ is odd, so let $y=2a+1$. Then $$2^{x-2}(2^{x+1}+1)=a(a+1).$$ Since $a$ and $a+1$ are relatively prime, then $a$ or $a+1$ is divisible by $2^{x-2}$. We proceed with casework.
Case 1 ($a=2^{x-2}b$): Then $$b(2^{x-2}b+1)=2^{x+1}+1 \implies b-1 = 2^{x-2}(8-b^2).$$ Since $b>0$, we have that $b-1$ is nonnegative, so $b=1,2$. Both of these cases don't work, so we get no solutions for this case. \newline
Case 2 ($a=2^{x-2}b-1$): Then $$b(2^{x-2}b-1)=2^{x+1}+1 \implies b+1 = (b^2-8) 2^{x-2} \implies \frac{b+1}{b^2-8}=2^{x-2}$$ This means $b \ge 3$, as $b+1$ and $2^{x-2}$ are nonnegative. Thus for $\frac{b+1}{b^2-8}$ to be an integer, we need $b=3$. Thus $x=4$, and $a=11$. \newline
Thus the solutions are $\boxed{(x,y)=(0,2),(0,-2),(4,23),(4,-23)}$.

We claim $(0, 2)$, $(0, -2)$, $(4, 23)$, and $(4, -23)$ are the only solutions. Clearly these work.

Clearly $x\geq -1$. By checking, we find that $x=-1$ does not work.

We claim that if $x\geq 1$, then $x=4$.

We have
$$(y+2^x+1)(y-2^x-1)=2^{2x}-2^x=2^x(2^x-1)$$
So, if we set $a=y+2^x+1$ and $b=y-2^x-1$, we have $ab=2^x(2^x-1)$ and $a-b=2^{x+1}+2$. If $a$ and $b$ satisfy this, then they are both positive or both negative. If they are both negative, then after negating them and switching $a$ and $b$ they will still satisfy the equations, so assume WLOG $a$ and $b$ are positive.

Clearly $a$ and $b$ are both even, so we can set $m=\frac{a}2$ and $n=\frac{b}2$.

Then, $mn=2^{x-2}(2^x-1)$ and $m=n+2^x+1$. Assume towards a contradiction $n$ is even. Then, $2^{x-2}|n$. Then, $n\geq 2^{x-2}$, so $m=n+2^x+1\leq 2^x-1$. This is a contradiction since then $n$ would have to be negative.

So, $n$ is odd, so $m=2^{x-2}k$ for some odd positive integer $k$. So, $2^{x-2}k=n+2^x+1$ and $kn=2^x-1$. So, $n=2^{x-2}k-2^x-1$, so $k(2^{x-2}k-2^x-1)=2^x-1$. If $k\leq 4$ then the RHS is negative, a contradiction. If $k\geq 8$, then $k(2^{x-2}k-2^x-1)\geq k(2^{x+1}-2^x-1)=k(2^x-1)>2^x-1$, a contradiction. So, $k$ is an odd number between $4$ and $8$, so $k$ is $5$ of $7$.

If $k=5$,
$$2^x-1=5(5\cdot 2^{x-2}-2^x-1)=5(2^{x-2}-1)=2^x+2^{x-2}-5$$
$$-1=2^{x-2}-5$$
$$4=2^{x-2}$$
$$x=4$$

If $k=7$,
$$2^x-1=7(7\cdot 2^{x-2}-2^x-1)=7(3\cdot 2^{x-2}-1)=21\cdot 2^{x-2}-7$$
$$4\cdot 2^{x-2}-1=21\cdot 2^{x-2}-7$$
$$6=17\cdot 2^{x-2}$$
a contradiction. So, $x=0$ or $x=4$, so the only pairs are $(0, 2)$, $(0, -2)$, $(4, 23)$, and $(4, -23)$.

We claim $(0, \pm 2), (4, \pm 23)$ are the only solutions.
It is obvious that $x$ is nonnegative. If $y$ is even, then $x=0$ since we need $2^x+1$ to be odd, which gives $(0, \pm 2)$ as a solution.
Now rearrange this to get $(y+1)(y-1) = 2^x(2^{x+1} +1)$. This means $\nu_2(y+1)=x-1$ OR $\nu_2(y-1) = x-1$. We take cases.
If $\nu_2(y-1)=x-1$, then set $y-1 = 2^{x-1} \cdot a$ for some odd $a$. Then $(y+1)(y-1) = 2^x \cdot a \cdot (a \cdot 2^{x-2} + 1)$, which implies that $a^2 \cdot 2^{x-2} +a = 2^{x+1}+1$. Clearly, $a\geq 3$ implies $a^2 \cdot 2^{x-2}  + a> 2^{x+1} + 1$ so the left hand side is greater than the right hand side. But then $a=1$, so $2^{x-2} = 2^{x+1}$, which is absurd.
If $\nu_2(y+1)=x-1$, then again set $y+1 = 2^{x-1} \cdot a$ for some odd $a$. Then $(y+1)(y-1) = 2^x \cdot a \cdot (a\cdot 2^{x-2}-1)$, so $a^2 \cdot 2^{x-2} - a = 2^{x+1} +1$. If $a\geq 5$, then it is easy to see the left hand side is much larger than the right hand side. This means $a=1,3$. But again, $a=1$ is absurd, so we get $a=3$. This becomes $9 \cdot 2^{x-2} -3 = 2^{x+1}+1$ and this becomes $2^{x-2} =4$, so $x=4$. This gives us $(4,\pm 23)$.

Having exhausted all the cases, we are done.

The answer is $(0,\pm 2),(4, \pm 23).$

Since LHS is an integer, we obtain $x\geq -1.$ Case $x=-1$ doesn't give any solution, and case $x=0$ gives the first answer. Henceforth $x>0;$ WLOG $y>0.$ Since $(y-1)(y+1)=2^x(2^{x+1}+1),$ factors in LHS are both even, and one of them is divisible by $2^{x-1}.$ Thus $y=2^{x-1}k\pm 1$ for odd positive integer $k,$ which yields $2^{x-2}(k^2-8)=1-(\pm k).$ If RHS is $1-k$ we obtain $k<3\implies k=1,$ which fails. If RHS is $1+k$ inequality $k\geq 3$ holds. But obviously RHS is bigger than LHS for $k\geq 5,$ so the only possible case - $k=3,$ which gives the second answer.

Denote $2^x=a$
We simplify the equation into $1+a+2a^2= y^{2}$
For $a<1$ ($x<0$), we have no solutions that maintain the conditions of being an integer and the problem statement.
For $a=1$ ($x=0$), we have solutions $(0, \pm 2)$ as $(x,y)$
For $a>1$, ($x>0$), all the $a$ are even and all the LHS are odd and therefore all $y$ are all odd. We can then express all $y=2n+1$ and get $1+2^{x}+2^{2x+1}= 4n^{2}+4n+1$.
Next, we do simplifying and get $2^{x}(1+2^{x+1}) = 4n(n+1)$
More TBA later

Clearly, $(0, \pm 2)$ are the only solutions for $x < 2$, so assume $x \geq 2$. We have $2^x + 2^{2x+1} = 2^x(2^{x+1}+1) = (y-1)(y+1)$. Since $y$ must be odd, $2^{x-1} \mid y-1$ or $2^{x-1} \mid y+1$.

Case 1: $2^{x-1} \mid y-1.$
Then $y = 2^{x-1}k + 1$. Substituting and dividing, we have $2^{x+1}+1 =  k(2^{x-2}k+1) \implies 2^{x-2}=\frac{k-1}{8-k^{2}}$, impossible for all $k$.

Case 2: $2^{x-1} \mid y+1.$
As before, this means $y = 2^{x-1}k - 1 \implies 2^{x+1}+1 =  k(2^{x-2}k-1) \implies 2^{x-2}=\frac{k+1}{k^{2}-8}$, which only has solutions if $k \leq 4$. Testing these cases, we have $k = 3 \implies (4, \pm 23)$ works.

To summarize, the only solutions are $\boxed{(0, \pm 2), (4, \pm 23)}$.

Again, storage from a while ago
We claim that the only solutions are $(x,y)=\{(0,\pm2), (4,\pm23)\}$. For simplicity in the future with positive exponents, let $x>2$. It's obvious that if not the only solutions are $(x,y)=(0,\pm2)$.
    We have that $$2^x(1+2^{x+1})=(y+1)(y-1)\implies 2^{x-1}|y\pm1$$ since the two factors differ by 2, meaning one of them has only one power of 2 in it.

    Case 1: $$y-1=k2^{x-1}\implies k2^{x-1}+2=y+1=2\frac{1+2^{x+1}}{k}\iff k(k2^{x-2}+1)=1+2^{x+1}\iff 2^{x-2}=\frac{k-1}{8-k^2},$$ which can be checked to not work for positive integers $k<3$ (otherwise denominator is negative).

    Case 2: Similar to before, $$y+1=k2^{x-1}\implies k(k2^{x-2}-1)=1+2^{x+1}\iff 2^{x-2}=\frac{k+1}{k^2-8}.$$ For this to be an integer, we need numerator greater or equal to denominator, which it is obvious that $k\le 3$. Checking, we find the only solutions are $(k,x,y)=(3,4,\pm23). \blacksquare$

My first ever IMO/ISL NT I think.

It is clear that $x$ cannot be negative (To see why taking LHS, when $x$ is negative as a fraction it is obvious that the numerator is odd and the denominator is even, and thus the answer is not an integer, let alone a perfect square). Then we consider the cases when $x=0,1$ for $x=0$ we see that $y=2,-2$ work and there are clearly no solutions for $x=1$.
\\
For $x\geq 2$, consider the following factorization; $$1+2^x+2^{2x+1}=y^2\implies 2^x(2^{x+1}+1)=(y-1)(y+1)$$Here, for any power of 2 greater than $2^1$, it is clear that it divides only one of the factors in the RHS. Then proceed based on two cases. \\
[b][i]Case 1 : [/i][/b] $y+1$ is only divisible by $2^1$ (max power of 2). Next, do the following substitution $y-1=2^{x-1}a$ and $y+1=2^{x-1}a+2$. Considering the case when $a=1$, gives us no solutions. Then for $a\geq 2$, notice that;
\begin{align*}
    (y-1)(y+1)&=(2^{x-1}a)(2^{x-1}a+2)\\
    2^x(2^{x+1}+1)&=2^x(a)(2^{x-2}a+1)\\
    2^{x+1}+1&=(a)(2^{x-2}a+1)\\
    2^{x-2}(a^2-8)&= 1-a\\
    2^{x-2}=\frac{1-a}{a^2-8}
\end{align*}
Now, the LHS is clearly a positive integer, thus the RHS must be so too. Since the numerator is most clearly negative, the denominator must be so too. This means that $a^2\leq 8\implies a\leq 2$, thus the only possibility is $2$. Checking this possibility gives us the solutions $(x,y)=(4,23);(4,-23)$\\
[b][i]Case 2 : [/i][/b] $y-1$ is only divisible by $2^1$ (max power of 2). Next, do the following substitution $y-1=2^{x-1}a$ and $y+1=2^{x-1}a+2$.
\begin{align*}
    (y-1)(y+1)&=(2^{x-1}a)(2^{x-1}a-2)\\
    2^x(2^{x+1}+1)&=2^x(a)(2^{x-2}a-1)\\
    2^{x+1}+1&=(a)(2^{x-2}a-1)\\
    2^{x-2}(a^2+8)&= a+1\\
    2^{x-2}=\frac{a+1}{a^2+8}
\end{align*}
It is clear that for all $a\geq 1$, $a+1<a^2+8$, thus the RHS is most clearly not an integer. But the LHS is, which means there are no solutions for this case. Thus, the set of all the solutions is, $(x,y)=(0,2);(0,-2);(4,23);(4,-23)$

We claim the only solutions are $(0, \pm 2)$ and $(4, \pm 23)$.

Consider rearranging the equation and factoring like so,

  $$2^x(2^{x+1} + 1) = (y-1)(y+1)$$

Clearly up to a factor of 2, $y-1$ and $y+1$ are coprime. Moreover $2^x$ and $2^{x+1} + 1$ are coprime. Also due to $\nu_2$ reasons we must have $ x \geq  0$. Thus take cases.

Assume that $x > 2$ and note then that either $y - 1$ or $y + 1$ is equal to $2^{x-1}m$ for some $m$ coprime to 2. Let $y - 1 = 2^{x-1}m \iff y = 2^{x-1}m + 1$. Then we have,
\begin{align*}
    2^x(2^{x+1} + 1) &= (2^{x-1}m)(2^{x-1}m+2)\\
    2^x(2^{x+1} + 1) &= 2^x(2^{x-2}m^2 + m)\\
    2^{x+1} - 2^{x-2}m^2 &= m - 1\\
    2^{x-2}(8 - m^2) &= m - 1\\
    2^{x} &= \frac{4(1-m)}{m^2-8}
\end{align*}
which has no solutions. Similarly take the second case, $y + 1 = 2^{x-1}m \iff y = 2^{x-1}m - 1$ which gives,
\begin{align*}
    2^x(2^{x+1} + 1) &= (2^{x-1}m -2)(2^{x-1}m)\\
    2^x(2^{x+1}+1) &= 2^x(2^{x-2}m^2 - m)\\
    2^{x+1} + 1 &= 2^{x-2}m^2-m\\
    2^{x-2}(8 - m^2) &= -1-m\\
    2^{x} &= \frac{4(m+1)}{m^2-8}
\end{align*}
By inspection $m = 3$ works giving $(4, \pm 23)$. To see that no other values work simply note the denominator grows exponentially, making the RHS a non-integer. Then checking when $x \leq 2$ gives us $(0, \pm 2)$.

We claim the only solutions are $\boxed{(0, \pm 2), (4, \pm 23)}$.

Rewrite the equation so that it becomes

\[2^x(2^{x+1} + 1) = (y-1)(y+1).\]

Up to a factor of $2$, $y-1$ and $y+1$ are coprime and $2^x$ and $2^{x+1}+1$ are also coprime. Clearly, $x \ge 0$, so we will do casework.

If $x>2$, note that either $y - 1$ or $y + 1$ is equal to $2^{x-1}m$ for some $m$ coprime to 2.

Let $y - 1 = 2^{x-1}m$. Then we have,

\begin{align*}
    2^x(2^{x+1} + 1) &= (2^{x-1}m)(2^{x-1}m+2)\\
    2^{x+1} - 2^{x-2}m^2 &= m - 1\\
    2^{x} &= \frac{4(1-m)}{m^2-8}
\end{align*}

which has no solutions.

Similarly take the second case, $y + 1 = 2^{x-1}m$ which gives

\begin{align*}
    2^x(2^{x+1} + 1) &= (2^{x-1}m -2)(2^{x-1}m)\\
    2^{x+1} + 1 &= 2^{x-2}m^2-m\\
    2^{x} &= \frac{4(m+1)}{m^2-8}
\end{align*}

From inspection, $m = 3$ works giving the pair $(x,y) = (4, \pm 23)$. Otherwise, notice that the denominator grows exponentially, making the right hand side a not an integer.

Finally, checking when $x \le 2$ gives us our other case $(x,y) = (0, \pm 2)$.

I claim that the only solutions are $\boxed{(0, \pm 2), (4, \pm 23)}$.

Observe first that the cases $x = 0$ and $x < 0$ are trivial, and yield two and no solutions respectively. Henceforth assume $x > 0$. Now observe $y$ is odd, so that $y = 2k + 1$ for $k \in \mathbb{Z}$. Now rearranging, factoring, and simplifying yields \[2^{x - 2}(1 + 2^{x + 1}) = k(k + 1).\] We now split into cases according to when $2^{x - 2} \mid k$ or $2^{x - 2} \mid k + 1$.

[b][color=#00f]Case 1:[/color][/b] ($2^{x - 2} \mid n$). Then $n = 2^{x - 2}N$ and $n + 1 = 2^{x -2}N + 1$, so that \[2^{x - 2}(1 + 2^{x + 1}) = (2^{x - 2}N)(2^{x - 2}N + 1) \implies 1 + 2^{x + 1} = 2^{x - 2}N^2  + N \implies 2^{x - 2} = \frac{N - 1}{8 - N^2}, \] which has no solutions for $N \in \mathbb{Z}$.

[b][color=#00f]Case 2:[/color][/b] ($2^{x - 2} \mid n + 1$). Then $n + 1 = 2^{x - 2}N$ and $n = 2^{x - 2} - 1$, so that \[2^{x - 2}(1 + 2^{x + 1}) = (2^{x - 2}N - 1)(2^{x - 2}N) \implies 1 + 2^{x + 1} = N(2^{x - 2}N - 1) \implies 2^{x - 2} = \frac{N + 1}{N^2 - 8} \implies \boxed{N = 3}.\] Hence $\boxed{x = 4}$ and this gives $\boxed{y = \pm 23}$ as the only other solution. $\blacksquare$

$1+2^x+ 2^{2x+1} = y^2$
$=>1+2^x+ 2^{2x}.2 = y^2$

Using Fermat's theorem,
$ 2^{2x} \equiv 2 (mod  2x) .... (1) $
Again,
$ 2^x \equiv 2 (mod x) $
$ => 2^{2x} \equiv 4 (mod x) ....(2) $
Looking at (1) we can clearly see that the 4 of (2) is easily reducible.
Hence, $x \mid 4 => x \leq 4 $

Again, y is an integer so  $x \geq -1$
So the bounding we achieve is $ -1 \leq x \leq 4$

Trying these x values gives us,
$ (x,y) = (0, \pm2), (4, \pm23) $


Somewhere in the FAQ, it mentioned that admitted applicants will be getting a list of things they need for Mathcamp. When do we get the list?

On the welcome page of the registration site it reads, "Additional information on arrival at camp, accommodations, academics, and extracurricular activities will be mailed out in early June."

. . . which last year meant mid-June.  I can post last year's list here if you want to see it.

[b][size=150]WHAT TO BRING TO MATHCAMP[/size][/b]

The camp will provide pillows and blankets, laundry detergent, and fabric softener.  If you do not know how to use a washer and dryer, learn before you come!

[irrelevant info regarding weather in Washington]

We recommend that you bring the following:

[b]CLOTHES[/b]

8-10 pairs of socks and underwear
8 light shirts or T-shirts
2 or 3 pairs of shorts and 2 or 3 pairs of long pants
1 waterproof jacket
2 sweaters/sweatshirts
1 or 2 pairs of sneakers and/or 1 pair of sandals
1 set of nicer clothes, with clothes hangers (nothing fancy or formal will be required)

[b]PERSONAL ITEMS[/b]

Pillowcase, sheets, towels (we will provide pillows and blankets, but not linens)
Alarm clock (very important!)
Toiletries, bathrobe, shower slippers (bathrooms are in the hall)
Long distance calling card or access number to call home. (make sure you know how to use it)
Optional: small reading lamp, small fan, umbrella, camera

[b]SCHOOL SUPPLIES[/b]

Notebook, pens/pencils, binder for handouts, and some paper (don't buy too much - you can easily buy more at the campus bookstore)
Scientific calculator only if you have one already - you can certainly do without it

[b]RECREATION[/b]

Sun block!! (There will be many outdoor activities and it is easy to get burned)
Good hiking or walking shoes, hat, sunglasses (Hikes are offered on a variety of levels.  Serious hikers, bring your boots!)
Swimsuit and extra towel (if you want to go swimming/kayaking)
Personal sports equipment (e.g. tennis racket)
Board games, chess sets, puzzles (the more the merrier)
Musical instruments
Water bottle (optional, but highly recommended, for hikes and daily hydration)

[b]RECOMMENDED SPENDING MONEY: 100-200 USD[/b]

For incidental expenses, extra field trips, etc.

[b]TRAVEL DOCUMENTS[/b]

Don't forget to carry your ticket, visa, and passport/ID safely!

[b]WHAT NOT TO BRING[/b]

Valuables or expensive electronic equipment: we cannot guarantee their safety

Do you think it's a good idea to bring a laptop?  Are computing resources (especially for independent projects) readily available or would we benefit from having our own computer?  Or would having a computer just be a distractor?  I'm guessing that there will be plenty of Mathematica to go around....

It sounds like a laptop has good and bad sides. It's good because you have a personal, readily-available platform to do work on, to takes notes, etc. It's bad because you have to watch out for it and make sure it never gets lost, damaged, stolen, etc.

I will likely be in a position in which I will order a laptop, but which can only arrive once I am at camp. I wonder how bad of an idea it is to have it be delivered to camp... What worries me is that it is a centralized mail pool. In all honesty, it's probably a really stupid plan to have it arrive there, but I fear that wanting to set it up and use it will make camp less interesting.

Hmm, do people actually use them in the classrooms? I tend to type much faster than I write, so it could be useful.

PS: I happen to be in Maine right now :D  (sailing and such)

As I understand it, a few people bring them every year. If you have one, it would likely be very useful in math classes (especially if we get Mathematica close to the beginning of camp, because then we can enter in equations and such as teachers write them on the board). You seem to be like me and type faster than write, so it would obviously be useful.

Is there any sort of safe or locker we can get so that no body can enter except with a key? Rooms don't work well, because lots of people besides you (e.g. cleaning staff) will likely enter it. Also, how exactly does the mail system work? Is all mail simply piled into one box and people take it? If not, and it is actually delivered in some semi-private way by the camp, that may be a deciding factor for having or not having my laptop delivered to camp.

Mail is delivered to the Colby College staff.  From there it is given to a Mathcamp staff member who brings it back to the dorms.  We then keep it in the office and post notices for all the campers to get their mail.  So it does go through non-Mathcamp staff, but is never in a public place for anyone to take it.

For reference, we don't have any sort of lockers or such for students; your room is the safest place to leave it, but you're right that a roommate, cleaning staff, etc. will still have access.

In general I don't really think a laptop is worth it; entering equations into Mathematica live is unlikely to provide much benefit. :)  You're welcome to bring it if you want, but Mathcamp can't take responsibility if something does happen.

I see, thanks for the info, Dan. I will probably not end up bringing the laptop (or rather, ordering it to arrive at camp) at least during the first year. Oh well, I'll just be very, very slightly less sad at the end of camp :-)

So will computers with Mathematica be readily available anyway?

(Speaking as an Alum, not staff)

Assuming that it's like last year...

There should be 5-6 computers that belong to Mathcamp available at all times between the dorms (they run Linux, and have some mathematical goodies, like LaTeX, on there, though I'm not sure if they have Mathematica on them (they're fairly old comps)).  These were available for the following reasons, in order:

1.  E-mailing your parents
2.  Doing math
3.  Everything else

If you need to use a comp, you'll certainly be able to (though you might need to kick someone playing some random flash game), but, from my experience, I think that I only used the comp twice for purposes other than e-mail.  I probably spent a grand total of 2 hours on the computers during the 5 weeks, there really isn't much need for them.

Additionally, last year, we occasionally had access to one of the campus computer labs.  I have no idea what Colby's policy is on this, however, so it might be different this year.

We're going to have Colby's computer lab open evenings, I think 7pm-9pm (but that may change).  The computers will indeed have Mathematica.

Thanks for the info, guys.  So if our independent project was comp-sci related, you don't think we'd have a problem?

(again, speaking as an Alum)

If we'll have access to Colby's comp labs, you definitely won't have a problem, since any moderately high powered stuff (running complex programs/evaluating nasty integrals/whatever) you need to do will be doable on them, while "everyday" stuff (writing programs/classes, "simple" math, etc) should be doable on Mathcamp's computers.  Last year, 3 projects were programming related, and there were a couple that focused on theoretical comp sci.  The full project list from '06 can be found at http://www.mathcamp.org/handouts2006/06projects.pdf.

In particular, the programming projects included understanding and implementing the Number Field Sieve algorithm (http://en.wikipedia.org/wiki/Number_field_sieve), working a problem from the 2003 International Conference for Functional Computing's programming contest, and writing a program to find solutions to the game "Ricochet Robots" (http://en.wikipedia.org/wiki/Ricochet_Robot), a fairly popular board game at camp.  In short, I'm sure you'll be able to do a comp-sci project.

Thanks a bunch eyefragment!  Can't wait to meet everyone.

Tjames, if you do some research project that requires computers, your project advisor will make sure that you have computer access and all the goodies needed for it.

Just curious, what is meant by "nicer clothes"? Like a dress shirt? I'm asking because, even after bringing a hanger, its going to get wrinkled in the luggage.

I'm not going to show up there with a tuxedo, but I don't want to come in wearing jeans and a t-shirt either on a more formal occasion. Or maybe I will  :lol:

Last minute packing (its tomorrow)! Hurray for procrastination.

I brought no nicer clothes last year, and never wanted any.  There are no formal occasions.  I'm not really sure why they put that in there.

I'd guess because of people who might want to go to church.

Yeah, you're probably right.  Although last year the people who went to church seemed to be wearing T-shirts and shorts (not that I was one of them).

Mathcamp starts TOMORROW! Yay!  :lol:

Yay! Less than 24 hours until we arrive!

Last year, I went to Seattle a few days early with my family, and so they dropped me off at MC. So, I'm in the dark as to how airport pickup works. How many people are generally on a shuttle to MC from the airport? (and on a related note, is anyone else arriving at Portland (as opposed to Bangor) at around noon?


Let $a; b; c$ be complexe numbers . If$(a+b)(a+c) = b; (b+c)(b+a) = c$and$(c+a)(c+b) = a$ then
$a; b; c \in R.$


[quote="Jimmy"]This might be too easy for this forum, and if so I apologize:

For non-negative real numbers $a$, $b$, and $c$, prove the following \[a^{2}(b+c)+b^{2}(a+c)+c^{2}(a+b) \ge 6abc\] [/quote]
Straight application of Muirhead: $\sum_{sym}a^{2}b\ge \sum_{sym}abc$
Another way:
By AM-GM, $a^{2}b+b^{2}c+c^{2}a\ge 3abc$, $ab^{2}+bc^{2}+ac^{2}\ge 3abc$, adding these get the result
Another way:
$a(b^{2}+c^{2})\ge 2abc$
$b(c^{2}+a^{2})\ge 2abc$
$c(b^{2}+a^{2})\ge 2abc$
Adding these get the result
Another way:
The ineq is equivalent to:
$b(a-c)^{2}+c(a-b)^{2}+a(c-b)^{2}\ge 0$
which is true

[quote="Jimmy"]And one more way (how I came up with the problem):

Consider the function $f(x) = \frac{1}{x}$, which is convex on the interval $(0,\infty)$, then by Jensen's we have \[\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \ge \frac{3}{a+b+c}\] From which the result follows.[/quote]
I think you need not use Jensen's to get the last inequality. You could just as well have used $AM-HM.$ Also, in the above form, we must have $a,b,c > 0$ -- we won't be able to say just non-negative numbers. But, of course, your orginal idea is a nice one, no doubt!

You're right in that we can't say "non-negative" from Jensen's, but once I saw the result, it was clearly true for non-negative numbers.  Anyway, I came up with the problem by accident, and just thought the result looked interesting. :)

Why must you expand and Muirhead's while AM-GM is simpler? \[(a+b)(b+c)(c+a) \ge 8abc\ \?\]


geometry
geometric transformation
homothety
parallelogram
reflection
trapezoid
power of a point

A circle with center $O$ intersects a circle with center $O^{\prime}$ at two points $A$ and $B$. Some line touches the first circle at $T$ and the second circle at $T^{\prime}$. The perpendiculars from the points $T$ and $T^{\prime}$ to the line $OO^{\prime}$ meet $OO^{\prime}$ at $S$ and $S^{\prime}$. Show that triangle $AST$ is similar to triangle $T^{\prime}S^{\prime}A$.

Let the center line OO' meet the common (external) tangent TT' of the circles (O), (O') at a point H. This is the external homothety center of the circles (O), (O') and thier homothety coefficient is

$h = \frac{HO}{HO'} = \frac{HT}{HT'} = \frac{OT}{O'T'} = \frac{r}{r'}$

Let the line HA intersect the circles (O), (O') at points C, A' different from the point A. The parallel lines $TS \parallel T'S'$ passing through the homothetic points T, T' are homothetic with the same homothety center H and coefficient h. The point pairs (A, A') and (C, A) lying on the homothetic circles (O), (O') and collinear with the homothety center H are also homothetic with the same homothety center H and coefficient h. Consequently, the triangle pairs $\triangle AST \sim \triangle A'S'T'$ and $\triangle TSC \sim \triangle T'S'A$ are centrally similar with the same homothety center H and coefficient h. We have to show that the triangles $\triangle AST \sim \triangle TSC$ are similar, because then we would be done.

Let N be the midpoint of the common tangent length TT'. Since NT = NT', the point N is on the radical axis AB of the circles (O), (O'). Since $TS \parallel T'S' \parallel AB$ (all these lines are perpendicular to the center line OO'), the radical axis AB cuts the segment SS' at its midpoint M. The lines $AS \parallel A'S'$ and the lines $CS \parallel AS'$ are parallel, because they connect homothetic pairs of points $(A, S) \sim (A', S')$ and $(C, S) \sim (A, S')$, hence, they form a parallelogram ASB'S' (the position of the point B' with respect to the circles (O), (O') is unknowm). One diagonal of this parallelogram is SS' with the midpoint M. The point M is also the midpoint of the other diagonal AB', which means that the points $B' \equiv B$ are identical. Moreover, since the diagonals $AB \perp SS' \equiv OO'$ of the parallelogram ASBS' are perpendicular to each other, it is a rhombus. Therefore, its diagonals AB, SS' bisect its internal angles. Since the lines $CS \equiv SB,\ AS \parallel A'S' \equiv S'B,\ TS \parallel AB$ are parallel to each other (the last 2 lines are both perpendicular to the center line OO'), it follows that the line TS bisects the angle $\angle ASC$. Therefore, $\angle AST = \angle CST$. Let U be the reflection of the point T in the center line OO'. Since the lines $AB \parallel TU$ are parallel, the quadrilateral ABUT is a cyclic trapezoid, which means that it is isosceles and its diagonals AU = BT are equal. Since the angles spannig the corresponding arcs on the equal chords AU = BT are equal,

$\angle CTS \equiv \angle ATU = \angle DAT \equiv \angle SAT$

Thus the triangles $\triangle AST \sim \triangle TSC$ are similar, having 2 angles equal. It follows that the triangles $\triangle AST \sim \triangle TSC \sim \triangle T'S'A$ are also similar.

:spider:


Ramsey Theory
real analysis
real analysis unsolved

Assume that $a_n, b_n$ are sequences of distinct integers such that $a_n$ contains arbitrarily long arithmetic progressions.
  a) If $a_n-b_n$ is bounded, then $b_n$ also contains arbitrarily long progressions.
  b) What about if $\sum_{n\geq 1}{\frac{1}{1+|a_n-b_n|}}$ diverges?

The first one follows rather easily from Van der Waerden's Theorem, but I have some questions about the second one (sorry to bug you like this; I always ask for clarifications, and this can be very annoying, I know :)):

Shouldn't that series diverge instead of converging? Because we can take $b_n$ as far away from $a_n$ as we want, so as to make the series converge and avoid any arithmetic progressions in the sequence $(b_n)_n$ (we could take $b_{n+1}>2b_n$, for instance), so it's clear that the problem analogous to the first one which arises like this has a negative answer :?.

Or maybe the question is whether it's possible for $b_n$ to have arbitrarily long arithmetic progressions if the series converges?

Oh, sorry, the series diverge in the second case. :blush:

It is still strange because for this series to diverge, it suffices to have $a_n=b_n$ for infinitely many $n$ but it is fairly clear that there exist sequences $a_n$ containing arbitrarily long arithmetic progressions such that $a_n=2^n$ for infinitely many $n$.


linear algebra
matrix
algebra
polynomial
vector
integration
calculus

How can I find the general solution to the homogenous problem

( X1 over X2 ) ' = ( 1 1 over -4 1 ) * ( X1 over X2 )

(read them as matrix)

also how can I find the particular solution (thus general) to the problem

( X1 over X2 ) ' = ( 1 1 over -4 1 ) * ( X1 over X2 ) + ( 5t over 2)

One suggestion: use Laplace transforms. You will have to assume initial conditions - but since you're looking for one particular solution, you can make any convenient assumptions about those. The Laplace transform of a matrix is a matrix, and the transform will turn the system of differential equations into a linear system of algebraic equations (with polynomial coefficients).

Second suggestion: for your given matrix $A$, find $e^{tA}.$ Then imitation the multiplying factor method for scalar first order equations.

Your equations reads $x'=Ax+b,$ where $x$ and $b$ are column vectors and $A$ is a $2\times2$ matrix.

$x'-Ax=b$ Now multiply by $e^{-tA}:$

$e^{-tA}x'-Ae^{-tA}x=e^{-tA}b$

$\left(e^{-tA}x\right)'=e^{-tA}b$

$x(t)=e^{tA}\int_{0}^{t}e^{-uA}b(u)\,du+x(0).$


$ 1>1/a\plus{}1/b\plus{}1/c$

So the interval clearly can't contain 2 or 3. Hence the required interval is easily seen to be $ [4,\inf)$

so for it to be true $ a,b,c \geq 4$?
and don't you mean $ 1 < 1/a \plus{} 1/b \plus{} 1/c$

no, for a can be 2, b can be 3, and c can be 100 and it could still work.

a+b+c>9

If a + b + c <9, considering that neither of the variables can be decimals, then the value of 1/a + 1/b +1/c is always greater than 1.

Misread, if $ 1<1/a\plus{}1/b\plus{}1/c$ then we can't have anything $ \ge3$ in the domain, so the interval has to $ a,b,c\equal{}2$.

(2,3,4)  :wink: (2,3,5) are the only solutions a< b < c

[quote="mathemonster"]a,b,c are natural numbers with a > b > c
Find on what interval c must be to have:
$ ac \plus{} bc \plus{} ab > abc$[/quote]

[hide="solution"]Dividing both sides by $ abc$, we have $ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}>1$. Now let's find all triplets of integers that do NOT work:
If c≥3, nothing works, since $ \frac{1}{3}\plus{}\frac{1}{4}\plus{}\frac{1}{5}\equal{}\frac{47}{60}$. If c=2, then b=3 and a=4,5. If c=1, then everything works. Thus $ c\in [1,2]$.[/hide]

[quote="1=2"][quote="mathemonster"]a,b,c are natural numbers with a > b > c
Find on what interval c must be to have:
$ ac \plus{} bc \plus{} ab > abc$[/quote]

[hide="solution"]Dividing both sides by $ abc$, we have $ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} > 1$. Now let's find all triplets of integers that do NOT work:
If c≥3, nothing works, since $ \frac {1}{3} \plus{} \frac {1}{4} \plus{} \frac {1}{5} \equal{} \frac {47}{60}$. If c=2, then b=3 and a=4,5. If c=1, then everything works. Thus $ c\in [1,2]$.[/hide][/quote]
Are you sure? :huh:
First, c is a natural number , so $ c\in${$ 1,2$}
If c is a positive real number, $ c\in (0,2]$


Dear all, I would like to solve this problem without calculus:

Find the maximum value of the expression

13 \sqrt{x^2-x^4} + 9 \sqrt{x^2 + x^4}

Thank you.

Well, I think that this problem should be moved to Pre-olympiad or Getting Started. ;)

[quote="pacoga"]Dear all, I would like to solve this problem without calculus:

Find the maximum value of the expression

$13 \sqrt{x^2-x^4} + 9 \sqrt{x^2 + x^4}$

Thank you.[/quote]

[hide="hint"]make a trig substitution![/hide]


modular arithmetic
number theory
relatively prime
Primitive Roots

Suppose that $m$ does not have a primitive root. Show that \[a^{ \frac{\phi(m)}{2}}\equiv 1 \; \pmod{m}\] for every $a$ relatively prime $m$.

If $m=xy$ with coprime $x,y>2$, then $2| \varphi(x), \varphi(y)$, giving:
\[a^\frac{\varphi(m)}{2}= a^{\varphi(x) \cdot \frac{\varphi(y)}{2}}\equiv 1^\frac{\varphi(y)}{2}=1 \mod x\]
and
\[a^\frac{\varphi(m)}{2}= a^{\varphi(y) \cdot \frac{\varphi(x)}{2}}\equiv 1^\frac{\varphi(x)}{2}=1 \mod y.\]

After handling the easy case $m=2^{k}\geq 8$, this now gives our result because for all other $m$, there are primitive roots.

[quote="ZetaX"]After handling the easy case $ m \equal{} 2^{k}\geq 8$, this now gives our result because for all other $ m$, there are primitive roots.[/quote]Is this something you can prove here or just a fact that you know?

It's near to trivial (and well known):
As there is no primitive root, we have $ k \geq 3$. Trivial checking gives that $ a^2 \equiv 1 \mod 8$ and inductively we have $ a^{2^{r\minus{}2}} \equiv 1 \mod 2^r$ implying (by $ a^{2^{r\minus{}2}} \equal{}1\plus{}2^r b$) that $ a^{2^{r\minus{}1}} \equal{} (1\plus{}2^r b)^2 \equal{} 1\plus{} 2^{r\plus{}1} b \plus{} 2^{2r} b \equiv 1 \mod 2^{r\plus{}1}$.

how do u have a=1+(2^r)b? for r = 3, that means a is 1 or 9 mod 16 (if we prove the r+1 case), but that missed the other residues in 16

we prove for any prime factor of $n$ that $||m||_p=k$ that $a^{ \frac{\varphi(m)}{2}}\equiv 1 \; \pmod{p^k}$

first of all it's easy to see that $\varphi(\frac{m}{p^k})$ is even because $\frac{m}{p^k} >2$

now let $x=\frac{\varphi(m)}{2} =\frac{\varphi(\frac{m}{p^k}) \times (p-1)p^{k-1}}{2} =\frac{\varphi(\frac{m}{p^k})}{2}  \times (p-1)p^{k-1} \implies \varphi(p^k) \mid x \implies a^x \equiv 1 \mod p^k$

because $p$ was an arbitrary prime factor then $a^x \equiv 1 \mod m$


1. Let n be a positive integer. Find the sum of all even integers between $ n^2-n+1$ and $ n^2+n+1$.
2. Show that $ 73 | 2^{2n+3}*3^{3n+2}+35^n$ for integers $ n >= 0$
3. Find the remainder of $ 5{}^7{}^n$ when divided by $ 31$
4. Show that $ 1977$ is not the sum of two cubes of integers
5. Show that$ ^3+y^3+z^3$ cannot equal $ 2011$, for all integers $ x,y,z$
6. Show that $ x^3+y^4$ cannot equal $ 7$, for all integers $ x,y$

[hide="1"]Note that $ n(n\minus{}1)$ is even for all $ n$. Then, we have that the even integers range from $ n^2\minus{}n\plus{}2 \longrightarrow n^2\plus{}n$.

The number of even integers is given by $ \frac{n^2\plus{}n\minus{}(n^2\minus{}n)}{2}\equal{}n$.
And the average of those integers is $ n^2\plus{}1$.

Thus the sum is $ n^3\plus{}n$[/hide]

[hide="3"]
Note that $ 5^{3}\equiv 1 \mod 31$.

Then $ 5^{7n}\equiv 5^{n} \equiv 5^{n'} \mod 31$ where $ n'\equal{} n\mod 3$
[/hide]

[hide="4"]$ 1977\equiv 6 \mod 9$

All cubes mod 9 are $ 1, \minus{}1,$ or $ 0$. Thus, the claim is proven.[/hide]

[hide="5"]$ 2011 \equiv 4 \mod 9$. The result follows from the fact that all cubes are $ 1, 0,$ or $ \minus{}1$ mod 9.[/hide]

[hide="6"]Looking at mod 13 (since $ 3*4\equal{}12$), we see that $ x^3\equiv 0, 1, 8, 12 \mod 13$ and $ y^4 \equiv 0, 1, 3, 9 \mod 13$

It is easy to check that taking one from both cannot sum to $ 7 \mod 13$[/hide]

[hide="2"]Verify that all that gobbledygook simplifies to $ 72\cdot 108^n\plus{}35^n$, which, under mod 73, is equivalent to $ \minus{}1\cdot 35^n\plus{}35^n\equal{}0$.[/hide]


calculus
integration
logarithms
derivative
function
calculus computations

$ \int r \log^2(r) dr \equal{} \frac {r^2}{2} \log^2(r) \minus{} \int r\log(r) dr$
$ \equal{} \frac {r^2}{2} \log^2(r) \minus{} \frac {r^2}{2}\log(r) \plus{} \frac {1}{2}\int r dr$
$ \equal{} \frac {r^2}{2} \log^2(r) \minus{} \frac {r^2}{2} \log(r) \plus{} \frac {r^2}{4} \plus{} C$

Checking the derivative...

$ r\log^2(r) \plus{} r\log(r) \minus{} (r\log(r) \plus{} r/2) \plus{} r/2 \equal{} r\log^2(r)$

What's wrong with that?

i am sorry, i lost completely, i still can't figure out.

Integration by parts:

$ uv \equal{} \int u dv \plus{} \int v du$

(This is essentially the product rule for derivatives, then integrated.)

Take $ u dv \equal{} r\log^2(r) dr$. To get rid of logs, let $ rdr \equal{} dv$ and $ u \equal{} \log^2(r)$. Apply the formula and you'll get the first line of what I wrote before. Then apply the same idea to the new integral and you'll be left with just an integral of $ r$, which you can do.

Is this clear?

$ \ln$ is the same as $ \log$ unless you have some other meaning for it besides "logarithm".

As far as I know in many books people uses $ \log$ to represent the natural logarithm, so I think you can understand Xevarion's solution.

I though ln has base e but log has base 10

In general, the notations $ \ln$ and $ \log_b$ are unambiguous. The definition of $ \log$ with no subscript is ambiguous, and depends on who you're talking to. Some possibilities:
Base 2: popular among computer scientists, or others dealing with algorithms.
Base $ e$: Popular among mathematicians, and anyone focusing on calculus applications.
Base 10: Popular in the days before cheap calculators, and fading now. Still common in chemistry and science/engineering, for dealing with scientific notation or logarithm-based measurement scales.
Unspecified base (the answer doesn't depend on which one you choose, as long as you're consistent): I use this one a lot.

Why do we tolerate this ambiguity? "log" is only one syllable when spoken, while every other notation for logarithms takes multiple syllables.

[quote]Why do we tolerate this ambiguity? "log" is only one syllable when spoken, while every other notation for logarithms takes multiple syllables.

[/quote]

You don't mean to say mathematicians are lazy, do you?  :(

I don't mean to say that [i]only[/i] mathematicians are lazy.

Also, it frequently doesn't matter. What does $ O(n\log n)$ mean? It might be base $ 2$, $ e$, or unspecified- and the mathematical meaning is the same in all cases.

Using $ \log_a(x) \equal{} \frac{\log_b(x)}{\log_b(a)}$, you can see that the only difference is a constant multiple. In a situation like this problem, the method is the same no matter what the base is; you might as well assume that it's a natural log.

I almost always use $ \log$ to refer to natural logarithm, not just out of laziness but because I almost never need any other kind of logarithm, and it's quite reasonable for the "simplest" notation to refer to the most commonly used mathematical objects.

[quote="cdymdcool"]True, but the differentiation methods for log and ln are completely different.[/quote]
Why? $ \log_{10}(x) \equal{} c\ln(x)$. So $ \frac{d}{dx} \log_{10}(x) \equal{} c\frac{d}{dx}\ln(x)$. The only difference is a constant multiple...

I have only one comment to add. The original poster wrote "In" in his first post. I see that on student papers - I see that more often than I'd like. I react to it like I just heard fingernails scraping on a blackboard. "It's an L, not an I! L is for logarithm! The function is a logarithm!" I tend to think that those you use this misspelling don't really understand what the function is, and don't associate the symbol $ \ln$ with the notion of "logarithm."

cdymcool hasn't done all that much to convince me otherwise. He did correct the spelling, but afterwards he reacted as if there was some vast gulf of difference in meaning between $ \ln$ and $ \log$ - which, as Xevarion and jmerry have been pointing out, there isn't, not really.

In a classroom setting, I think that fairly often I wind up saying "log" or "logarithm" out loud while simultaneously writing $ \ln.$

^^I think most professors do that.  In my differential equations class, the professor consistently writes $ \log$ but of course he is referring to the natural log.  I don't really find it that annoying, it just tends to confuse people.

[quote="Kent Merryfield"]

In a classroom setting, I think that fairly often I wind up saying "log" or "logarithm" out loud while simultaneously writing $ \ln.$[/quote]

I don't see what's so complicated in thinking of $ \ln(x)$ as "lawn x" (pronunciation), meaning the natural logarithm, ie $ \log_{e}(x)$, whereas $ \log$ is the general logarithm, meaning $ \log_{a}(x)$ where $ a$ can be any base you wish, though 2 and 10 are most common.

Why do people write out "$ \log{x}$"?  It's shorter and easier (for me, at least) to write/type "$ \ln{x}$" and just read it as "el en of x" or "lan of x" or "lawn of x"...  One of my math professors does that.


inequalities
search
inequalities proposed

Let a,b,c be positive number
Prove that
 $ \frac{a^4}{b^3\plus{}c^3}\plus{}\frac{b^4}{c^3\plus{}a^3}\plus{}\frac{c^4}{a^3\plus{}b^3} \geq \sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}\minus{}\frac{1}{2}(a\plus{}b\plus{}c)$

[quote="8826"]Let a,b,c be positive number
Prove that
 $ \frac {a^4}{b^3 \plus{} c^3} \plus{} \frac {b^4}{c^3 \plus{} a^3} \plus{} \frac {c^4}{a^3 \plus{} b^3} \geq \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca} \minus{} \frac {1}{2}(a \plus{} b \plus{} c)$[/quote]
Yes,it very easy.We can prove that:
$ \frac{a^6}{a^5\plus{}b^5}\plus{}\frac{b^6}{b^5\plus{}c^5}\plus{}\frac{c^6}{c^5\plus{}a^5} \ge \frac{a\plus{}b\plus{}c}{2}$

[quote="quykhtn-qa1"][quote="8826"]Let a,b,c be positive number
Prove that
 $ \frac {a^4}{b^3 \plus{} c^3} \plus{} \frac {b^4}{c^3 \plus{} a^3} \plus{} \frac {c^4}{a^3 \plus{} b^3} \geq \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca} \minus{} \frac {1}{2}(a \plus{} b \plus{} c)$[/quote]
Yes,it very easy.We can prove that:
$ \frac {a^6}{a^5 \plus{} b^5} \plus{} \frac {b^6}{b^5 \plus{} c^5} \plus{} \frac {c^6}{c^5 \plus{} a^5} \ge \frac {a \plus{} b \plus{} c}{2}$[/quote]

We can prove your inequality.But it doesn't relate to mine :)

[quote="Bugi"][quote="quykhtn-qa1"][quote="8826"]Let a,b,c be positive number
Prove that
 $ \frac {a^4}{b^3 \plus{} c^3} \plus{} \frac {b^4}{c^3 \plus{} a^3} \plus{} \frac {c^4}{a^3 \plus{} b^3} \geq \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca} \minus{} \frac {1}{2}(a \plus{} b \plus{} c)$[/quote]
Yes,it very easy.We can prove that:
$ \frac {a^6}{a^5 \plus{} b^5} \plus{} \frac {b^6}{b^5 \plus{} c^5} \plus{} \frac {c^6}{c^5 \plus{} a^5} \ge \frac {a \plus{} b \plus{} c}{2}$[/quote]

What does this have to do with the inequality?[/quote]

proving that $ \frac {a^6}{a^5 \plus{} b^5} \plus{} \frac {b^6}{b^5 \plus{} c^5} \plus{} \frac {c^6}{c^5 \plus{} a^5} \ge \frac {a \plus{} b \plus{} c}{2}$,the initial inequality will be indeed proved.it happens because of the rearrangement inequality.by the the way , to prove that $ \frac {a^6}{a^5 \plus{} b^5} \plus{} \frac {b^6}{b^5 \plus{} c^5} \plus{} \frac {c^6}{c^5 \plus{} a^5} \ge \frac {a \plus{} b \plus{} c}{2}$ just use cauchy´s inequality.

[quote="8826"]Let a,b,c be positive number
Prove that
 $ \frac {a^4}{b^3 \plus{} c^3} \plus{} \frac {b^4}{c^3 \plus{} a^3} \plus{} \frac {c^4}{a^3 \plus{} b^3} \geq \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca} \minus{} \frac {1}{2}(a \plus{} b \plus{} c)$[/quote]

Well, if you prove that $ \sum_{cyc} \frac {a^4}{b^3 \plus{} c^3} \geq \frac {a \plus{} b \plus{} c}{2}$ and applied
$ a \plus{} b \plus{} c  \geq  \sum_{cyc} \sqrt{ab}$ then the inequality can be solved.
The inequality $ \sum_{cyc} \frac {a^4}{b^3 \plus{} c^3} \geq \frac {a \plus{} b \plus{} c}{2}$ is very famous in this forum. You can search for an elegant proof.

Here is the much general one :D :
Let $ a,b,c$ be positive real numbers and $ x,y$ be the real numbers such that $ x\ge y \ge 0$. Prove that:
$ \frac {a^x}{b^y \plus{} c^y} \plus{} \frac {b^x}{a^y \plus{} c^y} \plus{} \frac {c^x}{a^y \plus{} b^y} \ge \frac {a^{x \minus{} y} \plus{} b^{x \minus{} y} \plus{} c^{x \minus{} y}}{2}$

[quote="dduclam"][quote="quykhtn-qa1"]
Yes,it very easy.We can prove that:
$ \frac {a^6}{a^5 \plus{} b^5} \plus{} \frac {b^6}{b^5 \plus{} c^5} \plus{} \frac {c^6}{c^5 \plus{} a^5} \ge \frac {a \plus{} b \plus{} c}{2}$[/quote]
Can you post your solution for this Inequality, quykhtn-qa1?  :wink:[/quote]
I think this is a hard inequality for the case $ a \leq b \leq c$ but I believe this is a true inequality. Anyway, what I found is that
if $ a \leq c \leq b$ then
$ \sum_{cyc} \frac{a^6}{a^5\plus{}b^5} \geq \sum_{cyc} \frac{a^3}{a^2\plus{}b^2} \geq \frac{a\plus{}b\plus{}c}{2}$


I've asked this before without really getting a satisfactory answer, but I'm not sure I asked it where stevem would see it:

In distribution theory, we have spaces of test functions: $ \mathfrak{D}$ for $ C_c^{\infty}$ and $ \mathfrak{S}$ for the Schwartz space of $ C^{\infty}$ functions that vanish rapidly at $ \infty.$ And then the distributions themselves are the dual spaces of these: $ \mathfrak{D}'$ for distributions, $ \mathfrak{S}'$ for tempered distributions.

I did those above as \mathfrak, but \mathfrak looks completely wrong to me. I see this D and this S (and there's an E for yet another space) printed in books in an ornate script-like font, with elaborate curlicues, especially on the S. Where can the desired symbols be obtained?

Do you think it's anything like these?
They are obtained by using the yfonts package which is included with MiKTeX. This adds Schwaber, Gothic and Fraktur text fonts with \textswab, \textgoth and \textfrak commands. I have also shown the \mathfrak version for comparison. You will find more details in the Latex Companion 2nd Edition pages 394-5 if you have a copy handy.

How about \usepackage{mathrsfs} which gives the result below? You get the same result with \usepackage{calrsfs} and \mathcal if you need \mathscr for anything else.

Thanks.  I was wondering how to denote an elliptic modular function.  I suppose it's:

\textswab{w}

Looks to be the same notation as Titchmarsh anyway.


Let $ a,b,c \in R$ and $ a\plus{}b\plus{}c\equal{}0$. Prove that $ \boxed{8^a\plus{}8^b\plus{}8^c\ge 2^a\plus{}2^b\plus{}2^c}$

[quote="thanhnam2902"]Let $ a,b,c \in R$ and $ a \plus{} b \plus{} c \equal{} 0$. Prove that $ \boxed{8^a \plus{} 8^b \plus{} 8^c\ge 2^a \plus{} 2^b \plus{} 2^c}$[/quote]
$ f(x) \equal{} 8^x \minus{} 2^x$ - concave then $ x > \minus{} Log_2(3)$ and arch then $ x < \minus{} Log_2(3)$
let $ a \le b \le c$ then for $ min(f(a) \plus{} f(b) \plus{} f(c))$ must  : $ b \equal{} c$ ;
=>$ Min(f(a) \plus{} f(b) \plus{} f(c)) \equal{} Min(f( \minus{} 2x) \plus{} 2f(x))$  will be then x = 0 $ \equal{} > f(a) \plus{} f(b) \plus{} f(c) \ge 0$;

[quote="thanhnam2902"]Let $ a,b,c \in R$ and $ a \plus{} b \plus{} c \equal{} 0$. Prove that $ \boxed{8^a \plus{} 8^b \plus{} 8^c\ge 2^a \plus{} 2^b \plus{} 2^c}$[/quote]
let $ 2^a \equal{}x, 2^b\equal{}y, 2^c\equal{}z$  then  we have : $ xyz\equal{}1$ and $ x,y,z$ posotive real numbers
ineq becomes :$ x^3\plus{}y^3\plus{}z^3 \geq x\plus{}y\plus{}z$
By  ineq AM-GM , we have :
 $ x^3\plus{}1\plus{}1 \geq 3x$
similar, $ LHS \geq x\plus{}y\plus{}z \plus{}2(x\plus{}y\plus{}z)\minus{}6 \geq x\plus{}y\plus{}z \plus{}6\sqrt[3]{xyz}\minus{}6 \geq x\plus{}y\plus{}z$

[quote="Hong Quy"]let $ 2^a \equal{} x, 2^b \equal{} y, 2^c \equal{} z$  then  we have : $ xyz \equal{} 1$ and $ x,y,z$ posotive real numbers
ineq becomes :$ x^3 \plus{} y^3 \plus{} z^3 \geq x \plus{} y \plus{} z$
By  ineq AM-GM , we have :
 $ x^3 \plus{} 1 \plus{} 1 \geq 3x$
similar, $ LHS \geq x \plus{} y \plus{} z \plus{} 2(x \plus{} y \plus{} z) \minus{} 6 \geq x \plus{} y \plus{} z \plus{} 6\sqrt [3]{xyz} \minus{} 6 \geq x \plus{} y \plus{} z$[/quote]

[color=darkblue]Thank HongQuy very much. Your solution is very nice for my problem.[/color]


That inequality is equivalent to $\sum (S-a_{i})\sum \frac{1}{S-a_{i}}\geq n^{2}$, now it is true by AM-GM.


quadratics
Vieta
algebra
quadratic formula

[hide]Using the quadratic formula for $a^{2}+19a+7=0$, we have

$a=\frac{-19\pm\sqrt{333}}{2}$. $x$ and $y$ must be the positive and negative parts of this. Now,

$x^{2}+y^{2}=\frac{333-38\cdot3\sqrt{37}+19^{2}-333-38\cdot3\sqrt{37}-19^{2}}{4}$

$=\frac{-19\cdot4\cdot3\sqrt{37}}{4}$
$=-19\cdot3\sqrt{37}$
$=\boxed{-57\sqrt{37}}$

Or, you can use newton's sums for a polynomial.[/hide]

[quote="vishalarul"][hide="The easy way"]Note that
$x^{2}+y^{2}=(x+y)^{2}-2xy$

Let $x=x_{1}$, $y=x_{2}$ (convenient notation)

Then, from Newton's sums,
$x_{1}+x_{2}=-19$
$(x_{1})(x_{2})=7$

Thus,
$x^{2}+y^{2}=19^{2}-2(7)=347$
[/hide]
Newton's sums make life a whole lot easier.  :P[/quote]

Those are [url=http://www.artofproblemsolving.com/Wiki/index.php/Vieta%27s_formulas]Vieta's Sums[/url] not [url=http://www.artofproblemsolving.com/Wiki/index.php/Newton_sums]Newton's Sums[/url]. But that is a good way to solve the problem.

[hide="Hint"]Use Root Coefficient relationship.[/hide]

I don't know what newton's sums and vieta's sums are but you don't need them here are a couple proofs of a couple formulas that I had in mind when I made the problem
[hide]If $ax^{2}+bx+c=0$ then x=$\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
So the product of the roots is
$(\frac{-b+\sqrt{b^{2}-4ac}}{2a})(\frac{-b-\sqrt{b^{2}-4ac}}{2a})=$a conjugate binomial=$\frac{b^{2}-b^{2}+4ac}{4a^{2}}=\frac{c}{a}$[/hide]
The other one is
[hide]
The sum of the roots
$\frac{-b+\sqrt{b^{2}-4ac}}{2a}+\frac{-b-\sqrt{b^{2}-4ac}}{2a}=\frac{-2b}{2a}=\frac{-b}{a}$[/hide]


geometry
geometric transformation
rotation
induction

How do the old 1990s to early 2000's tests compare to the "current" tests? (I'm talking about national sprint rounds.) Is it just me, or has the problem style changed drastically in the past few years?

Are the current tests easier than before? Harder? Require more insight? More "tricks"?

They have made an effort to take out a lot of the geometry problems since that is not considered a NORMAL middle school class.  The difficulty of the tests runs in cycles.  The last two or three have been on the harder side as have been all the tests from those years.  This year I think the tests are on the easier side so far so I expect the national test to be the same.  There should be a lot of ties and scores very close together both individually and teamwise.

I personally am hoping for the hardest test ever.  That way the kids and teams are differentiated a lot better.  On the easier tests the individual and the team are more severely punished for a careless mistake.

From what I have seen, the early 1990 tests tend to be really easy.

I agree with JBoyd that harder tests are better, but maybe this year state and chapter will be easy and nationals will be hard?

I really didn't notice much of a difference. I found that some of the older tests are easier and some of the older tests are harder

In general, they are getting harder. There are outliers (like 03 was very easy and 05 is very hard), but overall, the difficulty is slowly rising to compensate for the smarter kids, more hour of studying, etc. in these contests.

in general, what do u guys average for old compared to new...

old i guess is from 1990-1999 and new is 2000-2008......

I assume the older ones are the ones with 40 sprint and 10 target, and new is everything after that.

No, sorry, I meant the 1990-2002/2003 ish tests as compared to the 2004-2008 tests. I realize that there are lots of changes even in that time period, but many of the tests, in my opinion, were relatively similar in what they required. The current problems are sure getting more creative (Cats 'n fleas xD).

[quote="jonathanchou711"]1990 tests are a lot more easier than 2000 tests. The 84-90 tests were even easier and they had 40 sprint questions and 10 target questions in the same amount of time we do now.[/quote]

Correct me if I'm wrong, but I've always found it harder to do 40 problems in 40 minutes than to do 30 problems in 40 minutes? Also on the targets, they had even less time (4 minutes per pair as opposed to 6).

EDIT: Apparently on the sprint they had less time too, 30 as opposed to 40! So doing one problem per 45 seconds is easier than one problem per 80?

Firstly, for 1984-1989, I know that the time for the sprints was different. It was 40 questions in 30 minutes. You also must remember that back then, there was much more computation. I seem to remember this sprint problem from the 1990s that asked for 1/27 to the nearest thousandth. Of course, this is a bad example because it's just (1/9)/3, which is easy to do quickly, but still.

I haven't ever seen a 1984-1989 test, but judging from those team problems that were posted here, there were several legit problems, at least for team. For example, that rotation with the symmetries of the flower or whatever was, in my opinion, harder than most problems I've seen from 2000-2008 (and I've taken them all but one or two, which I should probably finish up).

The 1985 National Sprint #30 is a simple application of the Pythagorean theorem.  :|

Generally, I've found 1984 - 1989 were easier than now.

The older tests tend to be easier and they have different concepts

I think the main reason a lot of the older test are easier is that tons of different little shortcuts have been found recently that greatly lessen the work done.

Another reason:

People doing math in 1965, say, were marginally less smarter than people doing math in 1966, so we have marginally smarter test writers, so we get smarter as we have harder tests, and by induction, everything works

Bleh that would get a 0 in USAMO.

I wouldn't say that they were less smarter, I would say that they had less opportunity.

that too.

Also, those testwriters doing math in 1938 would have less opportunity than testwriters in 1939, so the tests would be slightly harder. = slightly smarter.

Well that depends where you draw the line between knowledge and intelligence.
(Did they even have math competitions in the 30s?  :maybe:

No, which is why there's a huge discrepancy between non-contest America and contest America.

Well you could say that about anything, couldn't you?

There is a big discrepancy between professional soccer player America and regular America.

There is a big discrepancy between contest science people and non-contest America.

There's a big discrepancy between the smell of people who rarely take showers and regular America.

The people who try hard (or in my last case, don't try at all) always make it out ahead, and in things such as math, this gap can get huge. (In any country, not just America.)

Anyways, getting back on topic. My opinion is that the older tests were more computation based and less trick/intuitiveness based. In addition as you go back, (in general) the tests get easier and easier.

[quote="jonathanchou711"]Cause you see nowadays, people are getting more beastly. Kids in like 5th grade are getting in Algebra and such.  :ninja:[/quote]

Although I agree with you that people are getting better at math, knowing advanced topics at a very shallow level does not equal beastliness. (Knowing advanced topics at a very deep level does, but in my opinion knowing simple topics at a very deep level also does.)

i'll give u some if u change ur avatar and location

[quote="james4l"]@Ihatepie: sorry, but I think you misinterpreted my comment. What I mean is a time difference - when America did not have contests, they were significantly worse at math than when they had contests.[/quote] sorry  :blush:

[quote="USAMO13"]Does anyone have old MC contests please!
I'll do anythin for 'em[/quote]

Wrong Topic.

Back on track, regarding pythag's comment, you have to know a broad amount of advanced topics at a deep level to equate beastliness.

Hm.. I've done even AMC's before and the older ones are exceptionally easier. Didn't they used to use calculators on AMC8's?  :maybe:


inequalities
trigonometry
ratio
real analysis
real analysis unsolved

you have a circle and two points A and B laying on the circle diametrically opposite to each other; you have to go from A to B;
if you travel on the diameter your speed is $ v_2$, while if you travel on the circle your speed is $ v_1$, with $ v_1 > v_2$; you can also mix the two paths (this should be longer, but I can't prove it);
which path do you choose (and why)?

I don't see how there is enough information to answer that question.  You say $ v_{1}>v_{2}$ but how much faster?  How large is the circle?  Not enough information.

the lenght of the circle shouldn't be necessary to minimize; about $ v_1$ and $ v_2$ that's what the man who gave me the problem told; I think too that the inequality should be something like $ \frac{v_1}{v_2}<2/\pi$ or similar, but I wanted a help to check.

[quote="Clausius"][color=darkred]We have a semicircle $ \mathcal S (O,R)$ with the diameter $ [AB]$. We want to go from $ A$ to $ B$ such : we travel on the semicircle from the point $ A$ until an ascertain point $ M$ with the speed $ v_1$ and after we travel on the segment $ [MB]$ with the speed $ v_2$ ([b][u]independent of the point[/u][/b] $ M$), where $ v_1 > v_2$. Ascertain the position of the point $ M$ so that the crossing time is minimum.[/color] [/quote]
[color=darkblue][b][u]Proof.[/u][/b] Suppose w.l.o.g. that $ R \equal{} 1$. Denote $ m(\widehat {AOM}) \equal{} x$. Thus, $ \stackrel{\frown}{AM} \equal{} x$ and $ MB \equal{} 2\cos\frac x2$. The crossing time is $ f(x) \equal{} \frac {x}{v_1} \plus{} \frac {2\cos\frac x2}{v_2}$, where $ 0\le x\le \pi$. Observe that $ f'(x) \equal{} \frac {1}{v_2}\cdot\left(\frac {v_2}{v_1} \minus{} \sin\frac x2\right)$, where $ 0 < \frac {v_2}{v_1} < 1$ and $ 0\le\frac x2\le\frac {\pi}{2}$. Therefore : if $ \frac {v_1}{v_2}\le \frac {\pi}{2}$, then we travel only the diameter ($ x \equal{} 0$) ; if $ \frac {v_1}{v_2} > \frac {\pi}{2}$, then we travel only the semicircle ($ x \equal{} \pi$ ) !

[b][u]Remark.[/u][/b] Here is an similar problem :[/color]

[quote="Virgil Nicula"][color=darkred]We have a semicircle $ \mathcal S (O,R)$ with the diameter $ [AB]$. We want to go from $ A$ to $ B$ such : we travel on a segment $ [AM]$ from the point $ A$ until an ascertain point $ M$ with the speed $ v_2$ and after we travel on the semicircle from the point $ M$ to the point $ B$ with the speed $ v_1$ ([b][u]independent of the point[/u][/b] $ M$), where $ v_1 > v_2$. Ascertain the position of the point $ M$ so that the crossing time is minimum.[/color] [/quote]

What exactly does it mean to "mix the two parts"? One possible interpretation is that you can move along a circle or any of its diameters. Going across a diameter will take you $ 2r/v_2$ units of time. Going around a half-circle takes $ \pi r/v_1$. If you travel along some part of a circle [b]and[/b] along a diameter, you will spend more than $ 2r/v_2$, so that's not going to be a winner. So you compare $ 2/v_2$ and $ \pi/v_1$ and choose whichever is smaller.

Another (harder) problem arises when you are allowed to move along any chords with speed $ v_2$. The answer stays the same, but requires more thought.

This sounds to me like a problem I solved for the efficient navigation of Budapest: suppose that you can travel along rays through the origin at speed $ v_1$ and along circular arcs centered at the origin at speed $ v_2$; what are the shortest paths between two arbitrary points, depending on the ratio $ v_1/v_2$?


let ABCD is a quadrilateral .Let M is a point in AB side,N,P in CD side
AN,AP intersect AM at A(1),A(2)
BN,BP intersect CN at B(1),B(2)
prove that: AC,BD,A(1)B(1),A(2)B(2) concurrent

Because of Papos theorem for points $A,M,B$ and $C,N,D$ we find out that lines $AC,BD,A_{1}B_{1}$ concur. Similar lines $AC,BD,A_{2}B_{2}$ concur, so all four concur in point $O$, which is intersection of diagonals of $ABCD$ .


See answers to the second question [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=312678]here[/url].

[quote="hellomynameis"]Does the USAMTS normally get harder with each successive round, or is it just me?[/quote]

I thought it should, but for some reason I did WAY better on Round 2 than Round 1.


[quote="Altheman"]are the boxes different?[/quote]

yes

[quote="Altheman"][hide="umm..."]Then wouldn't just be 11*13 because you can place 0 to 11 white balls in the first urn, and the remainder in the second urn and 0 to 12 red balls in the first urn, and the remainder in the second urn.

I am not that good at counting so this seems right to me, but I could still be wrong.[/hide][/quote]

[hide] As Altheman, I'm not sure if my resul is right, but I got $312$. I did it by counting how many different ways were to do it.[/hide]

[quote="Altheman"][hide="umm..."]Then wouldn't just be 11*13 because you can place 0 to 11 white balls in the first urn, and the remainder in the second urn and 0 to 12 red balls in the first urn, and the remainder in the second urn.

I am not that good at counting so this seems right to me, but I could still be wrong.[/hide][/quote]

[hide] As Altheman, I'm not sure if my result is right, but I got $312$. I did it by counting how many different ways were to do it.[/hide]

Seems less than 143.
[hide]Let $A$= the set of all of the ways to split up the 10 white balls into the two distinguishable boxes. Say $(a_1,a_2)$ represents placing $a_1$-many white balls into box one, and $a_2$-many balls into box 2. Then $A= \{ (0,10),(1,9),(2,8),(3,7),(4,6),(5,5),(6,4),(7,3),(8,2),(9,1),(10,0) \}$

Similarly, we get $B= \{ (0,12),(1,11),(2,10),(3,9),(4,8),(5,7),(6,6),(7,5),(8,4),(9,3),(10,2),(11,1),(12,0) \}$

$|A|=11$, and $|B|=13 \Rightarrow 11 \cdot 13=143$ distributions, but what about $(0,10)+(12,0)$ giving $(12,10)$, we also have $(5,5)+(7,5)$ giving $(12,10)$ as well; which is the same distribution. So it looks like there are over counts in this argument.[/hide]

[quote="RohanSingh"]In how many ways one can distribute 10White balls and 12Red Balls in to two boxes. Balls of same cplor are indistinguishable.[/quote]Break it into 2 problems, the first being distributing white, the second distributing red. The total will be the product of these (since it's an "and"). Therre are 11 ways to separate the white balls (0, 10), (1, 9), ... , (10, 0) and likewise, 13 ways to distribute the red. So multiply together and there are 143 ways to separate both. (Pretty sure this is right, correct me if I'm wrong though.)
To TheJerk: he said that balls of different color ARE distinguishable. If they weren't it would just be 22 balls into 2 boxes, i.e. 23 ways.


[hide]Actually, observing this with complementary counting

10^5-9^5=40951 is the answer, think about why this is true...correct me if I am wrong[/hide]

[quote="bpms"][quote="klutzycherry14"]if a password is 5 digits, and 7 is one of the digits, how many possible combinations are there for the password?? :?:  :huh:[/quote]
[hide]Well, the other four digits can have 10000 combos. The 7 can be in 5 spots so there are 50000 ways.[/hide][/quote]
This overcounts in cases with more than one 7.

ProtestanT, why are there 65 possibilities for each digit?? :maybe:

[quote="klutzycherry14"][quote="ProtestanT"][quote="klutzycherry14"]if a password is 5 digits, and 7 is one of the digits, how many possible combinations are there for the password?? :?:  :huh:[/quote]

[hide="solution"]
each password digit have 65 possible values.
the answer is
$65^{4}$ or $16777216$
[/hide][/quote]

each digit has 65 values?? explain please... :maybe:  :oops:[/quote]
I think he made mistake. 65 possibilities doesn't make any sense... (unless I am missing something)

[quote="ProtestanT"]omg someone deleted my list of the 65 possible values
they are
abcdefghijklmnopqrstuvwxyz0123456789`~!@#$\$$%^&*()-=_+[]\{}|;':",./<>?[/quote]
Generally, digits only means 0,1,2,3,4,5,6,7,8,9

[quote="ProtestanT"]omg someone deleted my list of the 65 possible values
they are
abcdefghijklmnopqrstuvwxyz0123456789`~!@#$\$$%^&*()-=_+[]\{}|;':",./<>?[/quote]
Don't be ridiculous.  :wink:  The problem stated
[quote="klutzycherry14"]if a password is 5 [b]digits[/b], and 7 is one of the [b]digits[/b], how many possible combinations are there for the password??[/quote]

[quote="ProtestanT"]then wouldnt the problem be a little too easy to belong here?[/quote]
No, this is perfectly at MathCounts level.
In fact, if the problem had even been what you thought it was, your answer and solution still would have both been wrong.  :wink:
And by the way, $65^{4}\neq 16777216$.  :P

[quote="ProtestanT"]but naturally in passwords, anything is allowed[/quote]
They've already given you all of the conditions in the problem statement. You can't just make up new conditions.  :wink:

but naturally in passwords, [i]anything[/i] is allowed

Okay, now assume that you can only use digits in other words (0,1,2,3,4,5,6,7,8,9)

Could we go from there?

[hide][b][color=red]ok people...since this is my problem...im gonna let u all know that i meant only digits as in numbers...(1,2,3,4,5,6,7,8,9,0,)
k? let's end the confusion![/color][/b][/hide]

[quote="ProtestanT"]$65^{4}$ or $16777216$[/quote]

u should know 5^anything =..............5

the last number is always 5

-jorian

oops lol i did $64^{4}$...anyways the correct answer is
[hide]
$5 * 10^{4}$
[/hide]

[quote="nutz_for2.718281828"][hide="Here is my solution."]We can use complementary counting. There are 100000 total arrangements, and there are 59049 arrangements where there is no 7. Thereore, there are 40951 combinations.[/hide][/quote]
he didnt say since one of teh digits is 7, the other digits cant be 7.

[quote="anirudh"][quote="klutzycherry14"]if a password is 5 digits, and 7 is one of the digits, how many possible combinations are there for the password?? :?:  :huh:[/quote]

[hide]I got $1\cdot 10\cdot 10\cdot 10\cdot 10$, which is $10^{4}$=$10000$[/hide][/quote]
You didn't consider when the seven could be in different spots and when there can be more than one seven.

No, $10^{5}-9^{5}$ is the correct answer. There are $10^{5}$ 5-digit strings of digits, and there are $9^{5}$ 5-digit strings without the digit 7.

jhredsox, "she", not "he".  :wink:


Do there exist $5$ points in the space, such that for all $n\in\{1,2,\ldots,10\}$ there exist two of them at distance between them $n$?

[quote="pco"][quote="Igor"]Do there existe $5$ points in the space such that for all $n\in\{1,2,\ldots,10\}$ there exist two points such that the distance between them is $n$?[/quote]

5 points $\Rightarrow $ exactly 10 distances point to point $\Rightarrow $ each ten distances 1 to 10 occur just once.
Then let $XY=1$ and Z another point. WLOG say $XZ>YZ$ (equality is impossible, as said above). Then, in triangle XYZ, $XZ\leq XY+YZ$ $\Rightarrow $ $0<XZ-YZ\leq XY=1$ and, since XZ and YZ are integers, $XZ=YZ+1$ $\Rightarrow $ X,Y and Z are colinear $\Rightarrow $ All five points, if they exist, are colinear.

Then, WLOG say the five colinear points are, in this order, A,B,C,D and E, with $AE=10$. The sum of the ten distances is :
$S=AB+(AB+BC)+(AB+BC+CD)+10+BC+(BC+CD)$ $+(10-AB)+CD+(10-AB-BC)+(10-AB-BC-CD)$
$S=40+2BC+2CD$, even, which can't be equal to $1+2+3+4+5+6+7+8+9+10=55$, odd.

So such 5 points don't exist.

--
Patrick[/quote]

My way for end:
$9$ can only be formed if $1$ is outside, also $2$ outside then or we can't form $8$ and then we have 2 cases with no solutions.

If possible, let the Pentagon formed be non-degenerate. Since $\binom{5}{2}=10$, every pair is connected. But, by triangle inequality, if one of the lengths is $1$ and the other sides of triangle are integers, then not possible. So some 3 points are collinear and dist between 2 of them is $1$. Let the points be $X,Y,Z$ such that $Y$ is between $X,Z$ and $XY=1$.
Now let the other 2 points be $P,Q$ not on line $XYZ$. Again $PXY$, $QXY$ don't satisfy triangle inequality so all points are collinear, say in the order $PQXYZ$.
Now $PQ \neq QX \neq XY \neq YX$ so their sum is at least $1+2+3+4+5=15>10$ not possible so done.


geometry perimeter AMC AIME trigonometry

Express the perimeter of the triangle $ ABC$ as a decimal rounded to the nearest hundredth.

[asy]real wedge(pair v, pair w){
  return (v.x*w.y) - (v.y*w.x);
}

pair intersect(pair A, pair v, pair B, pair w){
  real t = wedge(B-A,w)/wedge(v,w);
  return A + t*v;
}

pair degreePointer(real deg){
  return (cos(deg*pi/180),sin(deg*pi/180));
}

pair C = (0,0), B=(16,0);
pair A = intersect(B,degreePointer(135),C,degreePointer(60));
draw(A--B--C--cycle);
label("$A$",A,N);
label("$C$",C,W);
label("$B$",B,E);
label("12",(C+A)/2,NW);
label("$60^\circ$", C+(1,0), NE);
label("$75^\circ$",A+(0.5,-1),S);
label("$45^\circ$",B-(1,0),NW);[/asy]


function
induction
inequalities
number theory unsolved
number theory

find f:N->N so that f(x+y)<=f(x)*f(y)-1
                            f(x+y)=f(2x)-f(x-y) :lol:

put x-y=a, x+y=b, then f(a)+f(b)=f(a+b) so f(n)=f(1)n by induction
now, say f(1)=p, then from the inequality p(x+y)<=p 2 xy-1, which is equivalent to: 2<=(px-1)(py-1), since the rhs is increasing function in two variables, its minimun is at x=y=1, so we only need: 2<=(a-1) 2 , so a>=3 and we have infinitely many solutions for the functional ecuation...

Was it easy or i made a mistake?

A question: do you consider that N contains 0? It's quite important. A second observation: the solution is wrong. From $ f(x-y)+f(x+y)=f(2x)$ you cannot deduce that $ f(a)+f(b)=f(a+b)$. Remember you work with natural numbers.

Sorry, you are right :blush:  :blush:  :blush:  :blush:  :blush:

I was in a hurry when i did it


limit
logarithms
real analysis
real analysis unsolved

Let $(a_{n})$ be a real decreasing sequence st $\forall n \in \mathbb N : a_{n}>0$
study the nature of the serie $\sum \frac{a_{n}-a_{n+1}}{a_{n+1}}$

That depends on the value of $A=\lim_{n\to\infty}a_{n}.$

If $A>0,$ then the series converges.

$\sum_{n=1}^{\infty}\frac{a_{n}-a_{n+1}}{a_{n+1}}\le \frac1A\sum_{n=1}^{\infty}(a_{n}-a_{n+1})=\frac{a_{1}-A}{A}.$

Now suppose $A=0.$ I would conjecture that the series diverges. That holds for each of the examples I've tried, and I've tried $a_{n}=\frac1{n^{p}}$ for some $p>0, a_{n}=r^{n}$ for some $r<1,$ $a_{n}=\frac1{\ln n},$ and $a_{n}=\frac1{\ln\ln n}.$ I get divergence in each of those cases.

Aha! Now I have the proof.

Use the fact that $x-1>\ln x.$

$\sum_{n=1}^{N}\frac{a_{n}-a_{n+1}}{a_{n+1}}= \sum_{n=1}^{N}\left(\frac{a_{n}}{a_{n+1}}-1\right)> \sum_{n=1}^{N}\ln\left(\frac{a_{n}}{a_{n+1}}\right)$

$=\ln\left(\prod_{n=1}^{N}\frac{a_{n}}{a_{n+1}}\right)= \ln\left(\frac{a_{1}}{a_{N+1}}\right)=\ln(a_{1})-\ln(a_{N+1})$

But if $a_{N}\to0$ as $N\to\infty,$ then the limit as $N\to\infty$ of the sum is infinity.

nice Proof Kent  :lol:
I didn't think of using the fact that $x-1 > ln(x)$ to prove the divergence.


trigonometry
real analysis
real analysis unsolved

It is known that the equation $ x^2\equal{}cosx$ has two solutions. Using the taylor series of cosx approximate these solutions.

aty first you need to note that if $ |x| > 1$ then $ x^{2} > \cos x$ hence root of this equation is on interval $ ( \minus{} 1,1)$ thus we can write taylor expansion, say $ \cos x \approx 1 \minus{} \frac {x^{2}}{2}$ and we have
$ x^{2} \approx 1 \minus{} \frac {x^{2}}{2}$ then $ x \approx \sqrt {\frac {2}{3}}$ or $ x \approx \minus{} \sqrt {\frac {2}{3}}$
suppose
 $ x \approx \sqrt {\frac {2}{3}}$ then consider equation
$ x^{2} \equal{} \cos \left(\sqrt {\frac {2}{3}} \plus{} \alpha \right)$  where $ \alpha$ is very small , then say $ \cos(z \plus{} y) \equal{} \cos z \cos y \minus{} \sin z \sin y$
where $ z \equal{} \sqrt {\frac {2}{3}}$
and
$ y \equal{} \alpha$
and use same method  ($ \cos \alpha \approx taylor$) this approximation will be better since $ \alpha$ is small than our first $ x < 1$
after yourself .


Prove that $(a+b)/(c+d)$ is irreducible if $ad-bc=1$.

[hide="Proof"]Assume $a+b$ and $c+d$ have a common divisor $m>1$. Then $a+b=mx, c+d=my$ for some integer $x,y$. Therefore $b=mx-a, d=my-c$, giving

$1=ad-bc=a(my-c)-(mx-a)c=amy-ac-cmx+ac=m(ay-cx)$, which means that $m\mid 1$, and that's impossible.[/hide]

[quote="me@home"]Does only if hold also?[/quote]

No. $1 = \frac{1+5}{3+3}$, but $|5\cdot 1-3\cdot 3| = 4 \neq 1$ for example.

[hide="Really nice"] The quadrilateral formed by $(0, 0), (a, c), (b, d), (a+b, c+d)$ has area $1$, so it contains no interior lattice points by Pick's Theorem.  Hence there exists no point $(x, y)$ on the line between $(0, 0), (a+b, c+d)$.  QED. [/hide]

that's an amazing proof, congratulations t0rajir0u!!


Let $ a, b, c \ge 0$. Prove that: $ \boxed{\frac {a^{2008} \plus{} b^{2008} \plus{} c^{2008}}{3} \ge \displaystyle \left (\frac {a \plus{} b \plus{} c}{3} \right)^{2008}}$

[quote="delegat"]This inequality is equivalent to:

$ \sqrt [2008]{\frac {a^{2008} \plus{} b^{2008} \plus{} c^{2008}}{3}} \geq \frac {a \plus{} b \plus{} c}{3}$ which is true because of Power Mean inequality.[/quote]
 Let $ a\plus{}b\plus{}c\equal{}3$  then  solution becomes shorter,
We  need prove : $ a^{2008} \plus{}b^{2008} \plus{}c^{2008} \geq 3$
it's killed by AM-GM ineq

[quote="Hong Quy"][quote="delegat"]This inequality is equivalent to:

$ \sqrt [2008]{\frac {a^{2008} \plus{} b^{2008} \plus{} c^{2008}}{3}} \geq \frac {a \plus{} b \plus{} c}{3}$ which is true because of Power Mean inequality.[/quote]
 Let $ a \plus{} b \plus{} c \equal{} 3$  then  solution becomes shorter,
We  need prove : $ a^{2008} \plus{} b^{2008} \plus{} c^{2008} \geq 3$
it's killed by AM-GM ineq[/quote]

[size=150][color=darkblue]But $ a\plus{}b\plus{}c\neq 3$ ?[/color][/size]

[quote="thanhnam2902"][quote="Hong Quy"][quote="delegat"]This inequality is equivalent to:

$ \sqrt [2008]{\frac {a^{2008} \plus{} b^{2008} \plus{} c^{2008}}{3}} \geq \frac {a \plus{} b \plus{} c}{3}$ which is true because of Power Mean inequality.[/quote]
 Let $ a \plus{} b \plus{} c \equal{} 3$  then  solution becomes shorter,
We  need prove : $ a^{2008} \plus{} b^{2008} \plus{} c^{2008} \geq 3$
it's killed by AM-GM ineq[/quote]

[size=150][color=darkblue]But $ a \plus{} b \plus{} c\neq 3$ ?[/color][/size][/quote]
if let $ a \equal{} ka' , b \equal{} kb',c \equal{} kc'$  then  ineq's  right too when suppose that $ a \plus{}b\plus{}c\equal{}3$

You prove this problem: Let $ a, b, c \ge 0$. Prove that: $ \boxed{\frac {a^{n} \plus{} b^{n} \plus{} c^{n}}{3} \ge \displaystyle \left (\frac {a \plus{} b \plus{} c}{3} \right)^{n}}, \forall n \in N^*$

So, we let $ n\equal{}2008$ then OK!

[quote]Is it still true when a,b,c are real numbers?[/quote]

If $ n$ is an even number then it is obviously true but if $ n$ is an odd sometimes it is true sometimes not (for example if we take negative $ a$, $ b$ and $ c$ and odd $ n$ reverse inequality holds).


Find the number of ordered triples of non-negative integers (a,b,c) such that a+b+c=2007.

Find the number of ordered triples of positive integers (a,b,c) such that a+b+c=2007.

Find the number of ordered triples of positive integers (a,b,c) each $\ge2$ such that a+b+c=2007.

These are, essentially, the same problem, they all use the same method but #2 and #3 require an extra step.

It is not intended that you do casework on a, because that's a complete time-waster, and very computationally heavy.

You many express as ${n\choose r}$ hint hint

I'm sure about this one:
[hide="1"]
Using balls and urns, it's
$\binom{2009}{2}$.
[/hide]

Not so sure about this one:
[hide="2"]
Each number has to be at least 1, so leave the 1st ball on each end alone.  Then it's $\binom{2007}{2}$.  But we have to subtract the 2006 times when the 2 dividers are next to each other, so it's $\binom{2007}{2}-2006$.
[/hide]

Didn't get the 3rd one. :huh:

EDIT: Now I did (I hope)!  Oh, it's 2007 now?
[hide="3"]

Since each number has to be $\ge{2}$,the 1st and 2nd balls on each end have to be left alone.  Then it's $\binom{2005}{2}$.  But we have to subtract the 2004 times when the 2 dividers are next to each other and the 2003 times when the 2 dividers are one space apart.  So it's $\binom{2005}{2}-4007$
[/hide]

EDIT again: Corrected a error.
Those probably aren't right.

umm, i got [hide] 2008 * 2009 / 2 = 2017036 for the first one[/hide] im probably wrong, though. How did you use balls and urns for this?

The sum of the numbers is 2007.

Suppose there are 2007 balls, and each number is represented by a number of balls.
Put 2 dividers in there to separate the balls that represent the 3 numbers.

[quote="CircleSquared"]The sum of the numbers is 2007.
[hide]
Suppose there are 2007 balls, and each number is represented by a number of balls.
Put 2 dividers in there to separate the balls that represent the 3 numbers.[/quote]
oh, I guess I should have thought of that  :blush: thks for the explanation.
However, I got the same answer as madskillz for 2 and 3. I didn't really understand why you got 2007 C 2 for your explanation.[/hide]

[quote="CircleSquared"][hide]
If you look at the answer, $\binom{2007}{2}-2006$ is the same as $\binom{2006}{2}$.  However, my #3 doesn't match, what was your solution?

I got $\binom{2007}{2}$ in the beginning of the 2nd question because:
I used balls and urns, but you have to leave the 1st ball on each end alone (or else one of the end numbers would be 0). So we choose 2 from the remaining 2007 items to be the dividers.
[/hide]

EDIT: Forgot to hide!

Tell me if it isn't clear (which it probably isn't)[/quote]
oh i understand now, but I think madskillz's way is quicker, since that way you don' t need to calculate how many ways the two dividers are together. Thanks again for the explanations.  :)

[quote="madskillz1"]its just 2009C2, 2006C2, 2003C2 (you use the balls and urns method 3 times, however for the second time it becomes a+b+c=2004, third is a+b+c=2001)[/quote]

Your answer is correct. :)

[quote="CircleSquared"]I'm kinda confused as to why $\binom{2003}{2}$ works for the 3rd one, can anyone explain?[/quote]

Um,

\[a+b+c=2007 \]
now we have the annoying restriction $a,b,c \ge 2$.

To reduce this, we can subtract 2 from each:
\[(a-2)+(b-2)+(c-2)=2007 \]
So, whenever we find three nonnegative integers $a-2$, $b-2$, and $c-2$ that sums to 2007, they will always make $a$, $b$, and $c$ at least $2$.

However, the equation can be reduced as
\[a+b+c=2007-6=2001 \]
which gives $\binom{2003}{2}$.

That's a great solution, thanks!

EDIT:
In the second equation, shouldn't the right hand side be 2001?

BTW generalization:
If z numbers sum to x, and each is at least y, the number of ordered pairs of those numbers is
$\binom{x-yz+z-1}{z-1}$
Restrictions apply, such as zy<x, etc.

...sorry, that was kind of random. :dry:

How about this problem?

Find the number of ordered triples of positive integers $(a,b,c)$ with $a\geq 1,\ b\geq 10,\ c\geq 100$ such that a+b+c=2007.

By kohjhsd's method:

[hide]
$\binom{1898}{2}$
[/hide]

That's right, CircleSquared. :)

Challenge the following problem from Japanese university entrnace exam.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=51376&search_id=118914364[/url]


abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

I need help proving the following:

1. Let $ R$ be a Noetherian ring, $ I$ an ideal, and $ N \subseteq M$ be $ R$-modules. Let $ \frac{M}{N}$ be a finitely generated $ R$-module. Prove that $ \frac{IM}{IN}$ is a finitely generated $ R$-module.

2. Let $ R$ be a Noetherian ring, $ I$ an ideal of R. Prove that $ M$ is a finitely generated $ R$-module iff $ \frac{M}{\text{Nil}(R)M}$ is a finitely generated $ \frac{R}{\text{Nil}(R)}$-module.

The first problem seems to be straightforward. Can I just take the generators and multiply them by $ I$ to show this is finitely generated? Thanks for any suggestions with this one.

As for the second one, I am not seeing how to use the first part in this problem. I am guessing that I just let $ I\equal{}\text{Nil}(R)$ for one direction; but the other direction is confusing me.

Ad 2.: $ R$ is Noetherian, hence the ideal $ I\equal{}nil(R)$ is nilpotent, i.e. $ I^n\equal{}0$ for some $ n>0$. Now apply (1.) to the subsequent quotients of the finite chain of inclusions $ 0\equal{}I^nM\subseteq I^{n\minus{}1}M\subseteq \ldots\subseteq IM\subseteq M$ (using the assumption that $ M/IM$ is finitely generated over $ R$) to see that every step has a finitely generated quotient. This implies that $ M\equal{}M/I^nM$ is finitely generated over $ R$ as well.

Note that the technique shows that in many cases it suffices to assume that $ I\equal{}nil(R)$ has square 0: $ I^2\equal{}0$.


quadratics
algebra
polynomial
complex numbers

Let $ \alpha ,\ \beta$ be the roots of a quadratic equation : $ 2x^2\plus{}4x\plus{}3\equal{}0$. Find the values of the following expressions.

$ (\alpha \minus{}1)(\beta \minus{}1)$.

$ (\alpha \minus{}1)^4\plus{}(\beta \minus{}1)^4$.

I don't feel like using Greek symbols, so let $ a\equal{}\alpha$ and $ b\equal{}\beta$.
(a) $ (a\minus{}1)(b\minus{}1)\equal{}ab\minus{}(a\plus{}b)\plus{}1$.  We know that, from our quadratic,
$ a\plus{}b\equal{}\minus{}2$ and $ ab\equal{}\frac{3}{2}$. Therefore, we have that our expression
equals $ \frac{3}{2}\plus{}2\plus{}1\equal{}\boxed{\frac{9}{2}}$.
(b) Let $ e\equal{}a\minus{}1$ and $ d\equal{}b\minus{}1$.  We are looking for $ e^{4}\plus{}d^{4}$.
This equals $ (e^{2}\plus{}d^{2})^{2}\minus{}2(ed)^{2}$.  $ e^{2}\plus{}d^{2}\equal{}
(e\plus{}d)^{2}\minus{}2ed$.  From (a), $ ed\equal{}\frac{9}{2}$.  $ e\plus{}d$, meanwhile,
equals $ a\minus{}1\plus{}b\minus{}1\equal{}a\plus{}b\minus{}2\equal{}\minus{}4$.  Plugging these values in, we get that
our expression equals $ \boxed{\frac{17}{2}}$.

[hide="1st"]
$ f(x)\equal{}2x^2\plus{}4x\plus{}3$
$ \alpha\minus{}1$ and $ \beta\minus{}1$ will be roots of $ f(x\plus{}1)\equal{}2(x\plus{}1)^2\plus{}4(x\plus{}1) \plus{}3\equal{}2x^2\plus{}8x\plus{}9$

Thus, product of $ \alpha\minus{}1$ and $ \beta\minus{}1$ is $ \frac{9}{2}$
[/hide]

[hide="2nd - Can someone tell me what I did wrong?"]
Again, $ f(x)\equal{}2x^2\plus{}4x\plus{}3$
$ (\alpha\minus{}1)^4$ and $ (\beta\minus{}1)^4$ will be roots of $ f(\sqrt[4]{x}\plus{}1)\equal{}2(\sqrt[4]{x}\plus{}1)^2\plus{}4(\sqrt[4]{x}\plus{}1)\plus{}3\equal{}2\sqrt{x}\plus{}8\sqrt[4]{x}\plus{}9$

Let $ g(x)\equal{}2\sqrt{x}\plus{}8\sqrt[4]{x}\plus{}9$
Even if we substitute $ x^4$ for $ x$, the polynomial is the same.
$ g(x^4)\equal{}2x^2\plus{}8x\plus{}9$

Thus, the sum of the roots is $ \minus{}4$.[/hide]

[hide="Solution"]
Just... find the roots  :lol:

The roots are $ \frac{\minus{}4\pm\sqrt{16\minus{}4\cdot2\cdot3}}{4}\equal{}\frac{\minus{}4\pm2i\sqrt{2}}{4}\equal{}\frac{\minus{}2\pm i\sqrt{2}}{2}$.

So part one is:
$ \frac{\minus{}4\plus{}i\sqrt{2}}{2}\cdot\frac{\minus{}4\minus{}i\sqrt{2}}{2}\equal{}\frac{16\plus{}2}{4}\equal{}\boxed{\frac{9}{2}}$.

And part two is:
$ \frac{(\minus{}4\plus{}i\sqrt{2})^4\plus{}(\minus{}4\minus{}i\sqrt{2})^4}{16}$

Now $ (\minus{}4\plus{}i\sqrt{2})^4\equal{}(14\minus{}8i\sqrt{2})^2\equal{}68\minus{}224i\sqrt{2}$
And because of our friend, the property: $ \overline{z_1\cdot z_2}\equal{}\overline{z_1}\cdot\overline{z_2}$, we have the other root as $ 68\plus{}224i\sqrt{2}$.

So we have $ \frac{136}{16}\equal{}\boxed{\frac{17}{2}}$
[/hide]

Complex numbers? On a nursery/medical care exam?

[quote="FantasyLover"][hide="1st"]
$ f(x) \equal{} 2x^2 \plus{} 4x \plus{} 3$
$ \alpha \minus{} 1$ and $ \beta \minus{} 1$ will be roots of $ f(x \plus{} 1) \equal{} 2(x \plus{} 1)^2 \plus{} 4(x \plus{} 1) \plus{} 3 \equal{} 2x^2 \plus{} 8x \plus{} 9$

Thus, product of $ \alpha \minus{} 1$ and $ \beta \minus{} 1$ is $ \frac {9}{2}$
[/hide]

[hide="2nd - Can someone tell me what I did wrong?"]
Again, $ f(x) \equal{} 2x^2 \plus{} 4x \plus{} 3$
$ (\alpha \minus{} 1)^4$ and $ (\beta \minus{} 1)^4$ will be roots of $ f(\sqrt [4]{x} \plus{} 1) \equal{} 2(\sqrt [4]{x} \plus{} 1)^2 \plus{} 4(\sqrt [4]{x} \plus{} 1) \plus{} 3 \equal{} 2\sqrt {x} \plus{} 8\sqrt [4]{x} \plus{} 9$

Let $ g(x) \equal{} 2\sqrt {x} \plus{} 8\sqrt [4]{x} \plus{} 9$
Even if we substitute $ x^4$ for $ x$, the polynomial is the same.
$ g(x^4) \equal{} 2x^2 \plus{} 8x \plus{} 9$

Thus, the sum of the roots is $ \minus{} 4$.[/hide][/quote]It is easy to find the sum of the roots of a quadratic equation without solving it. The sum of the solutions of an equation that is not quadratic (even if it can be transformed into a quadratic equation by using some non-linear transformation) is a different story.
The values of $ y$ that are roots of $ y^2\minus{}3y\plus{}2\equal{}0$ are $ y\equal{}1$ and $ y\equal{}2$.
The sum of the roots of the quadratic equation is $ 1\plus{}2\equal{}3$, as expected.
The equation that results if we substitute $ y\equal{} \sqrt [4]{x}$ is
$ \sqrt {x} \minus{}3\sqrt [4]{x} \plus{} 2\equal{}0$
It is not a quadratic equation in $ x$, but we transform it into the quadratic equation in $ y$ above to solve it and find roots $ x\equal{}1$ and $ x\equal{}16$.
The sum $ \sqrt [4]{1}\plus{} \sqrt [4]{16}$ is still $ 3$, but $ 1\plus{}16\equal{}17$.
If you solve your equation, you might find
$ \sqrt [4]{x}\equal{}\minus{}2\plus{}i \sqrt{2} /2$ and $ \sqrt [4]{x}\equal{}\minus{}2\plus{}i \sqrt{2} /2$, which would mean
$ x\equal{}17/4 \plus{} 14\sqrt{2} i$ and $ x\equal{}17/4 \minus{} 14\sqrt{2} i$.
As you add those two solutions, you get $ \boxed{\frac{9}{2}}$. You have a novel way of solving the problem, but the path less traveled sometimes requires using a machete.

or substitute x with x + 1 and use newton sums for #2

All that for Nursery entrance exam! I wonder what type of test should an engineer or a mathematician take just to enter a university.  :o


APMO is tomorrow!

Good luck to all Canadian Mathlink/AoPSers! :)

Looking forward to an [u]even[/u] APMO~~

Good luck to everyone :)

[quote="Singular"]It is 516 AM.  I am awake.  Therefore, not going to be a good APMO for me.[/quote]
730 for me. :)

By the way, we should NOT discuss APMO on the forum until after the problems have been posted on the official website.

The solutions are up. let's discuss how everyone did.
I got about 19 - 25 pts. :(

What's the link to the official site??
 :)

Oh wow, the marking scheme for #5 is so ridiculous. 5 pts for proving $n \leq 48$??? (If you didn't try the problem, don't be fooled by the irritating notation used in the official soution, that part is very intuitive). I thought that the crux of the problem was finding the example, which I think should be worth at least 4 pts.

I'm angry I didn't write the APMO. For a few months a while back I was really into the golden ratio. I even wrote a blog entry on [url=http://en.wikipedia.org/wiki/Phinary]"Phinary"[/url]. . . .

Yeah, the hardest part about problem 5 was finding the case for n = 48, and the fact that n is less than or equal to 48 is really obvious.

[quote="rem"]Australia already posted the results of APMO for Australian students, but Canada still didn't. LOL :rotfl:[/quote]

We will get the results shortly after USAMO.

Does anybody know about the country scores? I am very curious about the ranking of HK this year.  :P

They still haven't posted the results. I am getting angry.

For Canadians, they've emailed the results already.


[url=http://community.boredofstudies.org/238/appreciating-beauty-elegance-extracurricular-topics/115068/elegant-proof-fermats-last-theorem.html]This url was too long so a moderator fixed it[/url]

What do you guys think? Has hyparzero really found an elegant proof of Fermat's Last Theorem that everyone has missed?

The only thing worth commenting on is that long URLs should be posted with the URL tag, as in [url=http://www.longurl.causes.horizontal.scrolling.edu]this example[/url]

Hundreds of "proofs" have arisen over the past few centuries.  I find it very hard to believe that so many people could have overlooked this particular "proof."

You have far more better chance to dig for gold in your yard than doing this.

[quote="BanishedTraitor"]LOL That's what I thought as well. I mean, at one time in history you probably have expert Number Theorist working on this problem. Is it really possible that they all missed this intriguing result? I really don't think so.[/quote]
This is not really the best reason for expecting that there would be no elementary solution. Of course, you could say, "if Gauss didn't think of something, there's no way I could" whenever you try to find a solution (or a more elementary solution) to a classical problem. But then you will never dare to try anything.

It is more appropriate to say "Fermat's Last Theorem turned out to be such a deep result that it seems extremely unlikely for a near-trivial elementary solution to settle it as well. Such a solution would completely give a miss to all of the deep ideas in Wiles' proof."

Of course I still find Wiles' proof to be a much better approach because it proved something more important and that is the Taniyama-Shimura Conjecture.


[quote="easyas3.14159..."]After looking at the posts on the classroom forum, just wondering what that was.[/quote]
Sorry,I don'y know that it's StewarT's teorem.
http://mathworld.wolfram.com/StewartsTheorem.html

[quote="fireemblem13"]I still don't quite get it. Where are the a and b coming from?[/quote]

The convention is that $a$ is the length of the side opposite $\angle A$ and $b$ is the length of the side opposite $\angle B$.

oh... it makes sense now. I might've seen and understood Stewart's Theorem but i don't think i could remember and apply it to a problem of some sort. How do u remember theorems like this?

Can someone post some problems that can be solved using Stewart's Theorem?

Use law of cosines twice to prove Stewarts Theorem.

You can use it for instance if you are given three median lengths of a triangle and you are trying to find the side of which the longest median goes to. This is a good time to use Stewarts Theorem since you already know that the centroid divides the segments of the medians into 2:1 ratios.

Can someone please explain law of cosines?  Maybe this is not the place...

sorry about the spam my computer is stupid!!


Hi!

Let A be symmetric. Given

\[ Ax \equal{} \lambda x,A \in {\mathbb{R}^{n,n}}\]

and eigenvectors and eigenvalues:

\[ \begin{gathered}
  {\lambda _1},...,{\lambda _n} \in \mathbb{R} \hfill \\
  {v^1},...,{v^n} \in {\mathbb{R}^n} \hfill \\
\end{gathered} \]

Let also be:

\[ \begin{gathered}
  \Delta A \in {\mathbb{R}^{n,n}}{\text{ symmetric}} \hfill \\
  \mu  \in \mathbb{R}{\text{ eigenvalue of }}A \plus{} \Delta A \hfill \\
\end{gathered} \]

How do I show that there is a \[ {\lambda _n} \in \sigma (A)
\]

so that

\[ \left| {{\lambda _n} \minus{} \mu } \right| \leqslant {\left\| {\Delta A} \right\|_2}\]

Regards,
\[ \Pi \Delta \Phi
\]

Why do you send 2 posts ?
Let $ u\not\equal{}0$ s.t. $ (A\plus{}\Delta{A})u\equal{}\mu{u}$ or $ (A\minus{}\mu{I})u\equal{}\minus{}\Delta{A}u\equal{}w$. According to Bauer Fike (your other post) there exists $ k$ s.t. $ |\lambda_k\minus{}\mu|\leq\frac{\parallel{}\Delta{A}u\parallel{}}{\parallel{}u\parallel{}}\leq\parallel{}\Delta{A}\parallel{}$

Hi!

I chose two posts, because often, when people see that a topic has two or more parts, they skip the topic. So I just wanted to raise my chances in getting an answer.
Thank you very much, you have been very helpful!

PiDeltaPhi


What percent of the first 200 prime numbers have reciprocals less than 0.05?

$ 0.05\equal{}\frac{1}{20} \implies n > 20$, so the answer is $ 100 \times \frac{200\minus{}8}{200}\equal{}\boxed{96 \%}$.

the only ones that do are 2,3,5,7,11,13,17,19, which is eight, or 100-4=96%.

EDIT: beaten.

$ 0.05\equal{}\frac{1}{20}$. Thus, we must find the number of primes less than $ 20$ and subtract that number from $ 200$. The number of these primes is $ 8$. $ 200\minus{}8\equal{}192$. The percent of $ \frac{192}{200}\equal{}\boxed{96}$.


What is the largest x for which
4^27 + 4^1000 + 4^x
equals the square of a whole number?

I will solve a more general problem first.

Let's determine all triples $(x, y, z)$ of nonnegative integers for which $4^x + 4^y + 4^z$ is a perfect square.

So, suppose $4^x + 4^y + 4^z = u^2$ and WLOG $x \geq y \geq z$.
We see that $4^z$ divides $u^2$ so $2^z$ divides $u$.
Hence there exists a positive integer $v$ such that $2^zv = u$.
Then our equation becomes $4^{x - z} + 4^{y - z} + 1 = u^2$.
Denote $x - z = p,\;y - z = q$. Of course $p$ and $q$ are nonnegative integers.
Notice that $u^2 > 4^p$ so $u > 2^p$. Hence $u \geq 2^p + 1$.
This gives $4^p + 4^q + 1 = u^2 \geq (2^p + 1)^2 = 4^p + 2^{p + 1} + 1$.
Hence $4^q \geq 2^{p + 1}$ or $p + 1 \leq 2q$ and hence $p - q \leq q - 1$.
Also, $u$ must be odd so we can write $u = 2w + 1$ where $w$ is an integer.
Then our equation becomes after some rewriting $w(w + 1) = 4^{q - 1}(4^{p - q} + 1).$
Now since $w$ and $w + 1$ are coprime, $4^{q - 1}$ has to divide exactly one of them.
This gives two cases to check.
But the condition $p - q \leq q - 1$ then gives $p - q = q - 1$ (can you see why?).

So, we get all solutions: $(x, y, z) = (r + 2s - 1, r + s, r)$ where $r,s \in \mathbb{N}$.
(We should also consider their permutations.)

===================================================

Now let's return to the original problem.
We can have
1) $r = 27$, $r + s = 1000$ which gives $x = 1973$;
2) $r = 27$, $r + 2s - 1 = 1000$ which gives $x = 487$;
3) $r + s = 27$, $r + 2s - 1 = 1000$ which gives no solution ($r < 0$).
The answer is 1973.

===================================================

In fact the solution above is way to long for this simple problem,
but I thought it would be interesting to show you the generalisation :)

So, the question is now : Which country has proposed this problem in 1973? :D

I bet on URSS....

Pierre.

Too easy... :D
http://www.kalva.demon.co.uk/soviet/sov72.html

I was quite right but the blame is to put on Arne :
Come on Arne! If $r=27$ and $r+s=1000$ then $x=r+2s-1=1972$!!!

Pierre.

I don't know which country has proposed it.I've taken these problems from this site: http://www.geocities.com/CapeCanaveral/Lab/4661/

Hahaha :D This is silly :) Well, I guess I was just a little bit too tired last night ;)