IMO
number theory
equation
IMO 1997
prime factorization

Find all pairs $ (a,b)$ of positive integers that satisfy the equation: $ a^{b^2} \equal{} b^a$.

I think I got this... hopefully...

We find the solutions (1, 1), (27, 3), and (16, 2), and we claim that no others exist.

First of all, a and b must have the same ratios of prime factors, so one must divide the other.  Note also that if b = xa, x > 1, then

$a^{x^2a^2} = (xa)^a \implies a^{x^2a} = xa \implies a^{x^2a - 1} = x$

Since x > 1, $a \ge 2$.  If x = 2, then $a^{2^2a-1} \ge 2^{2 \cdot 2^2-1} = 2^7 > 2$.  Then we can show easily by induction that if x > 2, $a^{x^2a-1} > x$.

We now deal with the case $a \ge b$.

If $a = b$, $a^{a^2} = a^a \implies a = 1$ gives us the solution (1, 1).

Let $a = kb, k > 1$.

$(kb)^{b^2} = b^{kb}$
$k^{b^2} = b^{kb-b^2}$
$k^b = b^{k-b}$

Note that $k-b > 0 \implies k > b \implies b | k$. Let $k = mb$.

$(mb)^b = b^{(m-1)b}$
$mb = b^{m-1}$
$m = b^{m-2}$

Now test cases m = 1, 2, 3, 4.

It turns out that m = 3, b = 3 works, giving us the solution (27, 3), and m = 4, b = 2 works, giving us the solution (16, 2).  If m = 5, since $b \ge 2$, we have $b^{5-2} \ge 8 > 5$.  From there on, it is easy to see by induction that $b^{m-2} > m$, so no other solutions exist.

My solution seems to be the same with some differences where I have no much number theory in it .. First we note that $(1,1)$ is a solution , so we can assume that $a \neq 1 $ , so we take the logarithms of base $a$ to have \[ b^2 = a \log_a b \] , but then clearly $\log_a b $ is a rational number. Hence we can write it in lowest terms $\log_a b = \frac{m}{n} $ which imply $b = a^{\frac{m}{n}}$ , so that we have \[ a^{2m/n} = \frac{am}{n} \] , We prove that $m=1$ , if not we treat the two cases $n\geq 2m $ and $ n < 2m $. If $n \geq 2m$ ,we raise the equation to the power $n$ and then divide by $a^n$ to have \[ 1 =  a^{n-2m} \frac{ m^n}{n^n} \] , but this is absurd as then $m^n$ can't cancel with any factor of the denominator.If $n < 2m $ ,one can obtain \[ a^{2m-n}=\frac{m^n}{n^n} \] , but then $n=1$ , and then $a^{2m-1}=m$ which is a contradiction. So we have $\log_a b = 1/n $ , and then $b^n = a $ , substituting in the original equation we have \[ b^{nb^2} = b^{b^n} \]. We take logarithms again to have \[ n = b^{n-2} \] , which obviously have solutions $(b,n)=(2,4),(3,4)$ only , and then original equation has $(a,b) \in \{(1,1),(2^4,2),(3^3,3)\}$

Take gcd(a,b)=d so that a=ud, b=vd and gcd(u,v)=1 ; and then use induction on variables to limit the possibilities of the values that the variables can take (a widely known technique in Number Theory)

@oiviluhatef can you express the technique in detail??

If I got to the equation $b^{k-1} = k+1,$ could I just say "it is obvious that the only solutions for $(b,k)$ are $(1,0),(2,3),(3,2),$ or would I have be rigorous by taking derivatives?

I would give at least an explanation on an olympiad (but probably take derivatives) to make sure I don't lose points.

So basically,this problem must be divided into 3 cases:
$ a \geqslant b > 1$ ,$b \geqslant a > 1$ and one of them is 1(the no-brainer one).
First we shall do $ a \geqslant b$:
$ a^{b^2} = b^a $,from this we see that $a = b^t$,inserting that back into our beginning equation we have:
$$ b^{tb^2}= b^{b^t} $$
$$ tb^2 = b^t $$
$$ t = b^{t-2} $$
When $ b \geqslant 4 $ we have the following $ b^{t-2} \geqslant 4^{t-2} > t $ This can be easily seen by induction:
Our base case $t=3$ works,thus let's say that for some $t$ the inequality holds then we have the following:
$4^{t-2} > t \implies 4^{t-1} > 4t > t+1 $, thus the inequality holds by induction.Thus we have that $1 < b  < 4$.
When $b=2$ we have $t = 2^{t-2}$,we see that $t=4$ (also can be seen by inequalities,which can be shown by induction).
When $b=3$ we have that $t=3^{t-2}$,we see that $t=3$ ( also seen by inequalities,which can be shown by induction).

Now let's tackle the case $b \geqslant a $:
Again we see that $b = a^t$,thus we have the following:
$$ a^{a^{2t}} = a^{at} $$
Thus we have that $a^{2t} = at$,this is simply not possible for $a > 1$.because by induction:
$$ a^{2t-1} > t $$
$$a^{2t} > at > t+1 $$
Of course the base case $a = 2$ works

Now the no-brainer $a =1 $
Now we have that $b^1 = 1$,thus $b=1$

So the solutions are:

$$\boxed{(a,b) \in \{(1,1);(16,2);(27,3) \}}$$

[quote=Birdews]Does this equation have negative roots?[/quote]

What about $a=b=-1$?

Solution from [i]Twitch Solves ISL[/i]: The answer is $(1,1)$, $(16,2)$ and $(27,3)$.
We assume $a,b > 1$ for convenience. Let $T$ denote the set of non perfect powers other than $1$.

[b][color=red]Claim:[/color][/b]  Every integer greater than $1$ is uniquely of the form $t^n$ for some $t \in T$, $n \in {\mathbb N}$.
[i]Proof.[/i]  	Clear. $\blacksquare$

Let $a = s^m$, $b = t^n$. \[ s^{m \cdot (t^n)^2} = t^{n \cdot s^m}. \] Hence $s = t$ and we have \[ m \cdot t^{2n} = n \cdot t^m 	\implies t^{2n-m} = \frac nm. \] Let $n = t^e m$ and $2 \cdot t^e m - m = e$, or \[ e + m = 2t^e \cdot m. \] We resolve this equation by casework [list] 	[*]If $e > 0$, then $2t^e \cdot m > 2e \cdot m > e+m$. 	[*]If $e=0$ we have $m=n$ and $m = 2m$, contradiction. 	[*]If $e = -1$ we apparently have 	\[ \frac{2}{t} \cdot m = m-1 \implies 		m = \frac{t}{t-2} \] 	so $(t,m) = (3,3)$ or $(t,m) = (4,2)$. 	[*]If $e = -2$ we apparently have 	\[ \frac{2}{t^2} \cdot m = m - 2 		\implies m = \frac{2}{1 - 2/t^2} = \frac{2t^2}{t^2-2}. \] 	This gives $(t,m) = (2,2)$. 	[*]If $e \le -3$ then let $k = -e \ge 3$, so the equation is 	\[ m-k = \frac{2m}{t^k} 	\iff m = \frac{k \cdot t^k}{t^k-2} 		= k + \frac{2k}{t^k-2}. \] 	However, for $k \ge 3$ and $t \ge 2$, 	we always have $2k \le t^k - 2$, 	with equality only when $(t,k) = (2,3)$; 	this means $m=4$, which is not a new solution. [/list]

The answer is $(27, 3), (16, 2), (1, 1)$.

It is easy to see that if $a^m=b^n$ and $a>b$ then $a=b^k$.

$\textbf{Case 1:}$ $a > b$

Then $a = b^k$. Substituting we get \[ b^{k^{b^2}}=b^{b^2k}=b^{b^k} \implies b^2k=b^k \implies k=b^{k-2} \] Notice that $1^{k-2} < k < 2^{k-2}$ if $k > 4$ which means that we only need to test out $k$ from $[2, 4]$. Testing out each each $k$ we find the only solutions are $(27, 3), (16, 2)$.

$\textbf{Case 2:}$ $a=b$

This is trivial, so I won't go into detail. In the end we get $(1, 1)$.

$\textbf{Case 3:}$ $a < b$

This means that $b = a^k$. Doing the same steps as above we get $k=a^{2k-1}$. Notice that $1^{2k - 1} < k < 2^{2k - 1}$ if $k > 1$. Now if $k=1$ then we have Case $2$ which we already covered.

Thus we can conlude that these are the only solutions.

The solutions are $(1,1)$, $(16,2)$, and $(27,3)$.

We have $a=1$ if and only if $b=1$. Notice that $a=2$ has no solutions. If $b=2$, then $a$ is a power of $2$. Quick casework gives $a=16$. From now on, assume that $a,b \geq 3$. Then, the equation implies that $a^b<b^a$. It's well known (by calculus) that this implies $a>b$ when $a,b \geq 3$.

Notice that $a^\frac{b^2}{a}=b$, so there must be nonnegative integers $n$, $c$, and $d$ such that $a=n^c$ and $b=n^d$. Notice that $c>d$. Then, we have $cn^{2d}=dn^c$ or $n^{c-2d}=\frac{c}{d}$. Now, we do casework on $c-2d$. Since $c>d$, we know that $c-2d$ is positive. Then, it's pretty easy to see by bounding that the only solution when $a,b \geq 3$ is $(27,3)$.

When in doubt, calcbash an inequality to prove that a finite number of solutions exist.

The answer is $(a,b)=(1,1),(16,2),(27,3)$ only, which can be verified to work, so we only have to show nothing else works. Hence suppose $a,b>1$.\\
Clearly for any prime $p$, we have $p \mid a \iff p \mid b$. Further, we can verify that $a=b^2$ doesn't work, as if $a=b^2$ then
$$a^a=b^a \implies a=b=b^2 \implies b=1,$$
which is impossible. Hence we can consider the following two cases:\\
[b]Case 1:[/b] $a>b^2$. Then, we have
$$b^a=a^{b^2}>b^{2b^2} \implies a>2b^2.$$
Next let $p$ be a prime dividing $a$ and $b$, so we have
$$b^2\nu_p(a)=a\nu_p(b) \implies \frac{\nu_p(a)}{\nu_p(b)}=\frac{a}{b^2}>2,$$
so we have $\nu_p(a)>2\nu_p(b) \implies b^2 \mid a$. Thus letting $\tfrac{a}{b^2}=k$ where $k>2$ is a positive integer, we have $\nu_p(a)=k\nu_p(b) \implies a=b^k$. Substituting this, we obtain
$$b^{kb^2}=b^{b^k} \implies b^{k-2}=k.$$
I claim that for $b \geq 4$, we actually have $b^{k-2}>k$ for all $k\geq 3$. This is true for $k=3$, and since we have $b\geq 4>2>\tfrac{k+1}{k}$, by induction the inequality holds for all $k$. Thus we only have to check $b=2,3$. For $b=2$ we get $2^k=4k$, which gives $k=4$ as the only solution—no solutions exist for $k<4$, and for $k>4$ we can use the fact that $2>\tfrac{k+1}{k}$ to prove that $2^k>4k$. Similarly, for $b=3$ we get $3^k=9k$ which gives $k=3$ as a solution, nothing for $k<3$, and $3^k>9k$ for $k>4$. These result in the solutions $(16,2),(27,3)$.\\
[b]Case 2:[/b] $a<b^2$. Then, we have
$$b^a=a^{b^2}>a^a \implies a<b.$$
Further, we have $b^a=a^{b^2}>a^b \implies \sqrt[b]{b}>\sqrt[a]{a}$. But we can verify that $f(x)=x^{1/x}$ is decreasing past $x>e$, since
$$\frac{d}{dx}\ln(f(x))=\frac{d}{dx}\frac{\ln(x)}{x}=\frac{1-\ln(x)}{x^2}<0$$
for $x>e$ by the quotient rule, so $\ln(f(x))$ is decreasing past that point and so is $f(x)$. Thus if $b>a\geq 3$ and $a<b$, we have $\sqrt[a]{a}>\sqrt[b]{b}$. Therefore we require $a=2$, in which case the original equation becomes $2^{b^2}=b^2$. On the other hand, for all positive integers $k$ we have $2^k>k$, since the inequality is true for $k=1$ and $2\geq \tfrac{k+1}{k}$ so the inequality holds for all $k$, so this case yields no solutions.\\
Combining these two cases implies that no solutions other than the three given exist, as desired. $\blacksquare$

The solutions are $(a,b)=(1,1),(27,3),(16,2)$.

It's obvious that $b$ is a rational power of $a$. Set $b=a^r$; we can rearrange to get $a^{r^2a^2}=r^aa^a$, or $a^{r^2a-1}=r$.

Here, it's clear again that $r$ is a rational power of $a$. If $a=1$, it's immediate that $b=1$ which gives the first pair of solutions. Now, assume $a\ge 2$; let $r=a^k$, and take the log base $a$ of both sides gives $a^{1+2k}=k+1$.

Here, we consider 3 cases:

If $k\ge 0$, we have $\frac{d}{dk} a^{1+2k}=2\cdot a^{1+2k}\ln(a)>1=\frac{d}{dk} (k+1)$ and $a^1>1$ when $k=0$, so there are no solutions in this case.

If $k\le -1$, a real power of a positive number would be nonpositive, which is impossible.

Now, suppose that $k\in (-1,0)$. Since $1+2k$ and $k+1$ are rational, the LHS and RHS necessarily have the form $\frac{1}{n}$ where $c=n^{-2k-1}$ is a positive integer and a rational power of $a$. Let $k=\frac{1-n}{n}$; we can derive $a^{\frac{2-n}{n}}=\frac{1}{n}$, and hence $n$ is a $|2-n|=n-2$th power. Via inspection, this has solutions at $n=3,4$ which correspond to the last two ordered pairs $(a,b)$.

Claim: $2^n>4n$ for $n\ge 5$.

Proof: We proceed by induction. The base case $n=5$ is straightforward. For the inductive step, assume that $2^{n-1}>4n-4$; after multiplying both sides by $2$, we get $2^n>8n-8$, and since $n\ge 2$ it holds that $8n-8\ge 4n$.

Therefore, the numeric solutions we have found are exhaustive and the proof is complete.

It is easy to see that $b=1\implies a=1$
\\$b=2 \implies a=16$ and $b=3\implies a=27$ (by considering the exponential functions and proving monotonicity by taking derivatives).

We will prove that those are the only solutions.

Assume $b>3$. It is not hard to check that $a\neq 1$ and $a\neq 2$. So, we have $a\geq 3$. If $a\leq b$, since $e<a\leq b$, $a^b\geq b^a=a^{b^2}>a^b$, which is impossible.
So, we have $a>b$. Then, there must exist  prime $p$ such that $v_p(a)>v_p(b)$. Takin vp of our equation, $b^2v_p(a)=av_p(b)$
\\ $\implies a>b^2$. But for any prime $q$, $b^2v_q(a)=av_q(b)$
\\ $\implies v_q(b)<v_q(a)$ for any prime $q$, equivalently $b\mid a$.
Let $a=kb, k>b$. The equation becomes $k^b=b^{k-b}$. Since $k>b$, $b<k-b$ or $k>2b$.
Then, $a>2b^2$ and for any prime $p$, $\frac{v_p(a)}{v_p(b)}=\frac{a}{b^2}>2$. This gives $b^2\mid a$ (as $2v_p(b)<v_p(a)$).
Let $d=\frac{a}{b^2}$. We have $\frac{v_p(a)}{v_p(b)}=d$ for any prime $p$ or $v_p(a)=v_p(b^d)$ for any prime $p$, which is equavalent to $a=b^d$.
So, substituting that into our equation, $(b^d)^{b^2}=b^{b^d}$. This reduces to $d=b^{d-2}$.
Clearly, $d\geq 3$. Finally $d=(1+(b-1))^{d-2}\geq 1+(b-1)(d-2)$ by Bernoulli's inequality. This gives $d\geq 1+(b-1)(d-2)$, which is equivalent to $$1\geq (d-2)(b-2)$$ This is a contradiction since $b-2>1$ and $d-2\geq 1$

The answer is $(a,b) = (1,1),(16,2),(27,3)$. Observe $a,b$ must have the same set of primes divisors. Assume $a,b \ne 1$. Write
$$a =  p_1^{\alpha_1} \cdots p_k^{\alpha_k} \qquad , \qquad b = p_1^{\beta_1} \cdots p_k^{\beta_k} $$
Then the given condition is equivalent to
$$ t = \frac{\alpha_1}{\beta_1} = \cdots = \frac{\alpha_k}{\beta_k} = \frac{a}{b^2} = p_1^{\alpha_1 - 2\beta_1} \cdots p_k^{\alpha_k - 2 \beta_k} $$
Note all of $\alpha_1 - 2 \beta_1, \ldots,\alpha^k - 2 \beta_k$ have the same sign. In particular either $\alpha_i \mid \beta_i$ for all $i$ OR $\beta_1 \mid \alpha_i$ for all $i$.

[b][color=#f00]Claim: [/color][/b] $\alpha_1 > \beta_1$.
[b][color=#00f]Proof: [/color][/b]FTSOC $\beta_1 \ge \alpha_1$. Then
$$ \beta_1 \ge \frac{\beta_1}{\alpha_1} = p_1^{2 \beta_1 - \alpha_1} \cdots p_k^{2 \beta_k - \alpha_k} \ge p_1^{2 \beta_1 - \alpha_1} \ge p_1^{\beta_1} \ge 2^{\beta_1} \ge \beta+1 $$
Thus we obtain a contradiction. $\square$

Now $t$ is an integer, and moreover $t \ge 2$. We in fact have $t \ge 3$.

[b][color=#f00]Claim: [/color][/b] $t \le 4$.
[b][color=#00f]Proof: [/color][/b] Observe
$$t = \frac{\alpha_1}{\beta_1} \ge p_1^{\alpha_1 - 2 \beta_1} \ge 2^{t - 2}   $$
Thus $t \le 4$. $\square$

Now it is easy to check that for $t=3,4$, the only solutions are $(a,b) = (27,3),(16,2)$, respectively. $\blacksquare$

[quote=ilikemath40]we only need to test out k from [2, 4]. Testing out each each k we find the only solutions are (27, 3), (16, 2).
[/quote]

I don't understand this part : How do you prove that the only solutions for k in ]1;4] are in [2;4] and that the solutions are then only the two cases  (27,3) and (16,2) ?

First, writting it as a logarithm we get $b^2 = a \log_a b$
Because $(a,b)$ are positive integers, we get that $\log_a b$ is a rational number
Let $\log_a b = \frac{c}{d}$, so $b = a^{\frac{c}{d}}$.
We get $a^{a^{\frac{2c}{d}}} = a^{\frac{ac}{d}}$ which implies $a^{\frac{2c}{d}} = \frac{ac}{d}$
From this, we can prove that $c=1$ and $b^d = a$
Plug it back in to get $b^{db^2} = b^{b^d}$, which implies $d=b^{d-2}$
We then have solutions of $(1,1)$, $(16,2)$, and $(27,3)$

The only solutions are $(1,1)$, $(16,2)$, and $(27,3)$, which all clearly work. Now we prove they are the only solutions.

Notice that \[\frac{\nu_p(b)}{\nu_p(a)} = \frac{b^2}{a},\]so either $a\mid b$ or $b\mid a$. If $a=b$, then $a = a^2$, so $a=b=1$, so now assume $a\ne b$, and also that $a,b>1$.

Case 1: $a\mid b$.
Let $b = k\cdot a$, where $k>1$. We have $a^{k^2a^2} = k^a\cdot a^a$, so $a^{k^2 a - 1} = k$. However, we have \[a^{k^2a - 1} > a^{2k^2-1} \ge 2^{k^2} >k,\]contradiction.

Case 2: $b\mid a$. \
Let $a = k\cdot b$, where $k>1$. We have $b^{b^2} k^{b^2} = b^{kb},$ so $b^{k-b} = k^b$. This implies that $k>b$ because $b^{k-b} = k^b>1$. Now, $k-b>b$ because otherwise $b^{k-b} < b^b < k^b$.

Notice that for any prime $p$, we have $\frac{\nu_p(k)}{\nu_p(b)} = \frac{k-b}{b}>1$, so $b\mid k$. Let $k = mb$, where $m>1$. We get \[b^{b(m-1)} = (mb)^b \implies b^{m-2} = m\]If $m>4$, then $b^{m-2} \ge 2^{m-2} > m$, contradiction, so $m\in \{2,3,4\}$.

If $m=2$, then $b^{m-2} = 1\ne m$, so no solutions.
If $m=3$, then $b=3$, so $(a,b) = (27,3)$.
If $m=4$, then $b=2$, so $(a,b) = (16,2)$.

We claim that the only possible solutions are $(1,1)$, $(16,2)$, and $(27,3)$.

Notice that for any prime $p$, if $\nu_p(a)=x$ and $\nu_p(b)=y$, we have that $xb^2=ay$. Taking this in two cases, if $b^2\geq{}a$, then we have that $y\geq{}x$ for all $p$, and if $b^2\leq{}a$, then we have that $y\leq{}x$ for all $p$, meaning that either $a\mid{}b$ or $b\mid{}a$. We now take this in cases.

Case 1: $b=ak$
We have that $a^{a^2k^2}=a^ak^a$, so we end up with $k=a^{ak^2-1}$. We now prove that for all $(a,k)$ for $a,k>1$, we must have $k<a^{ak^2-1}$. We can fix $a$ and then proceed with induction on $k$.

Notice that the base case of $k=2$ where $2<a^{4a-1}$ clearly works since $a>1$. Now notice that each time we add $1$ to $k$, the left side is multiplied by at most $2$, whereas the right side is multiplied by at least $a^a$, which is always at least $4$ since $a$ is an integer larger than $1$. Therefore no ordered pairs $(a,k)$ for $a,k>1$ work, and our only solution in this case is $(1,1)$.

Case 2: $a=bk$
We have that ${bk}^{b^2}=bk$, which gives us that $k^b=b^{k-b}$. We want to now prove that for all $b>3$, we have that no integer $k$ satisfies the equation. Notice that we must have $b\mid{}k$, so we let $k=xb$. We then have $x=b^{x-2}$, and for $b>3$, we have that $x<b^{x-2}$, which can be proved with induction similar to the previous case. We now have that $b$ is either $1$, $2$, or $3$.

Taking in cases and solving we see that our only solutions are now $(1,1)$, $(16,2)$, and $(27,3)$, and we are done.

Clearly, if $a=1\iff b=1$. When $a=2$ there are no solutions since $2^{b^2}>b^2$, and when $b=2$ this is $a^4=2^a$ which has $a=16$ as a solution. When $a=3$ this is $3^{b^2}=b^3$, which again has no solution since the left is much larger, and when $b=3$ this is $a^9=3^a$ which as $a=27$ as a solution. Noting these solutions, suppose $a,b\geq 4$ from now on.\\

Claim: $b< a$. Suppose FTSOC that $b\geq a$. Then, from $$b^2=a \log_a(b),$$ we have $$\frac{b^2}{\ln b}=\frac{a}{\ln a}.$$ However, $\frac{x}{\ln x}$ is strictly increasing on $[4,\infty)$ so $$\frac{b^2}{\ln b}>\frac{b}{\ln b}\geq \frac{a}{\ln a},$$ contradiction.\\

Taking the base $a$ logarithm, we have $$b^2=a \log_a(b),$$ so $b$ is a rational power of $a$. Thus, let $a=s^n$ and $b=s^m$ for positive integers $s,n,m$. Our equation then becomes $$s^{n\cdot s^{2m}}=s^{m\cdot s^n},$$ so $$n\cdot s^{2m}=m\cdot s^n.$$ This also means that $$s^{n-2m}=\frac{n}{m}.$$ Let $n=s^cm$. Since $b<a$, we know that $c\geq 1$. Thus, $$s^{s^cm-2m}=s^c$$ $$s^cm-2m=c$$ $$m(s^c-2)=c.$$ If $c=1$, then this is $$m(s-2)=1,$$ so clearly $m=1$ and $s=3$, giving the solution $(27,3).$ If $c\geq 2$, then $$m(s^c-2)\geq s^c-2\geq 2^c-2\geq c,$$ so the only possible case is all equalities, so $m=1$, $s=2$, and $c=2$, which gives the solution $(a,b)=(16,2)$ as expected earlier.\\

Thus, the solutions are $(1,1),(16,2),(27,3)$ and we are done.

The only answers are $(1, 1)$, $(27, 3)$, and $(16, 2)$. Assume $a, b > 1$ henceforth.

Let $d = \gcd(a, b)$ and let $a = dx$, $b = dy$, such that $$d^{b^2} \cdot x^{b^2} = d^a \cdot y^a.$$ However, $x$ and $y$ cannot share any prime factors, so upon canceling the $d$'s, we must have $1 \in \{x, y\}$. In particular, $a \mid b$ or $b \mid a$.

Suppose that $b \mid a$ first. Setting $a = db$, we have $$(db)^{b^2} = (b^d)^b \iff d^b = b^{d-b}.$$ Using similar logic, we must have $b \mid d$ (as $d > b$). Hence, setting $d = rb$, $r = b^{r-2}$.

Notice that for $b>5$, by monotonicity no solutions exist. So the only solutions here are $(b, r) = (3, 3) = (2, 4)$, which yield the two solutions above.

Now set $a \mid b$. Setting $b = da$, we have $d = a^{d^2 a - 1}$. However, this clearly yields no solutions by monotonicity again. Thus, the set of solutions is given by the set above.

Solutions: $(1,1)$, $(16,2)$, $(27,3)$.

Case 1: $b=a$. Then $a^2=a$ so $a=1$, and this leads to the solution $(1,1)$.

Case 2: $b>a$. Then taking $\nu_p$ of both sides and rearranging we get $\frac{\nu_pb}{\nu_pa}=\frac{b^2}{a}$ (where we consider $\frac00$ to be equal to anything). Then $\nu_pa\le \nu_pb$ for all $p$, so $a\mid b\mid b^2$, so $\frac{b^2}a$ is an integer constant, say $c$; then $b=a^c$. Then $a^{a^{2c}}=a^{ac}$, so $a^{2c-1}=c$. Now, $a^c=b>a$ so $a>1$, and hence $c=a^{2c-1}\ge 2^{2c-1}$. As $b>a$, $c>1$, so $2c-1>c$, and hence $c>2^c$, which is not possible. Hence no solutions in this case.

Case 3: $a>b$. Then since $a^{b^2}=b^a$ and we have $a>b$, we must have $b^2<a$, so $a>b^2$. Taking $\nu_p$ of both sides and rearranging we get $\frac{\nu_p a}{\nu_p b}=\frac{a}{b^2}$. Let the RHS in simplest form be $\frac cd$. Then for some integer $e\ge 2$, we must have $a=e^c$ and $b=e^d$. Then $e^{ce^{2d}}=e^{de^c}$. Then $ce^{2d}=de^c$. Now, we know $a>b^2$ so $c>2d$, and we get $\frac{c}{d}=e^{c-2d}$. Since the LHS is in simplest form we must have $d=1$ and so $c=e^{c-2}$. Now, since $e\ge 2$ we must have $c\ge 2^{c-2}$. Now, we know $c>2d=2$ so $c\ge 3$; then the possibilities are $c=3,4$. If $c=4$ then we have $4=e^2$ so $e=2$, which leads to the solution $(16,2)$. Otherwise $c=3$ so $3=e^{3-2}=e$, which leads to the solution $(27,3)$.

Answer: The only possible pairs are $(1, 1), (16, 2), (27, 3)$.

We can check the case $\min(a, b) \le 2$ by hand, so we can assume $\min(a, b) \ge 3$. Since $a^b < a^{b^2} = b^a$, so $a^b < b^a$, or equivalently $\frac{a}{\log(a)} > \frac{b}{\log(b)}$. Since $\frac{x}{\log(x)}$ increases on $(e, \infty)$, hence $a > b$. Therefore $a^{b^2} = b^a < a^a$, so $b^2 < a$. Note that for all primes $p$ dividing both of $a$ and $b$, we have $\nu_p(a)b^2 = \nu_p(b)a > \nu_p(b)b^2$, so $\nu_p(a) > \nu_p(b)$. Thus $b \mid a$. Let $k = \frac{a}{b}$. Since $b^2 < a$, so $b < k$. The given condition can be rewritten as $k^b = b^{k-b}$.

Since $k > b$, there exists a prime $p$ such that $\nu_p(k) > \nu_p(b)$. On the other hand, we have $\nu_p(k)b = \nu_p(b)(k-b)$, so we have $b < k - b$. Therefore for all primes $p$ dividing both of $b$ and $k$, we have $\nu_p(k)b = \nu_p(b)(k-b) > \nu_p(b)b$, thus $\nu_p(k) > \nu_p(b)$. Hence $b \mid k$. Let $k_1 = \frac{k}{b}$. The given condition can be rewritten as $k_1 = b^{k_1 - 2}$. Since $b \le 3$, we have $k_1 = b^{k_1 - 2} \ge 3^{k_1 - 2}$. Hence $2 < k_1 \le 3$, which forces $k_1 = 3$. Since $a = k_1b^2$, so we get $b = 3$ and $a = 27$. Therefore the solutions are $(1, 1), (16, 2), (27, 3)$. $\blacksquare$

Awful problem.

First if $a=1,b=1$ and if $a=2$ we have no solutions and if $b=2$ we have $a=16.$ Now if $a,b\ge 3$ we see $a^{b^2}>a^b$ so we have $a^b<b^a$ which gives $a>b.$ Take logs to get $\log_a b=\frac{b^2}a.$ Now the right side is rational so the left side is rational and thus there must be some $p$ such that $p^A=a,p^B=b,\gcd(A,B)=1.$ We then have $p^{2B-A}=\frac BA$ so let $B=A\cdot p^{-k}$ and we get $2A=p^k(A-k)$ or $A=\frac{kp^k}{p^k-2}$ and since $\gcd(p^k-2,p^k)\le 2$ we must have $p^k-2 \mid 2k$ since $A$ is an integer. Now if $k\ge 4$ we have $p^k-2 \ge 2^k-2 > 2k$ so there are no solutions. If $k=2$ we have $p^2-2\mid 4$ so $p=2$ which gives $A=4,B=1$ which gives $(16,2)$ which we already have. If $k=3$ we have $p^3-2\mid 6$ so $p=2$ which gives $A=4,B=\frac12$ which is impossible. If $k=1$ we have $p-2\mid 2$ so $p=3,4.$ We see $p=4$ gives $A=2,B=\frac12$ which is impossible and $p=3$ gives $A=3,B=1$ which gives $(27,3)$ and thus all solutions are $(1,1),(16,2),(27,3).$


limit
trigonometry
function
logarithms
calculus
calculus computations

I saw this in my book and dont know how to solve it

1) $\lim_{x\to 0^+}\left(\frac{\sin x}{x}\right)^{\frac{1}{x}}$

2)$\lim_{x\to \infty}\frac{\sin x}{x}$

3)$\lim_{x\to 0^+}\left(\frac{1}{x}-\frac{1}{\tan^{-1}x}\right)$

any help ?

what kind of proof do you want? i mean that we just allowed to use basic tools such that definition, arithmatic limit or squeze theorem.

for example $-\frac{1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x}$, then $\lim_{x \rightarrow \infty} \frac{\sin x}{x}=0$.

For 1 and 3 I would use power series.

For #3, I see that $\frac1x$ and $\frac1{\tan^{-1}x}$ are odd functions with (in complex language) a simple pole at zero. Write them as Laurent series around zero:

$\frac1x=\frac1x.$

$\frac1{\tan^{-1}x}=\frac1x+c_1x+c_3x^3+\cdots$

(I was using the fact that it's odd to get rid of even power terms.)

Subtract, and it's clear that $\frac1x-\frac1{\tan^{-1}x}=O(x)$ as $x\to0.$ Hence the limit is zero.

For #1:

$\frac{\sin x}x=1-\frac{x^2}6+O(x^4).$

Let $f(x)=\left(\frac{\sin x}x\right)^{\frac1x}.$ then

$\ln f(x)=\frac1x\ln\left(1-\frac{x^2}6+O(x^4)\right)$

$=\frac1x\left(-\frac{x^2}6+o(x^2)\right)=O(x)\to 0$ as $x\to 0.$

Thus $\lim_{x\to0^+}f(x)=1.$

[quote="Kent Merryfield"]
$\ln f(x)=\frac1x\ln\left(1-\frac{x^2}6+O(x^4)\right)$

$=\frac1x\left(-\frac{x^2}6+o(x^2)\right)=O(x)\to 0$ as $x\to 0.$
[/quote]

Sorry , I just started to learn the notation for big-oh and little-oh so Im not very clear how to proceed from this step
where the $O(x^4)$ suddenly change to $o(x^2)$ and how it changes again into $O(x)$ ?

Thanks .

Those steps were just me discarding details that I no longer needed - and probably never needed in the first place.

Roughly speaking, big-O means "no larger than some constant times" while little-o means "much smaller than" (in the limit).

As $x\to 0,$ something that is $O(x^4)$ is certainly $o(x^2).$ To say that is to give away information, since $x^{5/2}$ is also $o(x^2)$ but nowhere near $O(x^4).$

Then I decided that the exact coefficient ${-\frac16}$ in $-\frac{x}6+\cdots$  was unimportant - all I really needed was that it was something times $x,$ which is to say, $O(x).$

It would be poor practice for me to bury sensitive or delicate computations into this notation. I'm using the notation mostly to quickly dismiss things that turn out to be unimportant.

For number 2, this wouldn't be a proof, but

$\sin{x}$ goes between -1 and 1 no matter what x is, and $x$ will go to infinity. Thus the limit comes out to 0.


function
algebra
polynomial
calculus
calculus computations

We have a function $ f$ with $ f'''\uparrow$ for every $ x\in[a,b]$.Then prove that

$ f(x\plus{}1)\plus{}f(x\minus{}1)\minus{}2f(x)>f''(x)$  ,$ x\in(a\plus{}1,b\minus{}1)$

I have tried for a solution but.... :wink:

$ f(x\plus{}h) \equal{} f(x)\plus{}\frac{f'(x)}{1!}h\plus{}\frac{f''(x)}{2!}h^{2}\plus{}\frac{f'''(\zeta_{1})}{3!}h^{3}$
$ f(x\minus{}h) \equal{} f(x)\minus{}\frac{f'(x)}{1!}h\plus{}\frac{f''(x)}{2!}h^{2}\minus{}\frac{f'''(\zeta_{2})}{3!}h^{3}$.
Summing it up, pluging in $ h \equal{} 1$, and evaluating $ f'''$ you should get your result.

http://en.wikipedia.org/wiki/Taylor_polynomial


algebra
polynomial
complex numbers
advanced fields
advanced fields solved

I've got a 3 part problem here I can't solve, but probably a bit of help with the first part should get me on my way.

For each $n \in \mathbb{N}$, Let $P_{n}$ be the set of all monic polynomials of degree $n$ with rational coefficients \[a_{0}+a_{1}X+...+a_{n-1}X^{n-1}+X^{n}\in \mathbb{Q}[X]\]
(i) Show that $P_{n}$ is countable
(ii) Show that the set $P$ of all monic polynomials with rational coefficients is countable.
(iii) Let $A$ be the set of all complex numbers which are roots of monic polynomials with rational coefficients.  Show that $A$ is countable.
[i]Note: A is the set of all[/i] algebraic numbers [i]and it is a subfield of $\mathbb{C}$ which is algebraically closed.[/i]

$P_{n}$ is a finite union of countable sets (the coefficients), which is countable.  Think of $| P_{n}|$ as the size of the set of rational $n$-dimensional points $(a_{0}, a_{1}, ... a_{n-1})$.

The union of all $P_{n}$ is a countable union of countable sets, which is countable.  (I don't know how to prove this, admittedly :( )

$A$ is just the set $\{ p+qi \, | \, p, q \in \mathbb{Q}\}$, a finite union of countable sets, which is countable.

[quote="t0rajir0u"]$A$ is just the set $\{ p+qi \, | \, p, q \in \mathbb{Q}\}$, a finite union of countable sets, which is countable.[/quote]

I don't think this is true. The set of algebraic numbers contains many more elements than that, like $\sqrt{2}$.

If the set of polynomials with rational coefficients is countable and the number of roots of each such polynomial is countable, then clearly the set of all roots is countable (the algebraic numbers).

You can find more information about this [url=http://www.cut-the-knot.org/do_you_know/numbers.shtml#algebraic]here[/url].

[quote="t0rajir0u"]The union of all $P_{n}$ is a countable union of countable sets, which is countable.  (I don't know how to prove this, admittedly :( )[/quote]
If they're countable, they're well-ordered. So, do a diagonalization trick. Take the first element from the first set. Then the second element from the first set and the first element from teh second set. Then 3-1, 2-2, 1-3. And so on. You will eventually get everything, and this is clearly countable.


This is the question:
How many powers 9^n have leftmost digit 9, given that 9^4000 has 3817 digits and that its leftmost digit is 9?
Thanks.

Hello Gary!

I think this problem was discussed here not too long ago.  The attack is similar to #25 on the AMC12B.  (Have you seen the B exam?)

Ill take a shot at it.

[hide]

When a power of 9, say 9^n, has a leftmost digit of 9, then 9^(n-1) will have the same number of digits as 9^n. I'd guess you prolly start with the 3817 digits and count the number of times there is no # of digits increase going from 9^0 to 9^4000, so from 1 digit to 3817, when there are 4000 power increases.  4000-3816 = 184

[/hide]

Whats the answer?

note - i assumed less than or equal to 9^4000....


[color=darkblue]Denote $XYZ$- the area of the triangle $XYZ$. Let $ABC$ be an isosceles triangle $(AB=AC)$ and let $D\in (BC)$ and $E\in (AD)$ be two points so that $BD=AB$ and $[ABE]=[CDE]\ .$ Define the point $F\in AD$ so that $A\in (DF)$ and $DE=AF$. Prove that $m(\widehat{BAC})=2\cdot m(\widehat{AFB})\ .$
See http://www.mathlinks.ro/Forum/viewtopic.php?t=64311[/color]

Let $B',C'$ be the projections of $B,C$ onto $AD$.

Then $\triangle BB'D \sim \triangle CC'D \Rightarrow \frac{BB'}{CC'}= \frac{BD}{CD}$

$(ABE)=(CDE) \Leftrightarrow AE \cdot BB' = DE \cdot CC' \Leftrightarrow \frac{DE}{AE}= \frac{BB'}{CC'}$

Hence  $\frac{DE}{AE}= \frac{BD}{CD}\Rightarrow \boxed{\frac{DE}{AD}= \frac{AB}{BC}}\ \ (1)$
$\hline$

Consider a point $O$ such that $OA \parallel BC$ and $OA=OB$.
The isosceles triangles $OAB,ABC$ are similar hence $\boxed{\frac{OA}{AB}= \frac{AB}{BC}}\ \ (2)$

The line parallel to $AB$ passing through $O$ intersects the ray $DA$ at the point $M$.

We'll prove that $M\equiv F$

$\star$ We should be sure that $A \in (DM)$
[b]Indeed:[/b] We have $\angle ABO = \angle ABC$, but $O \not \in BC$, so $O,D$ are on different sides of $AB$. So the parallel projections $M,D$ of $O,D$ onto the line $AD$ are on different sides of $A$, hence $A\in (MD)$

The triangles $BAD,OMA$ are similar.
This yields that
$1) \ \triangle OMA$ is isosceles $(OA=OM)$ and

$2) \ \frac{AM}{AD}= \frac{OA}{BD}\Rightarrow \frac{AM}{AD}= \frac{OA}{AB}\ \ (3)$

From $(1),(2)$ and $(3)$ we get $AM=DE\Rightarrow$ the points $M,F$ coincide

Finally we have:
$OA=OM=OB \Rightarrow \angle BMA = \frac{\angle BOA}{2}= \frac{\angle BAC}{2}$


What are the last two digits of 1996^1996?

i knew i saw this before.  i think the answer is [hide] 96[/hide]

http://www.drmath.com/library/drmath/view/55787.html

the same question is on there, and they give you enough information so that you should be able to figure it out.  But if not, post again and i'll try to help.

Theres gotta be a more elegant way than this.

[hide]

1996^1996 = 6^1996 mod 10

We know the last digit is 6.

6^2 = 36

6^3 = 216

6^4 = 1296

6^5 = 7776

6^6 = 46656

6^7 = 279936

It's in cycles of 5. 1996 = 1 mod 5

So the last two digits are 56

[/hide]

[quote="JBL"]Be careful, though -- the 9 might cause you problems.  It affects the last 2 digits, as well.[/quote]

Right. I think if you use the same method but with 96 instead of just 6, SouthPaw, you will get the right answer.

[hide]Since we're talking about the last 2 digits, we must look at the last 2 digits of 1996.

In the following paragraphs, "~" means "has the last two digits of ..."

Notice that 966 ~ 96. Then,

 19961996

~ 961996

= (966)332 * 964

~ 96332 * 964

= 96336

= (966)56

~ 9656

= (966)9 * 962

~ 969 * 962

= 9611

= 966 * 965

~ 96 * 965

= 966

~ 96.[/hide]

But then, it's probably more efficient to just find 1996 mod 5. =|

By the way, if you want to show that 96^6 ends in 96, to make calculations easier you can just show (-4)^6 ends in (-4), or that 4^6 ends in 96 also.


Let $ A\in M_{n}\mathbb{(R)}$ and $ rang(A)=2$ :
         $ det(A+I)=1+Tr(A)+\det (\sum_{1=<i <j<=n}(\begin{array}{cc}a_{ii}&a_{ii+1}\\a_{jj-1}&a_{jj}\end{array} ))$
[b]???[/b]
 :blush:
[b] Love math must come from heart !!! [/b]

A suggestion:
Consider $ det(A\plus{}xI)$, and expand it as a polynomial in x. Try to interpret the coefficients of $ x^k$ (it is the sum of all the (n-k) by (n-k) principal minors, as well as the (n-k)-th symmetric sum of the eigenvalues.

What you have written are 1, trace of A (sum of all 1 by 1 principal minors) and sum of all 2 by 2 principal minors. The remaining terms are 0, because rank 2 implies that all but 2 eigenvalues are 0, which means that the k-th symmetric sum of eigenvalues = 0 for $ k \ge 3$.


Evaluate:

$ \sum_{i \equal{} 2}^{\plus{}\infty}\sum_{j \equal{} 2}^{\plus{}\infty}\sum_{k \equal{} 2}^{\plus{}\infty}{k^{\minus{}ij}}$

[quote="joao1"]Evaluate:

$ \sum_{i \equal{} 2}^{ \plus{} \infty}\sum_{j \equal{} 2}^{ \plus{} \infty}\sum_{k \equal{} 2}^{ \plus{} \infty}{k^{ \minus{} ij}}$[/quote]

Hello Joao1, are you sure there is a solution for this problem ? Where did you find it ?

Numeric calculus gives a result $ \approx 0.1303964...$ and I also showed that $ \sum_{i \equal{} 2}^{ \plus{} \infty}\sum_{j \equal{} 2}^{ \plus{} \infty}\sum_{k \equal{} 2}^{ \plus{} \infty}{k^{ \minus{} ij}}$ $ \equal{}(\sum_{i \equal{} 2}^{ \plus{} \infty}\sum_{k \equal{} 2}^{ \plus{} \infty}\frac{1}{k^i\minus{}1}) \minus{}1$

So it seems this is related to the computation of $ \sum_{k \equal{} 2}^{ \plus{} \infty}\frac{1}{k^i\minus{}1}$ which is not easy, according to me :

$ \sum_{k \equal{} 2}^{ \plus{} \infty}\frac{1}{k^2\minus{}1}\equal{}\frac 34$

$ \sum_{k \equal{} 2}^{ \plus{} \infty}\frac{1}{k^3\minus{}1}\equal{}???$

I don't  we could find an exact expression for this sum but we might be able to find an interval that this sum belongs to.
For example for an upper bound we could notice that:

   $ \frac {1}{k^3 \minus{} 1} < \frac {1}{k^3 \minus{} k} \equal{} \frac {1}{(k \minus{} 1)k(k \plus{} 1)}$.
We could use partial fractions(i guess,or some other trick) for the latter fraction which can be evaluated.
In a similar fashion we might be able to do this for the other sums too and get an upper bound.
I don't think an exact value is possible without the use of a computer.


Βρηκα μια ασκηση την οποια την ελυσα(ειναι μια ανισοτητα Jensen) αλλα θελω να ρωτισω αν υπαρχει καποιος αλλος τροπος να λυθει η ασκηση αυτη.

Εχουμε συναρτηση $f(x)=xlnx$ η οποια ειναι κυρτη στο $D_{f}=(0,-\infty)$

Η ασκηση ζηταει να αποδειξουμε οτι $\forall x_{1},x_{2}\in D_{f}$ με $x_{1}\neq x_{2}$ ισχυει

$\frac{f(x_{1})+f(x_{2})}{2}>f\left(\frac{x_{1}+x_{2}}{2}\right)$

Μπορείς να δώσεις αλλη μορφή στο ζητούμενο αλλά πάλι Jensen κάνεις Α) Να δείξεις ότι μια γνήσια κυρτή συνάρτηση έχει οποιαδήποτε χορδή της με τετμημένες άκρων  x_1,x_2 πάνω από την γραφική της παράσταση Β)Να κάνεις δυο ΘΜΤ στα [x_1,1/2(x_1,x_2] , [1/2(x_1,x_2,x_2[ σχολικός τρόπος Γ)Να θέσεις x_1=2a-2x , x_2=2x και να μελετήσεις την μονοτονία της f(2a-2x)+f(2x)-2f(a) προκειμένου να βρεις το πρόσημο της
R.B

apo genikeumenh Cauchy exoume
$\frac{x\cdot \frac{1}{x}+y\cdot \frac{1}{y}}{x+y}\geq  \sqrt[x+y]{\frac{1}{x^{x}\cdot y^{y}}}$ pou einai isodunamh me th zhtoumenh.

Εγω την ελυσα ετσι:

Για τυχαιο $x_{2}\in D_{f}$ θεωρουμε την συναρτηση

$g(x)=\frac{1}{2}f(x)+\frac{1}{2}f(x_{2})-f\left(\frac{1}{2}x+\frac{1}{2}x_{2}\right)$

και αρκει να δειξουμε οτι $g(x)>0$   $\forall x \in D_{f}$

$g'(x)=0 \Leftrightarrow x=x_{2}$       (η f' ειναι γν. αυξουσα)

$g'(x)>0 \Leftrightarrow x>x_{2}$

$g'(x)<0 \Leftrightarrow x<x_{2}$

Αρα η $g$ εχει ελαχιστη τιμη την $g(x_{2})$ και $\forall x \in D_{f}$ ισχυει

$g(x)\geq g(x_{2}) \Leftrightarrow g(x)\geq 0$ και για $x \neq x_{2}$  ειναι  $g(x)>0$


how many bit strings of legnth 10 contain no group of three consecutive 0s
I got 503.  Is this right?

Close  :wink:

[hide="Stone Stepping Solution"]
Add a 1 to the beginning and end of the string, to get a string of length 12 that starts and ends with 1.**

Let these be a path of stones, and let 1's represent stones that you step on. To get to any stone, the previous stone must've been the stone 1, 2, or 3 positions before it. Thus the number of ways to get to stone $ n$ is the number of ways to get to $ n \minus{} 1$ and then step to $ n$, plus the number of ways to get to $ n \minus{} 2$ and then jump to $ n$, plus the number of ways to get to $ n \minus{} 3$ and then jump to $ n$.

Thus, if $ f(n)$ is the number of valid ways to get to stone $ n$, then $ f(n) \equal{} f(n \minus{} 1) \plus{} f(n \minus{} 2) \plus{} f(n \minus{} 3)$.

There is one way to get to stone 1 (it was forced to be a 1), 1 way to get to stone 2 (we must go from stone 1), and 2 ways to get to stone 3 (we can go there either from stone 1 or stone 2.

Thus, $ f(1) \equal{} f(2) \equal{} 1$ and $ f(3) \equal{} 2$. Running the recursion, $ f(12) \equal{} 504$

** [i]We added 1's at the beginning and end to prevent casework -- this way, we must begin and end on a 1, and it does not introduce or take away any arrangements of the 10 numbers in between.[/i][/hide]
[hide="Computer Verification"][code]
Total[Table[
  Which[Product[
     IntegerDigits[n, 2, 10][[k]] + IntegerDigits[n, 2, 10][[k + 1]] +
       IntegerDigits[n, 2, 10][[k + 2]], {k, 1, 8}] == 0, 0, True,
   1], {n, 0, 1023}]]
[/code]

For each bitstring of length 10, multiplies each of the 8 sums of 3 consecutive digits. When this is nonzero, adds 1 to the total count.[/hide]
The stone steppng solution is one of the most common approaches to these types of problems. I'm curious -- how did you attempt to solve this?


For a plane set S define d(S) the diameter of S to be sup{PQ:P,Q in S}
let K be a triangle with sides 3,4,5 and its interior. If K=H1 \cup H2 \cup H3 \cup H4 what is the smallest value of max(d(H1),d(H2),d(H3),d(H4))?

I think answer is 25/13.

Is that correct?

I think the answer is correct.
do you have any proof for that?

I found 5 points within triangle such that they pairwise distances \geq 25/13 ==> max(d(H1),d(H2),d(H3),d(H4)) \geq 25/13. Finally I constructed appropriate sets.


solve:
1.A magician has 100 letters, numbered from the 1 to the 100. He distributes them in three boxes, a red one, a white and the third blue one, in such a way that each box contains, at least, a letter. Two ways to distribute letters will be considered different if at least one letter is located in different boxes, in both. A spectator, without being seen by the magician, chooses two of the three boxes, removes a letter from each one and aloud announces the sum of the numbers that was in extracted letters. Knowing this sum, the magician must identify the box from which letter has not removed. Of how many ways the magician can distribute letters in the boxes, so that he is able to always guess right?

IMO 2001 I would say exercise 3.

And my answer would be 12  Of course you can find the answer on any website so...


[hide]
The second one is pretty famous, $1681$. :D
The first one, $\sqrt{11(100a+c)}$ is an integer.
This means that $100a+c=11n^2$ for some $n \in \mathbb{N}$ Also, $11n^2$ has a tens digit of $0$. Keep trying $n$ until one has a tens digit of $0$. $n=8$ works, so $100a+c=704 \rightarrow (7,4) \rightarrow \boxed {7744}$
[/hide]

[quote="frt"]1. Find a 4-digit perfect square $abcd$ such that $a=b$ and $c=d$.

2. Find a 4-digit perfect square $abcd$ such that $ab$ and $cd$ are also perfect squares.

Note: What I mean by $abcd$ is a 4-digit number such that its thousandth digit is $a$, hundredth digit is $b$, and so on.[/quote]

[hide="Answer"]

1.  We can re-write the number as $aacc$ and let it be detonated by $x^2$ which can be simplified to $ 11(100a+c)$.  Thus,

$ 11(100a+c)=x^2$

  We must also observe that if a number is a square, its prime factorization must be in the form of $ p^{2n} \cdot r^{2n} \cdot s^{2n} $

  So the factorization of $ 100a +c$ must include one 11 (if it had 3 more our number would exceed 4 digits) and then simpy another number in the form of $n^2$.  Thus,

$100a+c=11n^2$

  Next we must notice that $100a+c$'s tens digit is 0.  Meaning that the 2- digit (we know this because $100a+c$ is a 3-dgit number) that was multiplied by 11, produced a 3-digit number with a 0 in the tens spot.  A special property assures us this 2-digit numbers digit must sum to 10.

  So our choices are:

19
28
37
46
55
64
82
91

  Of these only 64 is a square, and $n=8$.

  Going back we saw:

$106a+c=11n^2$

Now it becomes:

$106a+c=11 \cdot 64$

$106a+c=704$

  And $a=7$ and $c=4$.

  Going back to our orginal number $aacc$ we know see it is $\boxed{7744}$

2.  I think this is pretty well-known.  The answer is $\boxed{1681}$ [/hide]

[quote="frt"]1. Find a 4-digit perfect square $abcd$ such that $a=b$ and $c=d$.

2. Find a 4-digit perfect square $abcd$ such that $ab$ and $cd$ are also perfect squares.
[/quote]
[hide]1. The number is $aacc=11\times (100a+c)$ But this is a square, so $100a+c$ must be divisible by 11. Divisibility test of 11 says that $a-0+c=11k$. But $0<a+c<20$, so $a+c=11$. Also, $100a+c=11\times (10e+c)$ and this gives $e+c=10$ and $10e+c$ is square also. All 2-digit squares are 16, 25, 36, 49, 64, 81 and only 64 have $6+4=10$, $e=6$ and $c=4$. Now $a=e+1=7$. Number is $7744$

2. $100ab+cd$ is a square, $100ab+cd=(10e)^2+cd$. One way is look for $cd=2\times 10e+1$. $abcd=(10e)^2+2\times 10e+1=(10e+1)^2$. 2-digit squares are 16, 25, 36, 49, 64, 81 with $e=4,5,6,7,8,9$ and matching $cd=81,101,etc$ because they are all greater than 100. Only $e=4$ and $cd=81$, $ab=16$ and number is $1681$.[/hide]


Duke
college
algebra
polynomial
absolute value

$ A_1A_2A_3A_4A_5A_6A_7$ is a regular heptagon (7 sided-figure) centered at the origin where $ A_1 \equal{}(\sqrt[91]{6},0)$. $ B_1B_2B_3...B_{13}$ is a regular triskaidecagon (13 sided-figure) centered at the origin where $ B1\equal{}(0,\sqrt[91]{41})$.
Compute the product of all lengths $ A_iB_j$ , where $ i$ ranges between 1 and 7, inclusive, and $ j$ ranges between 1 and 13, inclusive.

That's a pretty slick solution.  I could not help but use it again.

[hide="Small Step Further"]

Here's the generalized result for side number, size, and angle displacement of two regular polygons.  Looks familiar:

Basically, followed gauss202's method, except replace $6^{1/91}$ by $a e^{i\theta}$ and $41^{1/91}$ by $be^{i\phi}$, so that A has a point whose polar angle is $\theta$ and B has a point whose angle is $\phi$.  In addition, regular polygon A has $n$ sides and circum-radius (?) of $a$, and regular polygon B has $m$ sides and circum-radius of $b$.  This gives as the final product:

$ | a^{nm}e^{inm\theta} - b^{nm}e^{inm\phi} | $

Split the exponents using Euler's formula, combine imaginary and real, then take the absolute value:

$ \sqrt{(a^{nm}\cos(nm\theta)-b^{nm}\cos(nm\phi))^2 + (a^{nm}\sin(nm\theta) - b^{nm}\sin(nm\phi))^2}$

$ \sqrt{a^{2nm} + b^{2nm} - 2a^{nm}b^{nm}\cos(nm(\theta - \phi))}$

This gives the same answer for the mentioned problem, as odd $n$ and $m$ with an angle of $\pi/2$ make the last term disappear. In addition, its obvious that it works for $m = n = 1$. [/hide]


[u]Geometry Revisted[/u]
I would give link but Amazon doesnt find it  :?:

[quote="Allan Z"]I am now preparing for USAMO and probably some IMO level questions, can anyone suggest me some geometry books?
Thanks[/quote]
Go to the blog of Gemath of mathlink to find that book [url]http://www.mathlinks.ro/Forum/weblog.php?w=659[/url]:D

and for geometry and others u may use IMO shortlists and the best choise for the shortlist is IMO compendium.
[url]http://www.imo.org.yu[/url]

http://www.amazon.com/Geometry-Revisted/dp/0394709195/ref=sr_1_1/105-9096820-5701211?ie=UTF8&s=books&qid=1183315021&sr=8-1

EDIT: moderators please move it to [b]Mathematical Books and Journals[/b]


two questions:

what is an easy way to find the sum of all two digit numbers with different digits?

what is the sum of all 3 digit numbers with different digits?

Complement argument and Inclusion-Exclusion.

Interesting problem!  :lol:

[hide]Ignoring the restriction that the first digit cannot be 0, there are $10(9)(8)$ three-digit numbers with no digit repeated.  Any digit $d$ will appear in the hundreds place one-tenth of the time.  Therefore, any digit $d$ will appear in the hundreds place $9\times 8=72$ times.  The same argument holds for tens and ones digits.  Therefore, the sum is $(100+10+1)(72)(0+1+2+\dots +9)=(111)(72)(45)=359640$.

Now we need to subtract the sum of all two digit numbers with different digits, neither of which are 0.  There are $(9)(8)$ such numbers.  Using the same argument as above, any digit $d$ will appear in the tens (or ones) place $8$ times.  Therefore, the sum is $(10+1)(8)(1+2+3+\dots +9)=(11)(8)(45)=3960$.

The answer is $359640-3960=\boxed{355680}$[/hide]


While I recognise that there are several ways to prove the Pythagorean Theorem, I have a problem with the way demonstrated in the proofs without words gallery.  It seems to me that it is too easy to, in morphing the two smaller squares into the hexagon, change the area.  In other words, I don't feel that the method used in the proofs without words gallery is a legitimate proof for the Pythagorean Theorem.

The ironic thing is that, as I type this, the proof that is in the side window is the one for the Pythagorean Theorem.

well there is a show on TV that uses that proof on the side window except it is explained furthered. i don't remember the show title but it could be on the NASA channel?

in the proof, what they are doing is turning the two squares in to parrellellograms that have the same base and height as the square, and thus the same area


calculus
derivative
inequalities unsolved
inequalities

Minimize $x^xy^y(1-x-y)^{1-x-y}$ w.r.t. the constraints $0 \leq x \leq \frac{1}{4}, 0 \leq y,1-x-y \leq 1$.

hello, i have found the minimum $ \approx 0.338850751353691804$ for $ x\equal{}\frac{1}{4}$ and $ y\equal{}\frac{3}{8}$.
Sonnhard.

hello, consider $ f(x,y)\equal{}x^xy^y(1\minus{}x\minus{}y)^{1\minus{}x\minus{}y}$ and compute the partial derivatives with respect to $ x$ and $ y$ and solve the system
$ f_x\equal{}0$
$ f_y\equal{}0$
Sonnhard.


The vertices of a right-angled triangle are on a circle of radius $R$ and the sides of the triangle are tangent to another circle of radius $r$ (this is the circle that is inside triangle). If the lengths of the sides about the right angles are 16 and 30, determine the value of $R+r$.

The sides about the right angles? Does that mean the two legs of the triangle? :huh:

I'm assuming it does...

[hide="Answer"]$\sqrt{16^2+30^2}=34\Rightarrow R=17$
$\frac{16\cdot 30}{2}=240=rs$
$s=\frac{16+30+34}{2}=40\Rightarrow r=6$
$R+r=\boxed{23}$[/hide]

[hide]Generally, we have $R+r=\frac{a+b}{2}$
[hide="because"]
$R=\frac{c}{2}$
$r=\frac{a+b-c}{2}$[/hide]

So $R+r=\frac{16+30}{2}=23$[/hide]


linear algebra
matrix
function
linear algebra unsolved

Let $A$ be $3\times3$ matrix,$detA=0$ such that it's algebraic complements $A_{ij}=a_{ij}^2$. Prove that $A=0$

[quote="eugene"]Let $A$ be $3\times3$ matrix it's algebraic complements $A_{ij}=a_{ij}^2$[/quote]

What is  algebraic complements ?

$A_{ij}=(-1)^{i+j}M_{ij}$, where $M_{ij}$-minor of the element in $(i,j)$ position.
If you know what is $adjA$, then $adjA=(A_{ij})^t$. I'm used to this notation beacause most of russian books uses it.
If i haven't fully explained, i'll give a definition

The matrix is not invertible $rank(A)\leq 2$

If $rank(A)=1$ then $Com(A)=0$ so $a_{ij}^2=0$ and $A=0$ contradict that rank of A is 1

If $rank(A)=2$ then $rank(Com A)=1$
...

Can someone continu the solution of this nice problem
thx

$n\geq 3$, $A\in M_n(K)$ such that $det(A)=0$

Let $f$:$K\rightarrow K$ where $K$ commutative field such that
i) $f(x)=0$ imply $x=0$
ii) $(f(x_1),...,f(x_n))$ colinear $(f(y_1),...,f(y_n))$ imply that
 $(x_1,...,x_n)$ colinear $(y_1,...,y_n)$

Suppose $Com(A)=(f(a_{ij}))$

Prove that $A=0$

Remind $Com $ is comatrix $M.^{t}Com(M)=det(M)I_n$

$detA=0$ implies $rank(ComA) \leq 1$. If $rank(ComA)=0$ we are done by the property 1) of our function $f$. If $rank(ComA)=1$, then all the columns of $ComA$ are collinear to (WLOG) it's first column, and now isung the property 2) of our function $f$ we can conclude that all the columns of $A$ are collinear to it's first column, thus $rankA=1$, from where $ComA=0$ if $n \geq 3$, which means that $A=0$ again using property 2).


geometry
rectangle
combinatorics unsolved
combinatorics

A 6*6 square is dissected into 9 rectangles by lines parallel to its sides such that all these rectangles have only integer sides. Prove that there are always [b]two[/b] congruent rectangles.

Consider any set of non-congruent rectangles with integer sides. Any prime area (including $ 1$) can only appear once, and the areas $ 4$ and $ 6$ can appear twice at most (a $ 2 \times 2$ rectangle and a $ 1 \times 4$, and a $ 2 \times 3$ rectangle and a $ 1 \times 6$). Thus the minimal total area of $ 9$ such rectangles is $ 1\plus{}2\plus{}3\plus{}4\plus{}4\plus{}5\plus{}6\plus{}6\plus{}7 \equal{} 38$, too much to fit in a $ 6 \times 6$ square.

[quote="randomgraph"]Consider any set of non-congruent rectangles with integer sides. Any prime area (including $ 1$) can only appear once, and the areas $ 4$ and $ 6$ can appear twice at most (a $ 2 \times 2$ rectangle and a $ 1 \times 4$, and a $ 2 \times 3$ rectangle and a $ 1 \times 6$). Thus the minimal total area of $ 9$ such rectangles is $ 1 \plus{} 2 \plus{} 3 \plus{} 4 \plus{} 4 \plus{} 5 \plus{} 6 \plus{} 6 \plus{} 7 \equal{} 38$, too much to fit in a $ 6 \times 6$ square.[/quote]

Thanks.  :)

However there is a mistake in your calculation because a rectangle with area $ 7\ (7 \times 1)$ is not possible. So I replace it with a rectangle of area $ 8\ (4 \times 2)$ to get a minimal total area of $ 1 \plus{} 2 \plus{} 3 \plus{} 4 \plus{} 4 \plus{} 5 \plus{} 6 \plus{} 6 \plus{} 8 \equal{} 39$, too much to fit in a $ 6 \times 6$ square.

Not that it matters much, but there's no mistake. I showed that any set of $ 9$ non-congruent rectangles with integer sides must have area at least $ 38$. This is independent of whether or not any of the rectangles needs to fit in a $ 6 \times 6$ square. If you put additional restrictions on the rectangles, it can only make matters worse (that is, it can only make the minimal area larger, as you did). In particular, such rectangles cannot tile anything that has area less than $ 38$, and that includes the $ 6 \times 6$ square.

It also includes, for instance, the $ 4 \times 9$ rectangle, which also cannot be tiled even though it does allow that $ 1 \times 7$ to fit.


The perfect squares from 1 through 2500, inclusive, are printed in a sequence of digits $ 1491625\ldots2500$. How many digits are in the sequence?

man i had this problem solved but I forgot to multiply each case by the total number of digits  :mad:

We will consider the number of digits by casework.

Case 1: The square has $ 1$ digit.
There are $ 3$ squares corresponding to this ($ 1$, $ 4$, and $ 9$). Thus, this case produces $ 3$ digits.

Case 2: The square has $ 2$ digits.
There are $ 6$ squares which fit this criteria (those from $ 4^2$ tp $ 9^2$). They contribute a total of $ 6(2)\equal{}12$ digits.

Case 3: The square has $ 3$ digits.
There are $ 22$ squares which fit the criteria (those from $ 10^2$ to $ 31^2$). They contribute $ 22(3)\equal{}66$ digits.

Case 4: The square has $ 4$ digits.
There are $ 19$ squares which fit this criteria (those from $ 32^2$ to $ 50^2$). They contribute $ 19(4)\equal{}76$ digits.

Thus, the grand total is $ 3\plus{}12\plus{}66\plus{}76\equal{}\boxed{157}$ digits.

[hide=Another Way]
3*1=number of squares that have 1 digit
6*2=number of squares that have 2 digits
22*3=number of squares that have 3 digits
19*4=number of squares that have 4 digits and are below 2500/equal to
$1\times3+2\times6+3\times22+4\times19=\boxed{157}$
[/hide]

I did the squares of 1...2500 instead!!!!! Ughhhhh

Ugh this is just bashing. These problems are kind of annoying

Heh....sometimes. But precalculus problems are EVEN MORE ANNOYING so beware! LOL


Hey, I often see questions that when solving, one needs to know how to do the following:

how many ways are there to split 50 up into four numbers summed up together, if 1 + 1 + 4 + 44 = 50 is the same as 4 + 1 + 44 + 1 = 50. It could also be like, how many ways are there to split 100, or 200 into six numbers, eight numbers, or ten numbers summed together.

Is there an easy way to do this or is it just a pain in the behind.

Some one please help. I would really appreciate it.

David Jia/Gainesville FLorida

Well, if you're including the number 0 as a possibility, it's quite easy. Say you want to break 50 up into 4 numbers. Represent this as
+++|+++|+++++++|+++....+++ where this is the ordered quadruple (3,3,7,37). All we need to do is rearrange the + and | and find the number of permutations. There are 4 | and 50 +, so 54 total. Thus choose 4 | out of 54 total slots, $\binom{54}{4}$ is the answer.

Oh, whoops. Made a mistake there. I did it for 5 variables and didn't see that order for the partition is irrelevant. So the answer then is $\frac{\binom{53}{3}}{4!}$. (Not 54.)

[quote="t3hggnore"]Hmm...I disagree with Hokkage's answer. The approach is definitely correct, but for partitioning into 4 groups, then you would use only 3 |'s. Also, the original problem stated that (1,1,4,44) was the same as (4,1,44,1), so I think you're actually overcounting by $4!$. Therefore, the answer should be \[ \displaystyle\frac{{54\choose 3}}{4!}  \][/quote]

this is not correct...suppose we have $(1,26,1,26)$, that is counted 6 times,
but $(1,2,3,48)$ is counted 24 times...

Hey guys, I really appreciate the help.

My first intuition was what Hokkage had in mind, and then I thought about what t3hggnore had said, because they are counted more than once, 54C4 could not be the answer because they are not ORDERED pairs and must divide out something, but different groups occur different times, as Altheman pointed out. So how does this work? Do you really have to do it the anal way or is there some correspondence that coiuld be found to do it easier.

You probably want to go look up partitions.  Here's the particular sequence you asked for:
http://www.research.att.com/~njas/sequences/A001400


On a line are given $2k-1$ white segments and $2k-1$ black ones. Assume that each white segment intersects at least $k$ black segments, and each black segment intersects at least $k$ white ones. Prove that there is a black segment intersecting all the white ones, and a white segment intersecting all the black ones.


How do i use the binomial theorem to find the sum of the infinite series
1 - 1/6 + ((1*3)/2!)(1/6)^2 - ((1*3*5)/3!)(1/6)^3 ...?
sorry for the messiness, i cant rreally use latex.

thank you very much in advance.

What does the binomial theorem tell you it is?  (Are you familiar with the general binomial theorem for non-integer exponents?)

[quote="iamhe"]How do i use the binomial theorem to find the sum of the infinite series

$ 1-\frac{1}{6}+\frac{1\cdot 3}{2!}\cdot \frac{1}{6^2}-\frac{1\cdot 3\cdot 5}{3!}\cdot\frac{1}{6^3}+\cdots$ ?

thank you very much in advance.[/quote]

$ \text{\LaTeX}$'d.

Anyway, as [b]t0rajir0u[/b] mentioned, you should be familiar with the generalized binomial theorem, which states that

$ (x+y)^r=\sum_{k=0}^{\infty}\binom{r}{k}x^{r-k}y^k$ for $ r\in\mathbb{R}$, where $ \binom{r}{k}=\frac{r(r-1)(r-2)\cdots(r-k+1)}{k!}$.


Find [tex]\displaystyle{\sum_{k=1}^n{k*k!}}[/tex] in terms of n.

I'm not quite sure what you want but here's what I got:

[hide]F(n)=1*(2+2(3+3(....+(n-2)+(n-2)((n-1)+n^2)[/hide] were N is the greatest integer in the summation.

You should have a very simple answer that is not arbitrarily long (i.e. something like [tex]2^n+n!-71[/tex], which isn't the right answer).

That would be pretty cool if it were the right answer, though ... I'd love to see -71 in an important place in an interesting identity.

Okay I will try to use mathematical induction, but I have never been taught this is school and I am just now reading it from the AoPS books and my textbook.

[hide]

(1) For n=1, the sum is:

1*1! =1 according to the summation notation

(1+1)! - 1 = 2-1 = 1 according to the formula I found out.

(2) For n=2, the sum is:

1*1! + 2*2! = 1+4 = 5 according to the summation notation

(2+1)! - 1 = 3*2*1 - 1 = 6-1 = 5 according to the formula I found out.

(3) Lets say that it is valid for some integer k.

sum from 1 to n of k*k! = (n+1)! - 1

(4) Now we have to prove its valid for (n+1)

sum of 1 to (n+1) of k*k! is:

1*1! + 2*2! + 3*3! + 4*4! + 5*5! + ... + k*k! + a final term which is (n+1)(n+1)!

and is also equals (n+2)! -1 (so we want to show)

up to k*k!, we already assumed that was equal to (n+1)! -1

So, for (n+1):

sum from 1 to (n+1) of k*k! =

(n+1)! -1 + (n+1)(n+1)!

factor

[(n+1)! (1+(n+1))] -1

(n+1)!(n+2) -1

(n+2)! - 1

So we have shown it is true.

[/hide]

[hide]Honestly though I didnt think of it myself.  Go to: http://mathworld.wolfram.com/FactorialSums.html

[/hide]

Very good - but have you tried using [hide]telescoping series[/hide]?

Since its just getting infinitely larger, can't it not converge to a sum?  I guess I don't know what telescoping is, but I assume it means it converges to a sum...

Can someone tell me if my proof is correct?  I have been practicing using this method...

edited: nevermind, thanks beta

Yes, that was a very good use of induction!  One complaint, though -- in your sums, you had a sum up to k equaling something in terms of n.  That is, you let your variables get mixed in together in a way they shouldn't have.

Okay, let me critique myself:

"(3) Lets say that it is valid for some integer k."

  ----> I meant "some integer n"

"1*1! + 2*2! + 3*3! + 4*4! + 5*5! + ... + k*k! + a final term which is (n+1)(n+1)! "

  ----> I meant n*n! instead of k*k!

Also, in my textbook it says you only have to show it works for 1, but to me I would not be convinced w/just 1, so I did 2 also.  I mean, most "formulas" that you just make up out of your head will randomly work for 1, and then will not work for any other numbers.  What is your take on this?  Am I just being skeptical?  Is it wrong to include why it works for n=2 also?

Testing for n=2 is a good idea just to make sure that what your testing isn't nonsense, but it isn't necessary.

[quote="JS1527"]
Also, in my textbook it says you only have to show it works for 1, but to me I would not be convinced w/just 1, so I did 2 also.  I mean, most "formulas" that you just make up out of your head will randomly work for 1, and then will not work for any other numbers.  What is your take on this?  Am I just being skeptical?  Is it wrong to include why it works for n=2 also?[/quote]

It's a little unnecessary to include n=2.  When you use induction, you show the case for n=1, and then you show that if it is true for n, it's true for n+1.  If you can show the both parts, then clearly it's true for 2 - let n=1, so n+1=2 (and so on.. that's why induction shows it's true for all naturals :) ).

Right -- when I think I have a formula for something, I will often check it up to n = 4 or 5, but when I'm writing up a proof of it, I will only include the first case.  The nice thing about induction is that if it works for the first case, and it works for the general part of the proof, it automatically works for every other small case that you might be inclined to check.  So, checking small cases is good to verify that your proof could be correct, but isn't at all necessary as part of the proof.

[quote="NAS"]can induction only be used for the naturals, could you for instance have a proof by induction that worked for all integers?  Or a proof that worked for all rationals?[/quote]

I'm not sure about integers, it may work for that.. definitely not rationals, though, because you can have infinitely small intervals with the rationals, so there's no "n+1" step, which defines induction.

[quote="ComplexZeta"]You can do an induction that works forwards and one that works backwards to get all integers. You can induct on the denominator and then on the numerator to get all rationals.[/quote]

Ooh, right.  Ignore my stupid comment before >.<

Though inductions on the reals are not possible, because they aren't countable.

Countable... *Cue Steven from MC*

(Hey Carol!)

Technically, you could use induction on any countable set, so long as you can set up a correspondance between that set and the natural numbers.  Anyhow, induction is usually not a good way to proceed for non-natural-number questions.

[quote="Caroline"][quote="Osiris"]Though induction on the reals are not possible, because they aren't countable.

Countable... *Cue Steven from MC*

(Hey Carol!)[/quote]

Haha, hey Tom =D  Although I wouldn't be all too surprised if one day Steven proves that the reals [i]are[/i] countable..[/quote]

That would be difficult.

There's a kid in my math class who proved that the reals are algebraicly closed.


Let $ x\equal{}\frac{a^3\plus{}b^3\plus{}c^3}{3abc},y\equal{}\frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}$.
Then our inequality can be rewritten as:

$ x\plus{}\frac{1}{y} \ge y\plus{}\frac{1}{x} \Longleftrightarrow x^2y\plus{}x \ge y^2x\plus{}y \Longleftrightarrow xy(x\minus{}y)\plus{}(x\minus{}y)\ge 0 \Longleftrightarrow  (x\minus{}y)(xy\plus{}1)\ge 0$.
So it suffices to prove that:

$ \frac{a^3\plus{}b^3\plus{}c^3}{3abc}\ge \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca} \Longleftrightarrow \sum_{sym}a^4b \ge 2\sum_{cyc}a^3bc$ which is true from muirihed :)

and another one solution we can prove that:
$ \frac {a^3 \plus{} b^3 \plus{} c^3}{3abc} \plus{} \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2} \geq 2$
and
$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {3abc}{a^3 \plus{} b^3 \plus{} c^3} \leq 2$
Thank for your solution! :)

[quote="AngelHolder"]
$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {3abc}{a^3 \plus{} b^3 \plus{} c^3} \leq 2$
[/quote]

this is not ture...

[quote="AngelHolder"]With $ a,b,c$ are positive numbers , Prove that:
$ \frac {a^3 \plus{} b^3 \plus{} c^3}{3abc} \plus{} \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2} \geq \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {3abc}{a^3 \plus{} b^3 \plus{} c^3}$[/quote]
\[ \frac {{a^3 \plus{} b^3 \plus{} c^3 }}{{3abc}} \plus{} \frac {{ab \plus{} bc \plus{} ca}}{{a^2 \plus{} b^2 \plus{} c^2 }} \ge \frac {{a^2 \plus{} b^2 \plus{} c^2 }}{{ab \plus{} bc \plus{} ca}} \plus{} \frac {{3abc}}{{a^3 \plus{} b^3 \plus{} c^3 }}\]
\[ \Leftrightarrow \sum {\left( {\frac {{a \plus{} b \plus{} c}}{{3abc}} \plus{} \frac {{a \plus{} b \plus{} c}}{{a^3 \plus{} b^3 \plus{} c^3 }} \minus{} \frac {1}{{a^2 \plus{} b^2 \plus{} c^2 }} \minus{} \frac {1}{{ab \plus{} bc \plus{} ca}}} \right)\left( {a \minus{} b} \right)^2 } \ge 0\]
it's easy to prove that
\[ \frac {{a \plus{} b \plus{} c}}{{3abc}} \ge \frac {1}{{ab \plus{} bc \plus{} ca}}\]and\[ \frac {{a \plus{} b \plus{} c}}{{a^3 \plus{} b^3 \plus{} c^3 }} \ge \frac {1}{{a^2 \plus{} b^2 \plus{} c^2 }}\]
so ... clear

[quote="AngelHolder"]With $ a,b,c$ are positive numbers , Prove that:
$ \frac {a^3 \plus{} b^3 \plus{} c^3}{3abc} \plus{} \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2} \geq \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {3abc}{a^3 \plus{} b^3 \plus{} c^3}$[/quote]

$ \leftrightarrow \frac {a^3 \plus{} b^3 \plus{} c^3}{3abc} \ge \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}$

$ \leftrightarrow [(ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c) \minus{} 3abc]\sum_{cyc} (a \minus{} b)^2\ge 0$

[b]which is true from AM-GM[/b]


Quote:For what real values of y does: |y + 2| = 2 |y - 2|

Give both an algebraic and a logical solution.[hide]

y = 2 and y = -2 produce zeros.

We'll consider solutions in the three intervals.

(1) If y > 2, then both quantities are positive.

The equation is: y + 2 = 2(y - 2) ... which gives us: y = 6

(2) If -2 < y < 2, then the right group is negative.

The equation is: y + 2 = 2(2 - y) ... which gives us: y = 2/3

(3) If y < -2, then both quantities are negative.

The equation is: -y - 2 = 2(2 - y) ... which gives us: y = 6 (?)

Since we began with y < -2, we must discard this answer.

There are no solutions less than -2.

Therefore, the answers are: y = 6, y = 2/3

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

|y + 2| is the undirected distance from y to -2.

|y - 2| is the undirected distance from y to 2.

We are seeking points whose distance from -2 is twice that from +2.

There is a point between -2 and 2: y = 2/3

and another point to the right of +2: y = 6.[/hide]

rrusczyk wrote:For what real values of y does

|y + 2| = 2 |y - 2|

Give both an algebraic and a logical solution.

algebraic:

[hide]

|y + 2| = 2 |y - 2|

so

y + 2 = 2 |y - 2|         or        y + 2 = -2 |y - 2|

y + 2 = 2 |y - 2|:

y + 2 = 2y - 4           or        y + 2 = -2y - 4

y=2y-6                                 y=-2y-6

-y=-6                                   3y=-6

y=6                                      y=-2

y + 2 = -2 |y - 2|:

y+2=-2y+4              or           y+2=2y-4

y=-2y+2                                 look above(-2)

3y=2

y=(2/3)

so y={-2,2/3,6}

how do you do the locical solution?[/hide]

$|y+2|=2|y-2|$
Squaring of both sides,we have $(y+2)^2=4(y-2)^2$,solving this ,
we obtain $y=\frac{2}{3},6$.

kunny

looking back on my answer, -2 is extraneous

[quote="LynnelleYe"]Um, sorry about my ignorance, but what do you mean by 'both an algebraic and logical solution'?[/quote]

Algebraic would mean using algebra and manipulation of the equation. A logical solution would be one that is more intuitive, like using reasoning to show that the answer is correct.

[quote="rrusczyk"]For what real values of y does

|y + 2| = 2 |y - 2|

Give both an algebraic and a logical solution.[/quote]

[hide="Click to reveal hidden content"]*On ]- :inf: ;-2], y <= -2 so y+2 <= 0 so |y+2|=-y-2
  On [-2;+ :inf: [, y >= -2 so y+2 >= 0 so |y+2|=y+2
*On ]- :inf: ;2], y <= 2 so y-2 <= 0 so |y-2|=-y+2
  On [2;+ :inf: [, y >= 2 so y-2 >= 0 so |y-2|=y-2

Therefore, we have:

*On ]- :inf: ;-2], -y-2=-2y+4 so y=6
*On [-2;2], y+2=-2y+4 so y=2/3
*On [2;+ :inf: [, y+2=2y-4 so y=6

Finally we find S={2/3;6}. But I think that kunny's method is better.
[/hide]


Πιστεύω να ασχοληθείτε με το επόμενο πρόβλημα .
Βρείτε τις λύσεις στους θετικούς ακεραίους της εξίσωσης
$f(x,y)=x^{-y}+y^{-x}+x+y-4=0$

Βιαστηκά καπως βρηκα οτι χ=0 ή χ=1 ή χ=2 αρα..... :wink:

Μπορείς να γράψεις τη διαδικασία ?Νομίζω ότι δεν είναι εύκολη .

Είσαι σίγουρος;

Νομίζω ότι είναι σχετικά απλή... Παίρνεις τις περιπτώσεις $(x,\, y) = (1,1), (1,2), (2,1)$ και κρατάς αυτές που σου κάνουν.

Τώρα αν $x,\, y \geq 2$ τότε $0 < x^{-y}+y^{-x}= \frac{1}{x^{y}}+\frac{1}{y^{x}}\leq \frac{1}{2^{2}}+\frac{1}{2^{2}}= \frac{1}{2}< 1$ άρα το $f(x,y)$ δεν πρόκειται να είναι ακέραιος (καθώς οι υπόλοιποι όροι του [i]είναι[/i] ακέραιοι και το $\mathbb{Z}$ είναι κλειστό ως προς την πρόσθεση).

Παιδιά συγγνώμη . Η εξίσωση είναι να λυθεί στους θετικούς πραγματικούς αριθμούς .Απλά αυτή την παραλλαγή την έδωσα να τη λύσει ο αδερφός μου και επηρεάστηκα και την έβαλα λάθος .Στους ακεραίους πάντως αυτή είναι η λύση .


I'm not sure if this problem hasn't been posted yet. If yes, I'm sorry.  :D

Well, the problem - it is given $ \triangle ABC$. Through the apexes $ B$ and $ C$ is constructed a circumference $ \omega$ such that the vertex $ A$ is an outside point for it. Let draw a circumference $ \lambda$ which will touch $ AB$, $ AC$ and also $ \omega$ in $ Q$ ($ \omega \cap \lambda \equal{} \{Q\}$). Prove that the bisector of $ \angle BQC$ pass  through the incenter ($ I$) of $ \triangle ABC$.

If the circle $ \lambda$ touches $ AB, AC$ at $ U, V,$ then $ BUIQ, CVIQ$ are both cyclic (see [url=http://www.mathlinks.ro/viewtopic.php?t=41667]Concyclic points with triangle incenter[/url]). The proposition immediately follows.

Dear Mathlinkers,
this nice result comes from Vladimir Protassov after my research.
For a synthetic proof :
http://perso.orange.fr/jl.ayme  vol. 2  Un remarquable résultat de Vladimir Protassov.
Sincerely
Jean-Louis


probability
floor function
AoPSwiki
geometric series
AMC

Let $ x$ be chosen at random from the interval $ (0,1)$. What is the probability that
$ \lfloor log_{10} 4x\rfloor \minus{} \lfloor log_{10} x\rfloor \equal{} 0$

A $ \dfrac{1}{8}$
B $ \dfrac{3}{20}$
C $ \dfrac{1}{6}$
D $ \dfrac{1}{5}$
E $ \dfrac{1}{4}$

The answer is $ C$. I checked the AoPSWiki, but there's no solution to this problem. Can anyone give me a hint? Thanks in advance.

4x and x have to lie within the same powers of 10. Use those to determine the bounds which give the probability.

[quote="AIME15"]Wait, do you want hints or a full solution?[/quote]
I used to think a hint is enough, but now I think I need a full solution.
[quote="aznlord1337"]4x and x have to lie within the same powers of 10. [/quote]
I've got that before posting, but I can't go on. Can you show me the following steps?

[quote="aznlord1337"]for example, .1 < x < 1, .1 < 4x < 1, .025 < x < .25. So on the interval .1 to 1, .1 < x < .25, a length of .15. And then we can keep dividing by 10 so u get a geometric series, .15 + .015... = .15/(1-.10) = 1/6.[/quote]
thanks dude


limit
integration
trigonometry
calculus
real analysis
real analysis solved

[u][b]The author of this posting is : liyi[/b][/u]
____________________________________________________________________

f(x) is piecewisely continous on $\left( -\pi ,\;\pi \right)$. It is continuous at u=0 and it has side derivatives(is it the right word?) there. Prove that

$\displaystyle \lim\limits_{\pi \rightarrow \infty }\int\limits_{-\pi }^{\pi }f\left( u\right) \frac{\cos \frac{u}{2}-\cos pu}{2\sin \frac{u}{2}}du=\frac{1}{2}\int\limits_{0}^{\pi }\left( f\left( u\right) -f\left( -u\right) \right) \cot \frac{u}{2}du$.

The left hand-side can be written with properties linearity of integral,
 and change variable

$1/2.\int_{0}^{\pi}(f(u)-f(-u))\cot(u/2)du + 1/2.\int_{0}^{\pi}(f(u)-f(-u))\frac{\cos(pu)}{\sin(u/2)}du$

With Riemann-Lebesgue the second integral tends to $0$ when $p\rightarrow +\infty $


<asy>
picture pic;
size(2in);
draw(pic, unitcircle);
size(6in)
draw(pic, unitsquare);
draw(pic, unitcircle);
</asy>

eh...

this was a test run...I phail...I guess I have to stick with LaTex... Geogebra doesn't work either...

There IS a test forum for this kind of stuff: http://www.artofproblemsolving.com/Forum/index.php?f=224

Do you have Java? You know you could just click the Geogebra button when you are posting something to be able to draw a picture. There is a test forum too, or you could just preview it and then not post it.

First, asymptote tags are [code][asy][/asy][/code] not <asy>. (The <> are for the wiki).

Second, your code doesn't work. Also, I'm not sure what you were trying to draw.

[asy]picture pic;
size(2in);
draw(pic, unitcircle);
size(6in)
draw(pic, unitsquare);
draw(pic, unitcircle);[/asy]

What is an asymptote?How useful is that? :blush:
And how can we post Geometric diagrams without using Geogebra?

Thank you. :)

muks, you can check AoPS wiki to find out what asymtote is. I was just too lazy to read the whole directions...


floor function
ceiling function
combinatorics proposed
combinatorics

[b]Problem.[/b] [i]Let $T_{m,n}$ be a complete $m$-partite graph on $n$ vertices such that each of the $m$ parts has either $\left[\frac{n}{m}\right]$ or $\left\lfloor\frac{n}{m}\right\rfloor$.  Prove that
\[\varepsilon(T_{m,n})=\binom{n-\left\lfloor\frac{n}{m}\right\rfloor}{2}+(m-1)\binom{\left\lfloor\frac{n}{m}\right\rfloor+1}{2}. \]
[/i]

Just a notational note: in the U.S., at least, $[x]$ and $\lfloor x\rfloor$ are generally taken to have the same meaning, "the greatest integer not exceeding $x$."  One uses instead $\lceil x \rceil$ for the ceiling function.

I used $[.]$ for rounding to the nearest integer.


I totally forget how to solve this. Could anyone give my some hints for this?

-0.59x + 0.2y + 0.28z = 0
-0.81x - 0.96y - 0.94z = 0
0.18y - 0.17z = 0

Thanks!

First step is to see if there is a non-trivial solution (the trivial solution is x = y = z = 0) by calculating the determinant of the system,

$ \left | \begin{array}{ccc} \minus{}0.59 & 0.2 & 0.28 \\ \minus{}0.81 & \minus{}0.96 & \minus{}0.94 \\ 0 & 0.18 & \minus{}0.17 \end{array} \right |$

For a non-trivial solution, this determinant must be zero. In this case, the value of the determinant is -0.264, and so the solution is x = y = z = 0.


MATHCOUNTS
AMC
AIME
LaTeX
complex numbers

I have decided to make a mock mathcounts(actually I already made one), since people are bored and AIME15's next round will not start until next week.

How to enter: Get together to form a group of 4 Aopsers, make team name and as levans said pm me the entries. No ridiculously long team names please. And no use of bad language in the team names. Choose a captain and the captain will pm me the entries which has to include the usernames of everyone in the group, and the team name.

Time line: Tuesday-Thursday is signups. Friday is when the testing will be posted.

Test: 30 question sprint-40 minutes;8 question target-I am going to do this differently and let you do all 8 questions in 24 minutes. There [b]might[/b] be a team round.

Results:There will be something for the top three individual finishers and top three teams(probably be smilies and a proud feeling). Individuals and teams will be scored based on the regular mathcounts system. Sprint-1 point each, Target-2 points each. Team total will be the average of everybody's score who are on the team. There [b]might[/b] be a team round.

No cheating of any sort. You may only use a pencil, paper, highlighter and erasers. You may use a calculator on the target. Problems may be taken from sources, but only a few. I will give credit to the sources after Friday. No spam in this topic. First I will give to a warning, then you will be eliminated.(spam as in having nothing to do with this topic or something that annoys others). If many people don't like this time I have chosen then I will postpone this to another time.

I don't think I missed anything....if I did I will edit.

[quote="AIME15"]Thank you for copying mine.

Anyway:

If isabella2296, vallon22, and johnnybeard consent, I would like to make a team with us four.

Team AIMEizzy22's Beard is:
isabella2296
vallon22
AIME15
johnnybeard[/quote]

Didn't he say to PM the entries

And what's the point of asking if so and so will consent include us

I could say if Darryl Wu says yes, he can be on my team XD

ummm...
shentang, could you plz edit the name, so we don't confuse yours with the one AIME made?

Yep, I will edit it to be Shen Mathcounts.  :P
Someone pmed me a good question. So I will add on to the rules. I don't want any to feel excluded so I am allowing 2, 3, or 4 person teams. If you can't find anyone then you can just go as an individual.

if a person goes as an individual, it might be a little hard to do the team round =)

There might be a team round, I am still not certain yet. I will probably decide tomorrow. The team round for the individual will not be scored offically, since he/she is not a "team". Btw, if a team does happen I will probably give a 35 minute time limit.

I'll join, but yeah it's hard to do a team unless everyone on a team have aim or something

Maybe a Team Round wouldn't be such a good idea, like moogra said. It'd be virtually impossible.

Ok, I have got it, I will just make the team round one question, 15 minute time limit. You guys can "talk" using whatever you want, but I am not responsible of on any policy, if you guys are breaking, or what you guys say. @Moogra please pm of if you are going as an individual or a group

Will anyone be a team with me?

[size=25]I hope someone will.[/size]

[quote="Quickster94"]Do u need any help with this?  I'd be happy to help grade, look over problems, etc[/quote]

I can help as well (if you need me).  I can make up some good probs.

I'll be a fill-in if anyone needs a member on their team, but I'm lazy and won't participate unless someone recruits me. :P

So far only 1 submission, and 5 hours left...

Can we all have an extension because only 1 person has turned in scores?

Fine, one last day. No more extra days. I have a feeling some people are cheating...

You say this test was wicked hard? :o
But I just finished my mock test and it was designed to be harder than shentang's! :D
Eh-oh, gotta edit my test!

By the way, I *think* I did all the problems on this test before, so I don't think I'm eligible to participate. :|

An extra day? Darn, I'm not going to win after all...

But this test really was hard, there's no way it's MathCounts level. It's way above Nats questions...

Hey Shen,

Keep making up those hard problems and thinking at that next level.  Good job.

Everyone else remember that if it is hard for MATHCOUNTS it is hard for everyone.  Just do your best, I am always trying to get my students to look at higher level problems!!

Go SHen!!

[quote="shentang"]Actually Yongyi, you only did some. I fused in some problems from mock test 1 with mock test 2.[/quote]

er...maybe you want to see those problems with your test and do something about it?

[quote="xoangieexo"]umm... theres a problem with number 21
following the pattern, it should be -666^2
because every root n taht is added (examples: 4, 5, 8, 9, 12, 13) is either divisible by 4 or n-1 is divisible by 4.
666 isn't and 665 isnt either...

oh wells im just gonna send in the answers

the test was wicked hard btw. definately not mathcounts lvl :-/[/quote]

Sorry yeah it is -666^2.
[size=200]
[b]Remember last day, everyone please send in their answer.[/b][/size]

[hide="Tied for first(individual)"]
xoangieexo
27r
sicarius1029

That's a three way tie so you guys will have to split these emotions:
 :winner_first:  :winner_second:  :winner_third: [/hide]

[hide="First place Team"]
No Name:
xoangieexo
27r
sicarius1029
 :winner_first:
[/hide]

[hide="Second place Team"]
AIMEizzy22's Beard:
[b]isabella2296[/b]
AIME15
vallon22
johnnybeard
I bolded isabella2296 since she was the only person on her team to send me her answers[/hide]

No third place.

Only 4 people sent me their answers...thats sad.

Anyways, congratulations to everyone who sent me their answers. Thanks for participating.

These are the answers I used:

interesting test
hm..... nearly got all of the sprint correct
but the target killed me
team wasn't too hard


Is it, in general, true that eigenvalues of hermitian matrices come as pairs that are reciprocal? (so that multiplication of eigenvalues of a 2x2 matrix is always 1). Or at least, does anyone knows such kind of matrix?

Thanks in advance!

The product of all the eigenvalues of a matrix is the determinant. So to have a $2\times 2$ matrix whose eigenvalues are reciprocals you would need a matrix whose determinant is $1.$ (All hermitian adds to this is the guarantee that the eigenvalues are real.)

Well, what i mean not just "all of them multiplies to 1", i mean reciprocal [i]pairs[/i]. So, if i have  a 4x4 matrix, then I have $\lambda_{1}\lambda_{4}= 1$ and $\lambda_{2}\lambda_{3}= 1$ (assuming that they're sorted). For a $2n \times 2n$ matrix, I'll have $n$ such pairs...


function
search
induction
algebra
functional equation
algebra unsolved

[quote="M4RI0"]Let $S$ be the set of nonnegative integers. Find all functions $f: S \rightarrow S$ such that:

$f(x^{2}+y^{2})=xf(x)+yf(y)$, for all $x,y \in S$[/quote]

$y=0\implies f(x^{2})=xf(x)\implies f(x^{2})+f(y^{2})=xf(x)+yf(y)=f(x^{2}+y^{2})$ so if $N^{2}$ is the set of perfect squares then $f(p+q)=f(p)+f(q)$ so $f(p)=a\cdot p$ for every $p\in N^{2}$.
$f(x^{2})=xf(x)$ for every $x \in N\implies xf(x)=ax^{2}\implies f(x)=ax$

[quote="harazi"]That's surely wrong: if $f(x+y)=f(x)+f(y)$ on a set $A$, it does not follow that $f$ is linear in $A$. By the way, use search button, the problem (or variants of it) have already been discussed.[/quote]
It is true if S is subset $xQ=\{xr|r\in Q\}$. In this case S is subset Z.

[quote="harazi"]Hmmm, I think I've said above that the solution is wrong...[/quote]
Can you solve it ? :)

[quote="M4RI0"]Let $S$ be the set of nonnegative integers. Find all functions $f: S \rightarrow S$ such that:

$f(x^{2}+y^{2})=xf(x)+yf(y)$, for all $x,y \in S$[/quote]
For that the function does not go of royal to royal?

for me it's the first time so I'm not sure. I think the only solutions are f(a)=0 or f(a)=a for every nonnegative integer a.
se x=y we have f(2x^2)=2xf(x)
if x=1 f(2)=2f(1)
if x=2 f(2^3)=2^3f(1)
if x=3 f(2^7)=2^7f(1)
.
.
.
if x=n f(z)=...
let's suppose that for every nonnegative integer n if x=n f(z)=zf(1)  (1)
if x=n+1 we have f(2z^2)=2z*(zf(1))=2z^2f(1); if 2z^2=t f(t)=tf(1)
so for induction the (1) is true.
if x^2+y^2=a we have f(a)=x^2f(1)+y^2f(1)=f(1)(x^2+y^2)=f(1)f(a)
so f(a)=f(1)f(a) and f(a)(1-f(1))=0 so 0r f(a)=0 or f(1)=1 and in the two cases we have
f(a)=a
f(a)=0

But the function $f(a)=ca$ clearly satisfies the equation for all positive integers $c$...

yes I made an error if x^2+y^2=a f(a)=af(1) and not f(a)f(1)
i complete my solution chaging it:
f(a)=af(1) and let'see that all the nonnegative integers a saatisfies the equation:
f(x^2+y^2)=a(x^2+y^2)
xf(x)+yf(y)=ax^2+ay^2
a=x^2+y^2 so a is a nonnegative integer

Your solution is incomplete and quite confusing, because not every nonnegative integer can be written as a sum of 2 squares.
And that's really the tricky part.

$f(0)=f(0^{2}+0^{2})=0\cdot f(0)+0\cdot f(0)=0$.
Let's set $f(1)=k$. Then we can show $f(n)=kn$ for all $n\in S$.
And the proof is quite similar to the one I gave [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=2890]here[/url].

First let's show $f(2^{n})=k\cdot 2^{n}$ for all $n\geq 0$:
This is clear for $n=0$, and if $n\geq 0$ with $f(2^{n})=k\cdot 2^{n}$, then
$f(2^{2n})=f((2^{n})^{2}+0^{2})=2^{n}f(2^{n})+0\cdot f(0)=k\cdot 2^{2n}$
$f(2^{2n+1})=f((2^{n})^{2}+(2^{n})^{2})=2^{n}f(2^{n})+2^{n}f(2^{n})=k\cdot 2^{2n+1}$

Next we show $f(2^{n}\cdot 3)=k\cdot 2^{n}\cdot 3$ for all $n\geq 0$:
Otherwise take $n\geq 0$ minimal with $f(2^{n}\cdot 3)\not=k\cdot 2^{n}\cdot 3$ .
If $n=2m$ even, then using above
 $f(2^{n}\cdot 5)=f((2^{m+1})^{2}+(2^{m})^{2})=2^{m+1}f(2^{m+1})+2^{m}f(2^{m})\\ \hspace*{0.7in}=k(2^{m+1})^{2}+k(2^{m})^{2}=k\cdot 2^{n}\cdot 5$
If $n=2m+1$ odd, then $n>m\geq 0$ and by minimality of $n$ we get
$f(2^{n}\cdot 5)=f((2^{m}\cdot 3)^{2}+(2^{m})^{2})=2^{m}\cdot 3f(2^{m}\cdot 3)+2^{m}f(2^{m})\\ \hspace*{0.7in}=k(2^{m}\cdot 3)^{2}+k(2^{m})^{2}=k\cdot 2^{n}\cdot 5$
In each case we get $f(2^{n}\cdot 5)=k\cdot 2^{n}\cdot 5$, and then
$2^{n}\cdot 3f(2^{n}\cdot 3)+2^{n+2}f(2^{n+2})=f((2^{n}\cdot 3)^{2}+(2^{n+2})^{2})=f((2^{n}\cdot 5)^{2})\\ \hspace*{2in}=2^{n}\cdot 5f(2^{n}\cdot 5)=k\cdot (2^{n}\cdot 5)^{2}$
Therefore
$2^{n}\cdot 3f(2^{n}\cdot 3)=k\cdot (2^{n}\cdot 5)^{2}-k\cdot (2^{n+2})^{2}=k\cdot (2^{n}\cdot 3)^{2}$ and $f(2^{n}\cdot 3)=k\cdot 2^{n}\cdot 5$.

And finally we can prove $f(n)=kn$ for all $n\in S$:
Let $n$ be minimal with $f(n)\not =kn$.
Then using above we must have $n>0$ and $n=2^{r}u$ for some $r\geq 0$ and some odd $u\geq 5$.
And then we have this identity
$u^{2}+(\frac{u-5}{2})^{2}=(u-2)^{2}+(\frac{u+3}{2})^{2}$
with $0\leq \frac{u-5}{2},\frac{u+3}{2},u-2<u$ for $u\geq 5$.
Multiplying with $(\frac{n}{u})^{2}=2^{2r}$ then leads to $n^{2}+m^{2}=p^{2}+q^{2}$ for some $0\leq m,p,q<n$
and by minimality of $n$ we then have $f(m)=km$,$f(p)=kp$ and $f(q)=kq$.
And so we get
$nf(n)+km^{2}=nf(n)+mf(m)=f(n^{2}+m^{2})=f(p^{2}+q^{2})\\ \hspace*{1in}=pf(p)+qf(q)=kp^{2}+kq^{2}=kn^{2}+km^{2}$
and $f(n)=kn$.

Conversely every funktion $f(x)=kx$ is a solution for all $k\geq 0$.

[quote="boxedexe"][quote="tmbtw"][quote="cckek"]

 $f(p+q)=f(p)+f(q)$ so $f(p)=a\cdot p$ for every $p\in N^{2}$.
[/quote]
Why like that ? :)[/quote]
It is Cauchy's functional equation.[/quote]
But $\mathbb{N}^{2}$ isn't closed under summation.
So $f(p+q)=f(p)+f(q)$ $\forall p,q\in\mathbb{N}^{2}$ doesn't help you prove $f(3)=f(1+1+1)=f(1)+f(1)+f(1)=3f(1)$, for example. This problem is much more complex.

I did not say that that proof was correct, but Cauchy's functional equation is that $f(x)=cx, c\in\mathbb{R}$ satisfies the functional equation $f(x+y)=f(x)+f(y)$.


inequalities
limit
complex numbers
real analysis
real analysis unsolved

We have in the plane the system of points $A_1,A_2,\ldots,A_n$ and $B_1,B_2,\ldots,B_n$, which have different centers of mass. Prove that there is a point $P$ such that \[ PA_1 + PA_2 + \ldots+ PA_n = PB_1 + PB_2 + \ldots + PB_n . \]

This was one of the cutest problems in the olympiad, I think :).

Let $G_A$ and $G_B$ be the two (distinct, according to the hypothesis) centers of gravity of the systems of points $(A_i)$ and $(B_i)$ respectively. Suppose WLOG they are on the $x$-axis of the plane, in this order ($G_A$, followed by $G_B$). First we make the point $P$ go to infinity on the ray $(G_AG_B$, and then on the ray $(G_BG_A$. If we prove that in the first case $\sum PA_i-nPG_A\to 0$, it will follow, by analogous reasoning, that $\sum PB_i-nPG_B\to 0$, and that the same relations hold for the second case as well. But in the first case we have $PG_A>PG_B$, while in the second we have $PG_A<PG_B$, which means that there is a $P$ such that $\sum PA_i<\sum PB_i$, and there is a $P$ such that the reversed inequality holds, so, by continuity, equality must hold for some $P$, and we are done.

All we need to prove now is the following assertion:

Let $a_1,\ldots,a_n$ be complex numbers, and let $g=\frac{\sum a_i}n$ be the center of mass of this system of points. Also let $x$ be an arbitrary real number (it works for complex $x$ as well, of course). Then $\lim_{t\to\infty}\left(\sum|a_i+tx|-n|g+tx|\right)=0$.

Fix $i\in\overline{1,n}$. We have $|a_i+tx|-|\Re(a_i+tx)|=\sqrt{\Re^2(a_i+tx)+\Im^2(a_i+tx)}-|\Re(a_i+tx)|$, which is equal to $\frac{\Im^2(a_i+tx)}{|a_i+tx|+|\Re(a_i+tx)|}$, and this goes to $0$ for $t\to\infty$ because the numerator is fixed ($t$ and $x$ are real, remember), while the denominator goes to $\infty$ at $t\to\infty$. For the same reasons, $n|g+tx|-n|\Re(g+tx)|\to 0$ as $t\to\infty$. But for large enough $t$ all $\Re(a_i+tx)$ and $\Re(g+tx)$ will have the same sign, hence $\sum|\Re(a_i+tx)|=n|\Re(g+tx)|$, and the conclusion follows.

Yes, this problem was proposed by Marius Cavachi. Yes, grobber you are right. This was one of the cutest problems given in this year Olympiad. ;)


If you made it and you're going to MOP, post here.

I don't want to go on the plane there alone. There's no direct flight, and I'll be lonely and stuff. Plus, I'm not an organized guy, so... I'd probably get lost or board the wrong plane or something.

If you're interested with maybe going with me (like meeting at the airport), send me a PM too.

Me, that's it (I'm a freshman so I get it easy. Plus, I think I got nice graders...) Why do you think I'm making a topic on this, Fanfan? XD


quadratics function Vieta algebra polynomial parameterization vector

How can we prove this:

The differential equation $y''+y'-6y = 0$ resembles the quadratic $t^{2}+t-6 = 0$ with roots $a$ and $b$

Then:

$(y''-ay')-b(y'-ay) = 0$
$(y''-by')-a(y'-by) = 0$

Question: Why do the diff. equations resemble the respective quadratic equations? In what way? How and why can we link the two equations together?
thanks  :)

Take a general second order homogenous differential equation:

$ay''+by'+cy=0$

Let $y=e^{mx}$, then we have

$am^{2}e^{mx}+bme^{mx}+ce^{mx}= e^{mx}\left(am^{2}+bm+c\right)=0$

The exponetial function is never zero, so we must have:

$am^{2}+bm+c=0$.

The two equations at the bottom of your post follow directly from Vieta's formula:

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

Is it enought to prove the two equations by letting $y = e^{mx}$? How about other function $y$ in general? (say, a polynomial)

Thanks  :D

A second-order differential equation without initial values has a solution with two parameters, which is unique up to parameter shifts.  (That last part is the most important.)  It just so happens that if the roots of that quadratic are different from each other, you'll end up with two solutions $y = e^{r_{1}x}$ and $y = e^{r_{2}x}$, from which you can determine that every single function of the form $y = A e^{r_{1}x}+B e^{r_{2}x}$ is a solution.  But this is a solution with two parameters, so by the uniqueness of solutions to ODEs, this is the only solution.

The short version: ODEs have this lovely property which says, more or less, that if you can guess the answers, you know that you've got all of them.  Jimmy's method is a good way to guess.

If you are interested in the relation between ordinary differential equations and algebraic equations in a more general context, the [url=http://www.efunda.com/math/laplace_transform/index.cfm]Laplace transform[/url] is what you want.

[quote="JBL"]ODEs have this lovely property which says, more or less, that if you can guess the answers, you know that you've got all of them.[/quote]
The name for this "lovely property": the existence-uniqueness theorem. The theorem is usually stated for first order equations, but it has a vector version for first order systems. And the second order equations we're talking about here can be rewritten as systems of two first order equations for two unknown functions.

I've gone so far as to tell a differential equations class, "There's really only one theorem in this course, and its corollaries." I exagerate, of course, but not by much.


Prove that if the polynomial $x^3 + ax^2 + bx + c$ has three distinct real roots then so does the polynomial $x^3 + ax^2 + (a^2 + b)x/4 + (ab - c)/8$.

Why do you consider this problem as very funny, Peter? ;)

the solution is so easy(i didn't actually check but it must be that way)...knowing vieta is something really helpful  :)

hm...
if we name p,q,r the roots of the first equation, by viete's formulas it's easy to check that (p+q)/2, (q+r)/2 and (r+p)/2 are the roots of the second equation...
in fact it's trivial, but you have to think about it :)
no strange inequalities, here :D


The length of one leg of a right triangle is 9 meters. The lengths of the other two sides are consecutive integer numbers of meters. What is the number of meters in the perimeter of the triangle?

You have to find the two consecutive integers whose squares are 81 apart. (or mem the triplet)
This happens to be 40 and 41.
P=9+40+41=90

[hide="Solution that shows how to get the answer"]
Let $ a=\text{the other leg}.$ That means $ a+1=\text{the hypotenuse}.$ By the Pythagorean Theorem, \begin{align*}a+1 &=\sqrt{9^2+a^2}\\a^2+2a+1 &=81+a^2\\ 2a+1&=81\\ a &=40.\end{align*} The hypotenuse is $ a+1,$ which is $ 40+1=41.$ Finally, the perimeter is $ 9+40+41=\boxed{90}.$[/hide]


Is there any references as to where I can find the proof of two positive defibnite matrices being positive matrices.

Basically if A is a positive definite matrix, is there any way of showing that A*A is also positive definite.

Thanks
Ingrid

We clearly need $A$ to be symmetric over $\mathbb{R}$ or Hermitian over $\mathbb{C}$ as well; otherwise $A=\begin{bmatrix}3&-4\\4&3\end{bmatrix}$ is a counterexample.

A real symmetric (Hermitian) matrix is orthogonally (unitarily) similar to a real diagonal matrix, and is positive definite if and only if all of the diagonal entries of that matrix are positive. Clearly, the square of any invertible real symmetric or Hermitian matrix is positive definite.

Another way to look at this: $\langle x,A^2x\rangle=\langle Ax,Ax\rangle>0$ whenever $A$ is self-adjoint and injective as a linear operator on an inner product space.


Whenever I go to the blog page it comes out blank. It is the same for the Comments. Can you fix this or is this part of upgrading the blogs?

I'm upgrading some stuff now. It shout be ok in a few minutes.


[hide]Note that $ n^{2}-n \equiv 0, 2, 2, 0 \pmod{4}$ for $ n \equiv 1, 2, 3, 4$

Then, $ i^{n^{2}-n}$ repeats $ 1,-1,-1,1$, and $ \frac{1-i^{n^{2}-n}}{2}$ repeats $ 0,1,1,0$

Replacing $ n$ with $ n+1$ will give the sequence we want:
$ \boxed{a_{n}= \frac{1}{2}-\frac{1}{2}i^{n^{2}+n}}$[/hide]

It is likely that the person @above will not see this post given that they wrote it almost $15$ years ago today, however, I believe that the sequence isn't specifically for integers.


Jen and Jerry need to carpet their L-shaped room, having dimensions as shown.  The carpet they chose is a solid color and is sold in rolls that are 12 feet wide.  How many square feet of wasted carpet will they have if the minimum length from the roll is used and only one seam is permitted?
[metapost=910de088ac96ecd4711b263b996a6f5b] save u;

 % Set the value of u to scale the entire image
 u:=.15cm;

 draw origin--(16u,0)--(16u,20u)--(4u,20u)--(4u,10u)--(0,10u)--cycle;

 label.lft(btex $ 10\'$ etex,(0,5u));
 label.rt(btex $ 10\'$ etex,(4u,15u));
 label.top(btex $ 4\'$ etex,(2u,10u));
 label.top(btex $ 12\'$ etex,(10u,20u));

 [/metapost]

Think of cutting such a shape out of a rectangle. The smallest possible rectangle you can cut such a shape out of is a 24*16=384. The area of the carpet is 280 so our answer is 384-280=104

Imagine having a roll 12x20, then having another roll 12x4.  If you arrange them correctly, you get a 2x4 piece of carpet that is not used, so the answer is $\boxed{8.}$