Divisibility
exponential
IMO Shortlist
Conceptual

The left hand side is even, so $ a^4\plus{}b^2$ is even, and $ a,b$ has the same parity.
However, if $ a,b$ are odd, we have that $ 4$ divides $ 7^a\minus{}3^b$, but $ 4$ doesn't divide $ a^4\plus{}b^2$.

So $ a,b$ are even, and let's say $ a\equal{}2a_1,b\equal{}2b_1$.

You'll get $ (7^{a_1}\minus{}3^{b_1})(7^{a_1}\plus{}3^{b_1})$ divides $ 4(4a_1^4\plus{}b_1^2)$

If $ b_1$ is odd, then $ 8$ divides $ (7^{a_1}\minus{}3^{b_1})(7^{a_1}\plus{}3^{b_1})$ but not $ 4(4a_1^4\plus{}b_1^2)$, so $ b_1$ is even.

Say $ b_1\equal{}2c_1$.

You'll get $ (7^{a_1}\minus{}3^{2c_1})(7^{a_1}\plus{}3^{2c_1})$ divides $ 16(a_1^4\plus{}c_1^2)$

So $ |(7^{a_1}\minus{}3^{2c_1})(7^{a_1}\plus{}3^{2c_1})|\leq |16(a_1^4\plus{}c_1^2)|$

But $ |(7^{a_1}\minus{}3^{2c_1})\geq 2$ (because both terms are even), and so $ 2(7^{a_1}\plus{}3^{2c_1})\leq 16(a_1^4\plus{}c_1^2)$.

But for $ a_1\equal{}1,2,3$ we easily get $ c_1\equal{}1,2$, and for $ a_1\geq 4$, we have that $ 2(7^{a_1})>16a_1^4$. Also, $ 2(3^{2c_1})>16c_1^2$ for all $ c_1\in \mathbb{N}$.

So it is enough to check $ (a_1,c_1)\equal{}(1,1),(2,1),(3,1),(1,2),(2,2),(3,2)$. We'll get $ (a_1,c_1)\equal{}1,1$, or $ (a,b)\equal{}(2,4)$ as the only solution.

By the way, Dida Drogbier, are you from Chelsea or AC Milan?  :D  :rotfl:

dig, dig, dig :P
alternative solution anybody? :D

This is my alternative solution.

The left hand side is even, so $ a^4\plus{}b^2$ is even, and $ a,b$ has the same parity.
However, if $ a,b$ are odd, we have that $ 4$ divides $ 7^a\minus{}3^b$, but $ 4$ doesn't divide $ a^4\plus{}b^2$.

So $ a,b$ are even, and let's say $ a\equal{}2x,b\equal{}2y$.

You'll get $ (7^{x}\minus{}3^{y})(7^{x}\plus{}3^{y})$ divides $ 4(4x^4\plus{}y^2)$

If $ y$ is odd, then $ 8$ divides $ (7^{x}\minus{}3^{y})(7^{x}\plus{}3^{y})$ but not $ 4(4x^4\plus{}y^2)$, so $ y$ is even.

Say $ y\equal{}2z$.

You'll get $ (7^{x}\minus{}3^{2z})(7^{x}\plus{}3^{2z})$ divides $ 16(x^4\plus{}z^2)$

So $ |(7^{x}\minus{}3^{2z})(7^{x}\plus{}3^{2z})|\leq |16(x^4\plus{}z^2)|$

But $ |(7^{x}\minus{}3^{2z})\geq 2$ (because both terms are even), and so $ 2(7^{x}\plus{}3^{2z})\leq 16(x^4\plus{}z^2)$.

But for $ x\equal{}1,2,3$ we easily get $ z\equal{}1,2$, and for $ x\geq 4$, we have that $ 2(7^{x})>16x^4$. Also, $ 2(3^{2z})>16z^2$ for all $ z\in \mathbb{N}$.

So it is enough to check $ (x,z)\equal{}(1,1),(2,1),(3,1),(1,2),(2,2),(3,2)$. We'll get $ (x,z)\equal{}1,1$, or $ (a,b)\equal{}(2,4)$ as the only solution.

tom: you shouldn't give solution to shortlist problem before IMO. you should talk to me (phone number below)

I believe that proving $b \equiv 0 \pmod{4}$ is unnecessary, the solution in that case is a bit more messy, but similar. The motivation for this solution is using size issues in divisibility, similar to the above posts.

[hide=solution]
We claim that $a, b$ are both even.
If $a\equiv1\pmod{2}, b\equiv1\pmod{2}$ then $a^4+b^2\equiv2\pmod{4}, 7^a-3^b\equiv(-1)^a-(-1)^b\equiv0\pmod{4}$, contradiction.
If $a\equiv1\pmod{2}, b\equiv0\pmod{2}$ then $a^4+b^2\equiv1\pmod{4}, 7^a-3^b\equiv(-1)^a-(-1)^b\equiv2\pmod{4}$, contradiction.
If $a\equiv0\pmod{2}, b\equiv1\pmod{2}$ then $a^4+b^2\equiv1\pmod{4}, 7^a-3^b\equiv(-1)^a-(-1)^b\equiv2\pmod{4}$, contradiction.

So we may write $a=2c, b=2d$. The given equation becomes
\[ (7^c-3^d)(7^c+3^d)|4(4c^4+d^2) \implies |7^c-3^d||7^c+3^d| \le 16c^4+4d^2 \]
$7^c-3^d=0$ clearly never occurs, so $|7^c-3^d| \ge 1$, hence $7^c+3^d \le 16c^4+4d^2$ since $7^c+3^d>0$.
Fix $d$. We must have
\[ 7^c-16c^4 \le 4d^2-3^d \]
When $c \ge 4$ this means $4d^2-3^d >0 \implies d=1 \implies 16c^4>7^c+1 \implies c \le 4$.
So consider the cases $c=1, 2, 3$ or $c=4, d=1$. The key is to keep using the inequality $7^c+3^d \le 4(4c^4+d^2)$.
If $c=1$, then $3^d \le 9+4d^2 \implies d=1, 2, 3$. Only $d=2$ works.
If $c=2$, then $d=1, 2, 3, 4, 5$. None of these work.
If $c=3$, then $d=1, 2, 3, 4, 5, 6$. None of these work.
Finally, $c=4, d=1$ does not work.

So our solution is $\boxed{(a, b)=(2, 4)}$.
[/hide]

[quote=AMN300]$4d^2-3^d >0 \implies d=1$[/quote]
why doesn't  $4d^2-3^d >0$  imply $d=1,2,or   3  $  ?

[hide=Solution][quote=2007 ISL N1]
Find all pairs of natural numbers $ (a, b)$ such that $ 7^a - 3^b$ divides $ a^4 + b^2$.
[/quote]

Note that $7^a-3^b \equiv 0 \pmod{2}$ so $a \equiv b \pmod{2}$. If $a \equiv b \equiv 1 \pmod{2}$ then $7^a - 3^b \equiv 0 \pmod{4}$ but $a^4+b^2 \equiv 2 \pmod{4}$. Therefore, $a \equiv b \equiv 0 \pmod{2}$ so let $a = 2x$ and $b=2y$.

Then we have $(7^x+3^y)(7^x-3^y) = 16x^4+4y^2$. Clearly $7^x-3^y \ge 1$ so $7^x+3^y \le 16x^4+4y^2$ which implies that $7^x-16x^4 \le 4y^2-3^y$ $\bigstar$.

Now consider the sign of the LHS and the RHS. Notice that $7^x-16x^2<0$ for all integers $1\le x \le 4$ and $7^x-16x^2 >0$ for all integers $x \ge 5$. Similarly, $4y^2-3^y > 0$ for all integers $1 \le y \le 3$ and $4y^2-3^y < 0$ for all integers $y \ge 4$.

If $x \ge 5$. Then the LHS is positive so the only possible values for $y$ is $1,2$ or $3$. Testing the three possibilities, we see that no ordered pairs satisfy this case.
If $x=4$ then $-1685 \le 4y^2-3^y$ so $y \in \{1,2,3,4,5,6\}$. None of these satisfy the original equation.
If $x=3$ then $-985 \le 4y^2-3^y$ so $y \in \{1,2,3,4,5,6\}$. None of these satisfy the original equation.
If $x=2$ then $-207 \le 4y^2-3^y$ so $y \in \{1,2,3,4,5 \}$. None of these satisfy the original equation.
Finally, if $x=1$ then $-9 \le 4y^2 - 3^y$ so $y \in \{1,2,3\}$. Out of these, only $(x,y) = (1,2)$ satisfies the originally equation giving our only solution as $(a,b) = \boxed{(2,4)}$.[/hide]

[hide=Solution][quote=ISL 2007 N1]Find all pairs of natural numbers $ (a, b)$ such that $ 7^a \minus{} 3^b$ divides $ a^4 \plus{} b^2$.

[i]Author: Stephan Wagner, Austria[/i][/quote]
[b][color=#000]Solution:[/color][/b] $7^a-3^b$ is even, so $a^4+b^2$ is also even. So, $a,b$ have same parity. Hence, $7^a-3^b \equiv 3^a-3^b \equiv 0 \pmod{4}$. For any $x$ is odd, $x^2 \equiv 1 \pmod{4}$ & Since, $a^4+b^2 \equiv 0 \pmod{4}$ $\implies$ $a,b$ are even. So, $8 \mid 7^a-3^b$ $\implies$ $a^4+b^2$ $\equiv$ $0$ $\pmod{8}$ $\implies$ $b \equiv 0 \pmod{4}$ $\implies$ $16 \mid 7^a-3^b$. So, if $a=2a_1$ & $b=4b_1$, where, $a_1,b_1 \in \mathbb{N}$
$$\implies (7^{a_1}+3^{2b_1})(7^{a_1}-3^{2b_1}) \mid 16(a_1^4+b_1^2)$$
$$\implies 2(7^{a_1}+9^{b_1}) \le \mid (7^{a_1}+9^{b_1})(7^{a_1}-9^{b_1}) \mid \le 16(a_1^4+b_1^2)$$
Note: $7^x \geq 8x^4$ for $x \geq 4$ and $9^x \geq 8x^4$ for $x \geq 3$. So, we have a good bound on $a_1,b_1$. If $b_1=1$ $\implies$ $a_1=1,2,3$. If $b_1=2$ $\implies$ $a_1=1,2,3$. If $b_1 \geq 3$ $\implies$ We have no Solution! So checking the permissible values, we get $\boxed{(a,b)=(2,4)}$ $\qquad \blacksquare$
[/hide]

It feels nice to get back to NT. Here's my solution that uses the same idea of bounding.
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[color=#00f][b]Proof: [/b][/color] Notice that clearly $7^a-3^b \equiv 0 \pmod 2$. Since $7,3$ have the same parity so this forces that either $(a,b)$ are both odd or are both even. We'll firstly work with the case when $(a,b)$ are both odd. Let $(a,b)=(2k+1,2l+1)$. Notice that $$7^a-3^b \equiv 7^{2k+1}-3^{2l+1} \equiv 0 \pmod 4$$ But $a^4+b^2 \equiv (2k+1)^4 +(2l+1)^2 \equiv 2 \pmod 4$ a contradiction. Now let $(a,b)=(2k,2l)$. Clearly we must have that $(7^k+3^l)(7^k-3^l) \mid 16k^4+4l^2$. But since $$2(7^k+3^l) \leq |(7^k+3^l)(7^k-3^l)| \leq 16k^4+4l^2$$ so we can estabilish that $7^k+3^l \leq 8k^4+2l^2$. This is equivalent to $3^l-2l^2 \leq 8k^4-7^k$. But notice that from induction $\forall l \geq 1$ we have that $3^l>2l^2$ and hence we have $0 \leq 3^l-2l^2 \leq 8k^4-7^k$ . But notice that $7^k \geq 8k^4$ $\forall k \geq 4$ hence we have to check for $k \leq 3$.
\\
Now notice that if $k=3$ then we have that $343+3^l \leq 648+2l^2 \implies 3^l-2l^2 \leq 305$. But notice that $3^l -2l^2 > 305$ for all $l\geq 6$ hence we have to check till $l=6$. But clearly none of the solution sets work. $\square$.
\\
Now if $k=2$ then we have $3^l-2l^2 \leq 128-49=79$ . But again notice that we that $3^l-2l^2 > 78$ for $l\geq 5$. Hence we check all the way uptil $5$. But clearly none of the solution sets work again $\square$.
\\
Now if $k=1$ then notice that we have $3^l-2l^2 \leq 1$ but notice that this is true only when $l \leq 2$ . So we check till $l=2$ and we have a solution set that woks which is $(k,l)=(1,2)$ or that $\boxed{(a,b)=(2,4)}$ $\blacksquare$.

[quote=tom_damrong]This is my alternative solution.
[/quote]

Ummm... You gave the exact solution (quite literally) as the one two posts above. How is yours an "alternative" solution?

Surprisingly, if you push hard enough this problem just falls.

We claim only $(2,4)$ works, and we can easily verify that it does.

Note that $2\mid 7^a-3^b,$ so $2\mid a^4+b^2,$ implying $a\equiv b\pmod{2}.$ This implies that $7^a\equiv 3^b\pmod {4},$ so $4\mid a^4+b^2,$ implying that $2\mid a$ and $2\mid b.$ Say that $a=2x$ and $b=2k.$ Then note $8\mid 49^{x}-9^k,$ so we must have $8\mid 4(4x^2+k^2),$ implying that $2\mid k.$ Let $k=2y,$ then we want to solve $(7^x-9^y)(7^x+9^y)\mid 16(x^4+y^2).$

Now we just bound. Note that $2(7^x+9^y)\leq |(7^x-9^y)(7^x+9^y)|\leq 16(x^4+y^2),$ implying $7^x+9^y\leq 8(x^4+y^2).$ Note that for all $x\geq 4,$ we have $7^x> 8x^4,$ and for all positive integers $y,$ $9^y>8y^2.$ So the only possible solutions are natural numbers $x\leq 3.$

In the case of $x=1,$ note that we must have
\[(7-9^y)(7+9^y)\mid 16(1+y^2),\]
or $81^y-49\leq 16y^2+16.$ Note this fails for any $y\geq 2,$ and that $y=1$ yields equality, so $(1,1)$ is the only solution for $x=1.$

In the case of $x=2,$ note that we must have
\[(49-9^y)(49+9^y)\mid 16(16+y^2).\]
We can easily verify that $y=1$ fails. For $y\geq 2,$ we want to ensure $81^y-7^4\leq 16y^2+16\cdot 16,$ which fails for all $y\geq 2.$

In the case of $x=3,$ note that we must have
\[(343-9^y)(343+9^y)\mid 16(81+y^2).\]
We can easily verify that $y=1$ and $y=2$ fail by size reasons, noting $343-9^y<16$ and $343+9^y>81+y^2$ for $y=1,2.$ For $y\geq 3,$ we want to ensure that $81^y-7^6\leq 16y^2+16\cdot 81,$ which fails for all $y\geq 3.$

So the only solution in $(x,y)$ is $(1,1),$ corresponding to $(2,4).$

If $a$ and $b$ don't have the same parity, then $7^a-3^b$ is even and $a^4+b^2$ is odd, absurd. Otherwise, $4\mid 7^a-3^b\mid a^4+b^2$ so $a$ and $b$ are both even. Let $a=2m,b=2n$. Then $(7^m-3^n)(7^m+3^n)\mid 16m^4+4n^2$. Now, if $m$ and $n$ are the same parity, the LHS is divisible by $8$ and otherwise it is still divisible by $8$, so either way $2\mid n$ and thus we can let $n=2k$ so $(7^m-9^k)(7^m+9^k)\mid 16(m^4+k^2)$. Now, we use the crowbar known as "size": it is clear that the LHS's absolute value is at least $2(7^m+9^k)$, so at least one of $8k^2\ge 9^k$ and $8m^4\ge 7^m$ occurs. The former can never occur and the latter can only occur for $m\le 3$. Now, at $m=3$ we get $(343-9^k)(343+9^k)\mid 16(81+k^2)$. But the value of $k$ that minimizes $|343-9^k|$ is $262$, which is bad because $16\cdot81/262<343$. If we have $m=2$, we get $(49-9^k)(49+9^k)\mid 16(16+k^2)$. Similarly to before, note $|49-9^k|$ is minimized at $32$, so since $16\cdot16/32=8<49$, this is bad. So finally $m=1$ and $49-81^k=(7-9^k)(7+9^k)\mid 16(1+k^2)$. Equivalently, $0<81^k-49\mid 16(1+k^2)$. Equality holds exactly at $k=1$. Moreover, $\log(81)\cdot 81>32$ and $\log(81)^2\cdot81>32$, so by differentiation the result follows: the divisibility holds only when $(a,b)=(2,4)$, which can easily be checked to be true.

Quite straightforward.

Note that $2\mid 7^a-3^b\implies 2\mid a^4+b^2$. Hence $a$ and $b$ have the same parity. If $a$ and $b$ are both odd, then $7^a\equiv (-1)^a\equiv -1\equiv (-1)^b\equiv 3^b$ (mod $4$) $\implies 4\mid 7^a-3^b$, which is contradictory to the congruence $7^a+3^b\equiv 2$ (mod $4$). Thus $a$ and $b$ are both even,let $a=2m,b=2n$. The divisibility condition becomes $7^{2m}-3^{2n}\mid 16m^4+4n^2\implies (7^m+3^n)(7^m-3^n)\mid 16m^4+4n^2$. Since $2\mid 7^m-3^n$,we have $7^m+3^n\le 8m^4+2n^2$. Then we can check case by case and get the unique solution $(a,b)=(2,4)$.

This took me way too long (expected to be cuter :(). This solution is essentially the same as m.essein's, but posted for fun.

We have by modulo $4$ reasons, that $a$ and $b$ are even. Taking also modulo $8$, we get that $a=2x$ and $b=4y$.
So essentially, we reduce our problem to the following:
$$(7^x-9^y)(7^x+9^y)\mid 16(x^4+y^2).$$
Thus, $7^x+9^y\leq 8(x^4+y^2)$ as $2\mid 7^x-9^y\implies 2\leq \vert 7^x-9^y \vert$ . Note that $7^x>8x^4$ for all $x>3$ and $9^y> 8y^2$ for all $y>0$. Therefore we must only consider values $x=1,2,3$. Note that if $x\leq 3$, then $y=1$ or $2$. Thus, we only need to consider pairs $(x,y)=(1,1),(1,2),(2,1),(2,2),(3,1),(3,2)$. Only solution is $(x,y)=(1,1)$.
Thus, we obtain our only solution pair $\boxed{(a,b)=(2,4)}$.

Notice that $7^a-3^b$ must be even, as both terms are odd. This implies that $a^4+b^2$ is even, so $a$ and $b$ have the same parity. However, if $a$ and $b$ were both odd, then we would have $$LHS\equiv (-1)^a-(-1)^b \equiv (-1)-(-1)\equiv 0 \pmod 4,$$ as $a$ and $b$ are odd, and $$RHS \equiv 1+1\equiv 2 \pmod 4$$ as an odd perfect square must be $1\pmod 4$, which is a contradiction. Therefore $a$ and $b$ are both even.

This means that if we let $a=2c$ and $b=2d$, then $$49^c-9^d | 16c^4+4d^2.$$ Then, since $49^c-9^d=(7^c+3^d)(7^c-3^d)$, we must also have $$7^c+3^d | 16c^4+4d^2,$$ in particular $7^c+3^d \le 16c^4+4d^2$. We must have $(7^c-16c^4)+(3^d-4d^2)<0$. The minimum of $7^c-16c^4$ is $-1695$ at $c=4$, and $3^d-4d^2>1695$ for $d\ge 7$, so $d<7$. Similarly, the minimum of $3^d-4d^2$ is $-9$ at $d=3$, and $7^c-16c^4>9$ for $c\ge 5$, so $c<5$. Checking the remaining cases, we get that $c=1, d=2$ is the only one that works, which corresponds to $(a,b)=\boxed{(2,4)}$.

Solved with [b]L567[/b] using the Catalan Conjecture (well technically not but yeah) - posting for storage.

Notice that $2 \mid 7^a-3^b \mid a^4+b^2$ meaning that $a \equiv b \pmod{2}$ and therefore $4 \mid 7^a-3^b \mid a^4+b^2$ and consequently $2 \mid a,b$, writing $a = 2x,b=2y$, we have $$(7^x-3^y)(7^x+3^y)=7^{2x}-3^{2y} \mid 4(4x^4+y^2)$$
Notice that $\mid 7^x-3^y \mid \geq 2$ (by the Catalan Conjecture) therefore $$7^x+3^y \leq 8x^4+2y^2$$
Notice that $7^x > 8x^4$ for all $x \geq 4$ and $3^y > 2y^2$ for all $y \geq 3$. \\
Checking all the cases for $x \in \{1,2,3\}$ and $y \in \{1,2\}$ gives us the only pair $(a,b) = (2,4)$. \\

Solution from [i]Twitch Solves ISL[/i]: The only answer is $(a,b) = (2,4)$, which works as $7^2 - 3^4 = -32$ divides $2^4 + 4^2 = 32$.
To show this is the only solution, we note the following: [list] 	[*]The left-hand side is always even, which implies $a$ and $b$ 	have the same parity. 	[*]This in turn implies the left-hand side is $0 \bmod 4$. 	[*]Hence $a^4+b^2 \equiv 0 \pmod 4$ which means $a$ and $b$ are even. 	[*]This means that $7^a -3^b \equiv 0 \pmod 8$, 	which further means $b \equiv 0 \pmod 4$. [/list]
This means we may let $a = 2x$ and $b = 2y$ with $y$ even to obtain \[ \left( 7^x - 3^y \right)\left( 7^x + 3^y \right) \mid (2x)^4 + (2y)^2. \]
To reduce to finite casework we contend:
[b][color=red]Claim:[/color][/b]  	$2 \cdot 3^a > 16a^4 + 4a^2$ once $a \ge 10$.
[i]Proof.[/i]  	$3^{10} = 243^2 = 2 \cdot59049 = 177147 > 160400$. $\blacksquare$
Hence we are done if $\max(x,y) \ge 10$. Next
[b][color=red]Claim:[/color][/b]  	$7^x > 16x^4 + 4 \cdot 10^2$ for $x \ge 6$
[i]Proof.[/i]  	$7^6 = 343^2 > 100000 > 16 \cdot 6^4 + 400$. $\blacksquare$
So we may assume $x \le 5$ as well.
This gives us the following cases. In all situations, note that $(2 \cdot 5)^4 + (2 \cdot 10)^2 = 10400$ and consequently if $(7^x-3^y)(7^x+3^y) > 10400$ there is nothing to verify. These cases are marked as ``big''  in the table. \[ \begin{array}{rr|ccccc} 	&& 7 & 49 & 343 & 2401 & 16807 \\ 	&& x=1 & x=2 & x=3 & x=4 & x=5 \\ \hline 	9 &y=2 & \text{OK} & 40 \cdot 58 \nmid 272 & \text{big} & \text{big} & \text{big} \\ 	81 &y=4 & 74 \cdot 88 \nmid 48 & 32 \cdot 130 \nmid 320 & \text{big} & \text{big} & \text{big} \\ 	729 &y=6 & \text{big} & \text{big} & \text{big} & \text{big} & \text{big} \\ 	6561 &y=8 &  \text{big} & \text{big} & \text{big} & \text{big} & \text{big} \\ \end{array} \]

LHS is even, so RHS is also even, so $a$ and $b$ have same parity. This implies $7^a-3^b\equiv 0\pmod 4$, so $a$ and $b$ are even, which implies $7^a-3^b\equiv 0\pmod 8$, so $b\equiv 0\pmod 4$.

Set $a=2x$ and $b=2y$ where $y$ is even.

So $(7^x-3^y)(7^x+3^y)\mid 16x^4+4y^2$.

We note that if $y>3$, then $3^y>4y^2$. If $x>4$, then $7^x> 16x^4$. So if $x>4$ and $y>3$, then $(7^x+3^y)> 16x^4+4y^2$, a contradiction.

Case 1: $x\le 4$.
If $x=1$, then $3^y+7\le 16+4y^2$. So $y<4$. Now $3^3+7=34\nmid (16+4\cdot 3^2)=52$. Thus, $y<3$. If $y=1$ or $y=2$, then $2(3^y+7)=16+4y^2$, which implies $7-3^y\mid 2$, so $y=2$.  Indeed, $\boxed{(a,b)=(2,4)}$ is valid.

If $x=2$, then $3^y+49\le 256+4y^2$. So $y<6$. If $y=5$, then LHS is $292$ and RHS is $356$, contradiction. If $y=4$, then LHS is $130$ and RHS is $320$, contradiction. Oops I forgot $y$ is even. Much less casework. If $y=2$, then we get $58\mid 272$, contradiction.

If $x=3$, then $3^y+343\le 1296+4y^2$. So $y<7$. If $y=6$, then $1072\mid 1440$, contradiction. If $y=4$, then $424\mid 1360$, contradiction. If $y=2$, then $352\mid 1312$, contradiction.

If $x=4$, then $2401+3^y\le 4096+4y^2$. So $y<7$. If $y=6$, then $3130\mid 4240$, contradiction. If $y=4$, then $2482\mid 4160$, contradiction. If $y=2$, then $2410\mid 4112$, contradiction.

Case 2: $x>4$ and $y=2$.
So we have $7^x+9\mid 16x^4+16$. So $x<5$, a contradiction.

We have exhausted all cases, so we are done.

[hide=Answer]$(a,b)=(2,4)$[/hide]
[hide=Solution][quote=ISL 2007 N1]Find all pairs of natural numbers $ (a, b)$ such that $ 7^a \minus{} 3^b$ divides $ a^4 \plus{} b^2$.

[i]Author: Stephan Wagner, Austria[/i][/quote]
$7^a-3^b$ is even,so $a^4+b^2$ is also even,so the parity of $a$ and $b$ must be same.
[b]Case 1:[/b]$(a,b)$ are both odd.
Note that $7^a-3^b \equiv 7-3 \equiv 4 \equiv 0 \pmod 4$,but $a^4+b^2\equiv 2 \not\equiv 0 \pmod 4 \implies $ Contradiction!
[b]Case 2:[/b]$(a,b)$ are both even.
Set $a=2k,b=2l$ to get $7^k+3^l|4(4k^4+l^2)$ and $7^k-3^l|4(4k^4+l^2)$
Now note that $7^k > 16k^4$ for all $k\ge 5$ and $3^l \ge 4l^2$ for all $l \ge 4$,this implies $7^k+3^l \ge a^4+b^2$.So we have a nice bound on $k$ and $l$.Next we bash every possible case in this bound.The only possible solutions are $(k,l)=(1,1)=(1,2) \implies (a,b)=(2,2),(2,4)$.
Check that $(a,b)=(2,2)$ doesn't work,so the only possible solution is $(a,b)=(2,4)\text{  } \blacksquare$[/hide]

Note that $7^a-3^b$ is even, so $a^4+b^2$ is even. Thus, $a^4,b^2$ have same parity which means $a,b$ have same parity.
\\$~$\\
If $a,b$ odd then $7^a-3^b$ is divisible by $4$ but $a^4,b^2\equiv 1\pmod 4$ so $a^4+b^2\equiv 2\pmod 4.$
\\$~$\\
Thus, $a,b$ are even. Let $a=2a_1,b=2b_1$ to get $49^{a_1}-9^{b_1}\equiv 0 \pmod 8$ which implies $16a_1^4+4b_1^2\equiv 0 \pmod 8$ so $b_1$ is even. Thus, let $b=4b_2.$
\\$~$\\
We have $49^{a_1}-81^{b_2}=(7^{a_1}-9^{b_2})(7^{a_1}+9^{b_2}).$ Note that the first factor is a difference of two odd but distinct integers, so it's at least two. Thus, $7^{a_1}+9^{b_2}\le 8(a_1^2+b_2^2)$
\\$~$\\
Easy bounding gives $9^k>8k^2$ and so $7^{a_1}\le 8a_1^2$ implying that $a_1=1,2$ or $3.$ Then, more bounding gives the solution $(1,1).$ Thus, $(a,b)=(2,4)$ is the only solution

Solved with [hide=hints]Missed difference of squares even though I got $2\mid a,b$. :blush:[/hide].

Clearly $a$ and $b$ have the same parity. Then, suppose that $a,b\equiv 1\pmod 2$. Then we have that $a^4+b^2\equiv 2\pmod 4$ while $7^a-3^b\equiv 0\pmod 4$, a contradiction. Then we may set $a=2x$ and $b=2y$ to obtain that
\[7^{2x}-3^{2y}\mid 16x^4+4y^2\implies 7^x+3^y\mid 16x^4+4y^2\implies 7^x+3^y\le 16x^4+4y^2.\]
Then by bounding we obtain that $x\le 4$ and $y\le 6$. Finally, plugging into our original equation gives that only $(a,b)=\boxed{(2,4)}$ works (we omit the calculations).

Obviously $a,b$ are the same parity. If they are both odd then $a^4+b^2 \equiv 2 \pmod 4$ and $7^a-3^b \equiv 0 \pmod 4$, a contradiction. Thus $a=2x$ and then $b=2y$. Thus $$7^{2x}-3^{2y}|16x^4+4y^2 \implies (7^x+3^y)(7^x-3^y)|16x^4+4y^2 \implies |(7^x-3^y)|(7^x+3^y) \le 16x^4+4y^2 \implies 7^x+3^y \le 8x^4+2y^2.$$ Next note that $3^y \ge 2y^2+1$ for all $y \ge 1$, so we have that $7^x+1 \le 8x^4$. This inequality gives $x=1,2,3$. If $x=1$ then we have that $3^y \le 2y^2+1$ or $y=1,2$, and we get the solution $(x,y)=(1,2)$ so $(a,b)=(2,4)$. If $x=2$, then $3^y \le 79+2y^2$, so $y=1,2,3,4$ upon checking gives no solution. If $x=3$, then $3^y \le 305+2y^2$ which gives $y=1,2,3,4,5$ which upon checking gives no solutions, so the sole solution is $(a,b)=(2,4)$.

Missed the obvious bounding at the end :c (oops this looks like asdf's sol)

The key claim is that

[color=blue][b]Claim: [/b][/color]$2|a$ and $4|b$.
[i]Proof: [/i] Since $7^a - 3^b$ is even, $a$ and $b$ must have the same parity.  Thus $7^a - 3^b \equiv 0 \pmod{4}$ and so $2|a, b$.  Then $7^a-3^b \equiv 1 - 1 = 0 \pmod{8}$ and so $4|b$.

Let $a = 2c$ and $b = 4d$.  We can rewrite the original equation as $$7^{2c}-3^{4d} \mid (2c)^4 + (4d)^2,$$ or, $$(7^c+3^{2d})(7^c-3^{2d}) \mid 16(c^4+d^2) \implies (7^c+3^{2d}) \leq 16(c^4+d^2).$$
Bound.  This is left as an exercise to the reader.  The answer is $(a, b) = \boxed{(2, 4)}$. $\square$

ISL Marabot solve

Clearly, $7^a-3^b$ is even. So, parity of $a,b$ must be same. Therefore, $4|7^a-3^b$. Which gives, $a,b$ are even.
Let, $a= 2x, b = 2y$. So, $(7^x+3^y)(7^x-3^y)|16x^4+4y^2$
Here, $7^x- 3^y$ is even and not $0$.
So, $(7^x+3^y)|8x^4+2y^2$
$\implies 7^x+3^y \leq 8x^4+2y^2$
But, $3^y > 2y^2 \ \forall y\in \mathbb N$. So, $7^x > 8x^4$ which gives $x\leq 3$.
And, for each $x$ we get $y < 3$. And by checking all of the pairs we can see that only $(x,y) = (1,2)$ works. Which gives the only solution $(a,b) = (2,4).  \ \ \ \blacksquare$

We have that $7^a-3^b$ is even so we need the parities of $a,b$ to be the same. So we have that $4|7^a-3^b$ which implies that $a,b$ are even. We now let $a=2x$ and $b=2y$ so we have that $7^2x-3^2y=(7^x+3^y)(7^x-3^y)|16x^4+4y^2$  this implies that $7^x>8x^4$ else there is a contradiction so we have that $x \leq 3$ and $y<3$ so the only pair is $(x,y)=(1,2)$ which means that $(a,b)=(2,4)$ QED

Take mod $2$
$7^a-3^b \equiv 0 \pmod 2$ so $a^4+b^2 \equiv 0 \pmod 2$
$a+b \equiv 0 \pmod 2$
$a \equiv b \pmod 2$
Take mod $4$
$(-1)^a-(-1)^b \equiv 0 \pmod 4$ if $a$ and $b$ have the same parity, so $a^4+b^2 \equiv 0 \pmod 4$ and $a$ and $b$ are all even
Let $a=2c, b=2d$
$49^c-9^d$ divides $16c^4+4d^2$
Take mod $8$
$49^c-9^d \equiv 0 \pmod 8$
So $16c^4+4d^2 \equiv 0 \pmod 8$
$4d^2 \equiv 0 \pmod 8$
$d^2 \equiv 0 \pmod 2$
$d \equiv 0 \pmod 2$
Let $d=2e$
$49^c-81^e$ divides $16c^4+16e^2$
$7^{2c}-9^{2e}$ divides $16c^4+16e^2$
$(7^c-9^e)(7^c+9^e)$ divides $16c^4+16e^2$
$|7^c-9^e|(7^c+9^e) \le 16c^4+16e^2$
$7^c+9^e \le 16c^4+16e^2$
For $e \ge 2$, $9^e > 16e^2$ because $e=2$ satisfies it and multiplying by $9$ is more than multiplying by $\frac{(e+1)^2}{e^2}$ so it will always satisfy
For $c \ge 5$, $7^c > 16c^4$ because $c=5$ satisfies it and multiplying by $7$ is more than multiplying by $\frac{(c+1)^4}{c^4}$ so it will always satisfy
So for $e \ge 2, c \ge 5$, $7^c+9^e > 16c^4+16e^2$
So to satisfy $7^c+9^e \le 16c^4+16e^2$, $e=1$ and $c=1,2,3,4$
Plug in all possible values of $c$ and $e$ to find that only $c=1, e=1$ works and results in $\boxed{a,b=(2,4)}$

The answer is $(a,b)=\boxed{(2,4)}$ only, which clearly works.

Note that the LHS is a multiple of $4$, so the RHS must also be a multiple of $4$, so $a$ and $b$ are even. However, this makes the LHS a multiple of $8$, so $b$ is a multiple of $4$. Let $a=2m$ and $b=4n$. Then
\[(7^m-9^n)(7^m+9^n)\mid 16(m^4+n^2).\]
Since the first factor is even, we have
\[7^m+9^n\mid 8(m^4+n^2)\rightarrow (7^m-8m^4)+(9^n-8n^2)\le 0.\]

Let $f(m)=7^m-8m^4$ and $g(n)=9^n-8n^2$. Then we can see that $f(1)=-1$, $f(2)=-79$, $f(3)=-305$, $f(4)=353$, and $f(n)>353$ for all positive integers $n\ge 5$.
We can also see that $g(1)=1$, $g(2)=49$, $g(3)=657$, and $g(n)>657$ for all positive integers $n\ge 4$.
Therefore, we have to check $(m,n)=(1,1),(2,1),(2,2),(3,1),(3,2)$.
If $(m,n)=(2,1)$, then $58\mid 8(17)$ which is not true.
If $(m,n)=(2,2)$, then $130\mid 8(20)$ which is not true.
If $(m,n)=(3,1)$, then $352\mid 8(82)$ which is not true.
If $(m,n)=(3,2)$, then $424\mid 8(85)$ which is not true.
If $(m,n)=(1,1)$, then $(a,b)=(2,4)$. QED.

This one's a few minute headsolve.

Note that the LHS is divisible by 2, so a,b have same parity; in particular, if a,b odd then LHS is 0 mod 4, while RHS is 2 mod 4, contradiction, so we deduce a=2c,b=2d. Now the problem is evident because $7^c+3^d\mid 7^{2c}-3^{2d}\mid 16c^4+4d^2$, which it's obvious LHS increases faster so by obvious bounding we can manually check solutions that's only $2,4$.

As $7^a-3^b$ is even, $a, b$ are of the same parity; on the other hand, this implies $8 \mid 7^a - 3^b$, which forces $2 \mid a$ and $4 \mid b$ for mod $8$ reasons.

On the other hand we must now have $7^{a/2} + 3^{b/2} \leq a^4+b^2$. By comparing the $a$ and $b$-terms, it suffices to check $(a, b) = (2, 4), (4, 4), (6, 4), (8, 4)$, of which only $(2, 4)$ works.


$ u_{n \plus{} 1} \equal{} 2u_{n} \plus{} 3 \plus{} n^2 \plus{} 3.2^n$

$ \Longleftrightarrow \frac {u_{n \plus{} 1} \plus{} (n \plus{} 1)^2\plus{}2(n \plus{} 1)\plus{}6}{2^{n \plus{} 1}} \minus{} \frac {3}{2}(n \plus{} 1) \equal{} \frac {u_n \plus{} n^2\plus{}2n\plus{}6}{2^n} \minus{} \frac {3}{2}n$

$ \equal{} \cdots \equal{} \frac {u_1 \plus{} 1^2\plus{}2\cdot 1\plus{}6}{2} \minus{} \frac {3}{2}\cdot 1 \equal{} \frac{7}{2}$.

$ \therefore u_n \equal{} (3n \plus{} 7)\cdot 2^{n \minus{} 1} \minus{} (n^2\plus{}2n\plus{}6)\ (n\geq 1)$.

[color=red]Last Edited[/color]

[quote="kunny"]$ u_{n \plus{} 1} \equal{} 2u_{n} \plus{} 3 \plus{} n^2 \plus{} 3.2^n$

$ \Longleftrightarrow \frac {u_{n \plus{} 1} \plus{} (n \plus{} 3)(n \plus{} 4)}{2^{n \plus{} 1}} \minus{} \frac {3}{2}(n \plus{} 1) \equal{} \frac {u_n \plus{} (n \plus{} 2)(n \plus{} 3)}{2^n} \minus{} \frac {3}{2}n$

$ \equal{} \cdots \equal{} \frac {u_1 \plus{} (1 \plus{} 2)(1 \plus{} 3)}{2} \minus{} \frac {3}{2}\cdot 1 \equal{} 5$.

$ \therefore u_n \equal{} (3n \plus{} 10)\cdot 2^{n \minus{} 1} \minus{} (n \plus{} 2)(n \plus{} 3)\ (n\geq 1)$.[/quote]

Thank you for your intelitence solution but $ u_n\equal{}2^{n\minus{}1}.(3n\plus{}7)\minus{}(n^2\plus{}2n\plus{}6)$  :wink:

I think kunny's original solution works for 3n instead of 3 as the second term on the right-hand-side of the recurrence relation.

^^ I use "Phương trình sai phân tuyến tính"!!! (Please translate to English^^. I cann't do it! :( )


i have found that when i do the famous ineq3 of manlio
but i don't sure can anyone prove that, or show that this is not right
let a,b,c>0
prove that
[/tex]8abc(a+b+c)+7(a^4+b^4+c^4)+6(a^2b^2+b^2c^2+c^2+a^2)>=
16(a^3b+b^3c+c^3a)+4(a^3c+c^3b+b^3a)[tex/][/tex]

[quote="darij grinberg"][quote="Master"]let a,b,c>0
prove that
[/tex]8abc(a+b+c)+7(a^4+b^4+c^4)+6(a^2b^2+b^2c^2+c^2+a^2)>=
16(a^3b+b^3c+c^3a)+4(a^3c+c^3b+b^3a)[tex/][/tex][/quote]

I guess that 6(a^2b^2+b^2c^2+c^2+a^2) should be 6(a^2b^2+b^2c^2+c^2a^2) instead. If this is the case, your inequality doesn't hold generally; a counterexample is a = 4, b = 3, c = 1, for instance.  :(

  Darij[/quote]
Thanks a lot  darij grinberg
But i am very interest in this ineq
Can you make any condition to repair it ;)

hmm... I think it will be hard to find strong relations between a cyclic non-symmetric side and a symmetric side...

I think the first thing you need to do is making 4 and 16 the same number in [quote]16(a^3b+b^3c+c^3a)+4(a^3c+c^3b+b^3a)[/quote]


induction
combinatorics proposed
combinatorics

Given two integers $ m,n$ satisfying $ 4 < m < n.$ Let $ A_{1}A_{2}\cdots A_{2n \plus{} 1}$ be a regular $ 2n\plus{}1$ polygon. Denote by $ P$ the set of its vertices. Find the number of convex $ m$ polygon whose vertices belongs to $ P$ and exactly has two acute angles.

Related questions are http://www.artofproblemsolving.com/Forum/viewtopic.php?t=89764 and a continuous version http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64437  (I searched for "acute convex.")

Notice that if a regular $ m$ polygon has exactly two acute angles, they must be at consecutive vertices: for otherwise there would be two disjoint pairs of sides that take up more than half of the circle each. Now assume that the last vertex, clockwise, of these four vertices that make up two acute angles is f\mbox ixed; this reduces the total number of regular $ m$ polygons $ 2n \plus{} 1$ times and we will later multiply by this factor.

Suppose the larger arc that the first and the last of these four vertices make contains $ k$ points, and the other arc contains $ 2n \minus{} 1 \minus{} k$ points. For each $ k$, the vertices of the $ m$ polygon on the smaller arc may be arranged in $ {2n \minus{} 1 \minus{} k \choose m \minus{} 4}$ ways, and the two vertices on the larger arc may be arranged in $ (k \minus{} n \minus{} 1)^2$ ways (so that the two angles cut off more than half of the circle).

The total number of polygons given by $ k$ is thus $ (k \minus{} n \minus{} 1) ^ 2 \times {2n \minus{} 1 \minus{} k \choose m \minus{} 4}$. Summation over all $ k$ and change of variable gives that the total number of polygons (divided by a factor of $ 2n \plus{} 1$) is
\[ \sum_{k \geq 0} k^2 {n \minus{} k \minus{} 2 \choose m \minus{} 4}\]
This can be proven to be exactly $ {n \choose m \minus{} 1} \plus{} {n \minus{} 1 \choose m \minus{} 1}$ by double induction on $ n > m$ and $ m > 4$. The base cases $ n \equal{} m \plus{} 1$ and $ m \equal{} 5$ are readily calculated. The induction step is
\[ \sum_{k \geq 0} k^2 {n \minus{} k \minus{} 2 \choose m \minus{} 4} \equal{} \sum_{k \geq 0} k^2 {(n \minus{} 1) \minus{} k \minus{} 2 \choose m \minus{} 4} \plus{} \sum_{k \geq 0} k^2 {(n \minus{} 1) \minus{} k \minus{} 2 \choose (m \minus{} 1) \minus{} 4} \] \[ \null \quad \equal{} {n \minus{} 1 \choose m \minus{} 1} \plus{} {n \minus{} 2 \choose m \minus{} 1} \plus{} {n \minus{} 1 \choose m \minus{} 2} \plus{} {n \minus{} 2 \choose m \minus{} 2} \] \[ \null \quad \equal{} {n \choose m \minus{} 1} \plus{} {n \minus{} 1 \choose m \minus{} 1}
\]
So the total number of $ m$ polygons is \[ (2n \plus{} 1) \times \left [{n \choose m \minus{} 1} \plus{} {n \minus{} 1 \choose m \minus{} 1}\right]\]

I think above answer is not clearly correct.
Because when n=6, m=5  the answer is 650
but above answer is 260
can it correct??

Yes, there is a small mistake here:
[quote="bodan"] . . . and the two vertices on the larger arc may be arranged in $ (k \minus{} n \minus{} 1)^2$ ways (so that the two angles cut off more than half of the circle) . . .
[/quote]
Instead of $ (k \minus{} n \minus{} 1)^2$ it should be $ (k \minus{} n)^2$. Then the sum is \[ \sum_{k \geq 0} k^2 {n \minus{} k \minus{} 1 \choose m \minus{} 4} \equal{} (2n\plus{}1) \left[{n \choose m \minus{} 1} \plus{} {n \plus{} 1 \choose m \minus{} 1}\right]\]
Then for $ n\equal{}6$ and $ m\equal{}5$ we get $ 650$.

Just for completeness, here is the correct version:

[hide="Correct version"]
Notice that if a regular $ m$ polygon has exactly two acute angles, they must be at consecutive vertices: for otherwise there would be two disjoint pairs of sides that take up more than half of the circle each. Now assume that the last vertex, clockwise, of these four vertices that make up two acute angles is fixed; this reduces the total number of regular $ m$ polygons $ 2n \plus{} 1$ times and we will later multiply by this factor. Suppose the larger arc that the first and the last of these four vertices make contains $ k$ points, and the other arc contains $ 2n \minus{} 1 \minus{} k$ points. For each $ k$, the vertices of the $ m$ polygon on the smaller arc may be arranged in $ {2n \minus{} 1 \minus{} k \choose m \minus{} 4}$ ways, and the two vertices on the larger arc may be arranged in $ (k \minus{} n)^2$ ways (so that the two angles cut off more than half of the circle).

The total number of polygons given by $ k$ is thus $ (k \minus{} n) ^ 2 \times {2n \minus{} 1 \minus{} k \choose m \minus{} 4}$. Summation over all $ k$ and change of variable gives that the total number of polygons (divided by a factor of $ 2n \plus{} 1$) is
\[ \sum_{k \geq 0} k^2 {n \minus{} k \minus{} 1 \choose m \minus{} 4}
\]
This can be proven to be exactly $ {n \choose m \minus{} 1} \plus{} {n \plus{} 1 \choose m \minus{} 1}$ by double induction on $ n > m$ and $ m > 4$. The base cases $ n \equal{} m \plus{} 1$ and $ m \equal{} 5$ are readily calculated. The induction step is
\[ \sum_{k \geq 0} k^2 {n \minus{} k \minus{} 1 \choose m \minus{} 4} \equal{} \sum_{k \geq 0} k^2 {(n \minus{} 1) \minus{} k \minus{} 1 \choose m \minus{} 4} \plus{} \sum_{k \geq 0} k^2 {(n \minus{} 1) \minus{} k \minus{} 1 \choose (m \minus{} 1) \minus{} 4}
\]
\[ \null \quad \equal{} {n \minus{} 1 \choose m \minus{} 1} \plus{} {n \choose m \minus{} 1} \plus{} {n \minus{} 1 \choose m \minus{} 2} \plus{} {n \choose m \minus{} 2} \equal{} {n \choose m \minus{} 1} \plus{} {n \plus{} 1 \choose m \minus{} 1}
\]
So the total number of $ m$ polygons is
\[ (2n \plus{} 1) \times \left [{n \choose m \minus{} 1} \plus{} {n \plus{} 1 \choose m \minus{} 1}\right]
\][/hide]

that's correct thank you
your solution is so simple and great!!


modular arithmetic
Divisibility
number theory
IMO Shortlist
generating functions
roots of unity
complex numbers

Let $m,n\geq2$ be positive integers, and let $a_1,a_2,\ldots ,a_n$ be integers, none of which is a multiple of $m^{n-1}$.  Show that there exist integers $e_1,e_2,\ldots,e_n$, not all zero, with $\left|{\,e}_i\,\right|<m$ for all $i$, such that $e_1a_1+e_2a_2+\,\ldots\,+e_na_n$ is a multiple of $m^n$.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

Consider each of the $ m^n$ integers $ \sum e_ia_i$ with each of the $ e_i$ varying in $ \{0,1,\ldots,m\minus{}1\}$. If any two have the same residue $ \pmod{m^n}$ then we take their difference and we are done. Otherwise, they form a complete residue class $ \pmod{m^n}$ (1) (with the all-zeroes case corresponding to $ 0\pmod{m^n}$). We show this is a contradiction.

Let $ b_i$ be the number of these numbers $ i \pmod{m^n}$ Take $ \epsilon\equal{}e^{2i\pi/m^n}$ and consider $ \sum b_i\epsilon^i\equal{}\sum_{e_i\in\{0,1,\ldots,m\minus{}1\}} \epsilon^{e_1a_1\plus{}\ldots\plus{}e_{m^n}a_{m^n}}\equal{}\prod \frac{\epsilon^{ma_i}\minus{}1}{\epsilon^{a_i}\minus{}1}\equal{}0$ by assumption (1) (note that the denominators cannot be zero by the problem condition). But then $ \epsilon^{ma_i}\equal{}1$ for some $ i$, and hence $ m^{n\minus{}1}|a_i$; contradiction.

Everywhere (Compendium, PfTB) there's solution using root of unity, but is there any solution using [hide = this] Let $R_k$ be the set of all $n$-tuples $\{(x_1, \cdots, x_n) | x_i \in \mathbb{Z},  \sum x_i = km^n\}$ and consider the set $S = \{ R_k : k \in \mathbb{Z} \}$. Interpret $a_i$ as basis vectors in $n$ dimensional space (i.e a $n$-tuple $(0,\cdots,0,a_i,0,\cdots,0)$), and let $J = \sum e_i \vec{a_i}$. Now the condition is equivalent to $J \in S$  [/hide] interpretation ?

Similar problem was appeared in Romanian TST
https://artofproblemsolving.com/community/c6h53557p335760

I really enjoyed this problem! For simplicity, let $[a,b]$ be the set of integers $a\leq k\leq b.$

For each vector $v=(x_1,x_2,...,x_n)\in\mathbb{R}^n$ define $f(v):=x_1\cdot a_1+x_2\cdot a_2+...+x_n\cdot a_n.$ In what follows, we will only work in $\mathbb{F}_{m^n}.$

Begin by assuming that no non-zero vector $v\in[-(m-1),m-1]^n$ satisfies $f(v)=0.$ Let $u,v\in [0,m-1]^n$ be two distinct, non-zero vectors. If $f(v)=f(u)$ then, by the linearity of $f,$ we get that for $w=u-v\in [-(m-1),m-1]^n,$ $f(w)=0.$ This is a contradiction to our assumption. Therefore, $\{0\}\cup\{f(v): v\neq 0, \ v\in [0,m-1]^n\}$ is a complete residue set mod $m^n.$

However, since $f(0)=0$ we can more simply say that $\{f(v): v\in[0,m-1]^n\}$ is a complete residue set modulo $m^n.$ Thus \[\prod_{i=1}^n\frac{X^{a_i\cdot m}-1}{X^{a_i}-1}=\prod_{i=1}^n\big(X^{a_i\cdot 0}+X^{a_i\cdot 1}+...+X^{a_i\cdot (m-1)}\big)=\sum_{v\in [0,m-1]^n}X^{f(v)}=\sum_{i=0}^{m^n-1}X^i=\frac{X^{m^n}-1}{X-1}.\]By letting $X$ be the $m^n$th root of unity and using the fact that $m^{n-1}\nmid a_i,$ we get a contradiction.

Solved with a hint
It suffices to show that there exist two choices of $(b_1,\ldots,b_n)$ with $0 \leq b_i<m$ for all $i$ such that $b_1a_1+\cdots+b_na_n$ collide modulo $m$, whence we can take $e_i$ to be the termwise difference between these two sequences. If no collisions occur, because there are $m^n$ choices of a sequence, $b_1a_1+\cdots+b_na_n$ must cover every residue modulo $m^n$ exactly once. Thus if we let $z=e^{2\pi i/m^n}$ be a primitive $m^n$-th root of unity, we should have
$$0=\sum_{(b_1,\ldots,b_n)} z^{b_1a_1+\cdots+b_na_n}=\prod_{i=1}^n \sum_{j=0}^{m-1} z^{ja_i}=\prod_{i=1} \frac{z^{ma_i}-1}{z^{a_i}-1}.$$
Thus some $i$ must exist with $m^n \mid ma_i \implies m^{n-1} \mid a_i$, which is illegal. Thus a collision must occur, so we are done. $\blacksquare$

interesting
[hide = solution]
Consider the $m^n$ tuples $(e_1, e_2, \dots, e_n)$ where $0 \leq e_i < m$. If two of these are identical modulo $m^n$, we're done so assume that the sums $\sum e_ia_i$ forma a complete residue class modulo $m^n$. Let $\zeta$ be a primitive $m^n$th root of unity. It follows that: \[0 = \sum_{0 \leq i < m^n} \zeta^i = \sum_{\text{tuple} e_i} \zeta^{\sum_{e_i a_i}} = \prod_{i = 1}^n \frac{\zeta^{ma_i}-1}{\zeta-1}\]
It follows that $\zeta^{ma_i}-1 = 0$ for some $i$, implying $m^{n-1} \mid a_i$, contradicting the problem's conditions.
[/hide]


geometry
3D geometry
calculus
integration

For the equation \[ (3x^2+y^2-4y-17)^3-(2x^2+2y^2-4y-6)^3=(x^2-y^2-11)^3, \]
determine its solutions $(x, y)$ where both $x$ and $y$ are integers. Prove that your answer lists all the integer solutions.

More thorough solution
[hide="Solution"]We can say that $ (3x^2 \plus{} y^2 \minus{} 4y \minus{} 17) \equal{} a$, $ (2x^2 \plus{} 2y^2 \minus{} 4y \minus{} 6)^3 \equal{} b$, and $ (x^2 \minus{} y^2 \minus{} 11)^3 \equal{} c$. Then $ a^3 \minus{} b^3 \equal{} c^3$ and $ a \minus{} b \equal{} c$. $ a^3 \minus{} b^3 \equal{} (a \minus{} b)(a^2 \plus{} ab \plus{} b^2)$ so we wand to find $ a^2 \plus{} b^2$. Since $ a \minus{} b \equal{} c$, $ a^2 \plus{} b^2 \minus{} 2ab \equal{} c^2$ and $ a^2 \plus{} b^2 \equal{} c^2 \plus{} 2ab$. Now we can manipulate our product of cubes so that $ c^3 \equal{} (a \minus{} b)(c^2 \plus{} 3ab)$ and since $ a \minus{} b \equal{} c$, we can manipulate it even further: $ c^3 \equal{} c(c^2 \plus{} 3ab)$.
$ c^3 \equal{} c^3 \plus{} 3abc$
$ abc \equal{} 0$
Now we have different cases to check.  Either $ a \equal{} 0$, $ b \equal{} 0$ or $ c \equal{} 0 \equal{} > a \equal{} b$

[b]Case 1:[/b] $ a \equal{} 0$
$ 3x^2 \plus{} y^2 \minus{} 4y \minus{} 17 \equal{} 0$
$ 3x^2 \plus{} (y \minus{} 2)^2 \equal{} 21$
We know that since $ 3|21$ then $ (y \minus{} 2)^2|3$. So the possibilities of $ y$ are $ 5,2, \minus{} 1$. The middle number doesnt yield an integral value of $ x$, but the other values yield the solutions $ (2,5),$ $ ( \minus{} 2,5),$ $ (2, \minus{} 1),$ $ ( \minus{} 2, \minus{} 1)$.

[b]Case 2:[/b] $ b \equal{} 0$
$ 2x^2 \plus{} 2y^2 \minus{} 4y \minus{} 6 \equal{} 0$
$ x^2 \plus{} y^2 \minus{} 2y \minus{} 3 \equal{} 0$
$ x^2 \plus{} (y \minus{} 1)^2 \equal{} 4$
There arent that many cases to check at all, and we quickly see the solution set is $ (0,3)$, $ (0, \minus{} 1)$, $ (2,1)$, $ ( \minus{} 2,1)$

[b]Case 3:[/b]$ c \equal{} 0 \equal{} > a \equal{} b$
$ x^2 \minus{} y^2 \minus{} 11 \equal{} 0$
$ (x \plus{} y)(x \minus{} y) \equal{} 11$
$ (x \plus{} y) \equal{} \frac {11}{x \minus{} y}$
Therefore, $ x \minus{} y \equal{} 11,1, \minus{} 1, \minus{} 11$ for a total of 4 solutions: $ (6,5)$, $ (6, \minus{} 5)$, $ ( \minus{} 6,5)$, $ ( \minus{} 6,5)$.

We see in total there are 12 solutions: $ \boxed{(6,5)}$, $ \boxed{(6, \minus{} 5)}$, $ \boxed{( \minus{} 6,5)}$, $ \boxed{( \minus{} 6,5)}$ $ \boxed{(2,5)}$, $ \boxed{( \minus{} 2,5)},$ $ \boxed{(2, \minus{} 1)},$ $ \boxed{( \minus{} 2, \minus{} 1)}$, $ \boxed{(0,3)}$, $ \boxed{(0, \minus{} 1)}$, $ \boxed{(2,1)}$, $ \boxed{( \minus{} 2,1)}$[/hide]

Sry for reviving

[quote]Fermats Last theorem tells us something![/quote]
how is this?

For any $ n>2$ ($ n$ is an integer), there are no positive integer solutions to $ x^n\plus{}y^n\equal{}z^n$.

Note the requirement that they must be positive. Where did we say that this must be the case here?


function
algebra
polynomial
inequalities
limit
logarithms
complex analysis

For arbitrary $a_1,\ldots,a_n \in \mathbb{C} \setminus \{0\}$, does there necessarily exist a $z \in \mathbb{C}$ such that $1+e^{a_1z} + \cdots + e^{a_nz} = 0$?

Yes, because there is a theorem (due to Hadamard, if I don't mistake) stating that any non-constant entire function $F(z)$ of order $1$ can be represented as $ce^{\beta z}$ times a canonical product $z^m\prod_j \left(1-\frac z{z_j}\right)e^{z/z_j}$ over its zeroes. In your situation, the function is, clearly, not a pure exponent, so the canonical product must be non-trivial. Moreover, in this particular case zeroes are infinitely many and one can say a lot about their location.

The proof of the Hadamard factorization theorem can be found in any decent textbook on entire functions.

Thank you! But, come to think of it, I just found a completely elementary solution that does not neither Hadamard's result nor Rouche's theorem -- but only the fundamental theorem of algebra.

For simplicity, we assume in the sequel that the $a_i$ are reals; the proof easily extends to the general case. Of couse, we may then assume that $a_1 \leq a_2 \leq \cdots \leq a_s < a_{s+1} \leq \cdots \leq a_n = a$, with $a_n > 0$.

 For every $k$, we consider the polynomial
\[
f_k(x) := 1 + x^{[ka_1]} + \cdots + x^{[ka_n]} \in \mathbb{C}[x].
\]
Let $x_k$ be any root of $f_k$, and define $z_k$ by $e^{z_k} = x_k^k$ and $\text{Im} \, z_k \in [0,2\pi)$ (that is, $z_k = k \, \textrm{Log}_0 \, x_k$).

We claim that the sequence $\{ |z_k| \}$ is bounded. Assuming the contrary, for every $M>0$ there is a $k=k(M)$ with $| \Re  (z_k)| > M$ (because the imaginary part of $z_k$ is bounded by our choice of the principal branch of the logarithm). For this $k$, we either have $|e^{z_k}|>e^M$ or $|e^{z_k}| < e^{-M}$, hence either $|x_k|^k>e^{M}$ or $|x_k|^k < e^{-M}$. We see that both cases are impossible when $M$ is large enough. Indeed, for the first case (the second one is treated analogously), $f_k(x_k)=0$ entails
\[
(n-s+1)x_k^{[ak]} = - x_k^{[a_sk]} -  x_k^{[a_{s-1}k]} \cdots - 1
\]
By the triangle inequality, $|x_k|^{[ak]} \leq 1 + |x_k|^{[a_{s-1}k]} + \cdots + |x_k|^{[a_1k]}$, so
\[
|x_k|^{-b_s} + |x_k|^{-b_{s-1}} + \cdots + |x_k|^{-b_1} \geq 1,
\]
where $b_s = [ka]; b_{s-1} = [ka] - [ka_{s-1}], \ldots, b_1 = [ka] - [ka_1]$ are all $> ck$ for some $c>0$ and large enough $k$. But then, $|x_k|^{-b_i} < |x_k|^{-ck}< e^{-cM}$ for all $i$, giving $se^{-cM} > 1$. Clearly, this cannot hold when $M$ is large, a contradiction. Hence, we have the desired boundedness of $\{ |z_k| \}$.

In this manner, we proved that the all points $\{z_k\}_{k \geq 1}$ are contained in a bounded, and hence in a  compact subset of the plane (notice that our choices of $x_k$ as roots of $f_k$ were [i]arbitrary[/i]). By the compactness, the set $\{z_k\}_{k \geq 1}$ has at least one limit point, and let $z$ be such point. In other words, there is a subsequence $z_{k_1},z_{k_2},\ldots$ of $\{z_k\}$ converging to $z$. For this $z$, we have
\[
0 = \lim_{i \to \infty} f_{k_i} (e^{\frac{z}{k_i}}) = \lim_{i \to \infty} \big( 1 + e^{z\frac{[k_ia_1]}{k_i}} + \cdots + e^{z \frac{[k_ia_n]}{k_i}} \big) = 1 + e^{a_1z} + \cdots + a^{a_nz},
\]
as desired.

The general case when $a_i \in \mathbb{C}$ can be treated in exactly the same way.

--Vesselin

Furthermore, in the case when the $a_i$ are reals that we described above, and when $a_n>a_{n-1}$ strictly (that is, the maximal $a_i$ occurs just once, or equivalently, $s=n$), each $f_k$ for $k$-large enough has a root of absolute value $\geq 1$ (because the roots multiply to $\pm 1$); choosing $x_k$ as this root ($|x_k| \geq 1$), we get that $\Re (z_k) \geq 0$, i.e. we bound the $z_k$ in the right half-plane, so we have $\Re (z) \geq 0$. In this manner, we obtain from the above proof the following:

[b]Result. [/b] Let $0 < a_1 < a_2 < \cdots < a_n$ be distinct positive reals, and let $b_0,b_1,\ldots,b_{n-1} \in \mathbb{C}$ be arbitrary complex numbers with $|b_0| \geq 1$. Then, the exponential sum
\[
f(z) := e^{a_nz} + b_{n-1}e^{a_{n-1}z} + \cdots + b_1e^{a_1z} + b_0
\]
has a zero $z$ with $\Re (z) \geq 0$. Moreover, there is a $B>0$ such that $|\Re (z)| < B$ for any zero $z$ of $f$.

In particular, there is a $z$ with $\Re (z) > 0$ satisfying $1 + e^{z \ln{2}} + e^{z \ln{3}} = 0$, providing another solution to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=19406]this[/url] problem. This result also implies the following

[b]Corollary. [/b] Let $a_1,\ldots,a_n, n \geq 2,$ be arbitrary distinct positive reals. Then, there exists a non-trivial continuous function $f: \mathbb{R} \to \mathbb{R}$ satisfying
\[
 f(a_1x) + f(a_2x) + \cdots + f(a_nx) = 0
\]
for all $x$.

--Vesselin

What do you think, does this simple polynomial approximation stand a chance of being correct? :o  It's 3a.m. local time here and I'm very sleepy, so I may have overlooked something ... but it seems clear to me, anyway.

Actually, it is clear that if $f$ is an entire function with no zeros, then $f(z)=e^{g(z)}$, where $g$ is entire. In the case when $f$ has order $1$, $g$ must be a linear function, and it is clear that our exponential sum does not have the form $e^{az+b}$. So, that every non-trivial exponential polynomial (not of the form $ce^{\beta z}$) has a zero over the complex numbers (infinitely many, in fact) can be shown without Hadamard's theorem (or Rouche's theorem).


Suppose there's a single terraced street that contains 19 houses. The postman notices that no 2 adjacent houses ever receive mail on the same day, and that no more than 2 adjacent houses don't receive mail on any day. What is the maximum number of days he can deliver before repeating the pattern?

Any help with this problem would be greatly appreciated!

[quote="sawyerpip"]single terraced street that contains 19 houses.[/quote]  Don't know what that means.  If it means "a row of 19 houses," then the question is asking for the number of subsets of $ \{1, 2, \ldots, 19\}$ that contain no two consecutive elements and whose complement contains no three consecutive elements.  I suggest you begin with a recursive approach: you'll want three sequences, one to count subsets that contain the last element, one to count subsets that don't contain either of the last two elements, and one to count subsets that don't contain the last element but do contain the next-to-last element.


superior algebra
superior algebra unsolved

I am reading the book [i]Rational Points on Elliptic Curves[/i] by Silverman and Tate. On page 42 I have a question about group.

The last paragraph mentions something like this:
Suppose $ G\equal{}\mathbb T^1 \times \mathbb Z_2$, where $ \mathbb T^1\equal{}\{z\in\mathbb C: |z|\equal{}1\}$.
For $ m\in\mathbb N$, let $ G[m]\equal{}\{g\in G: \text{ord }(g)|m\}$. Then $ G[m]$ is also a group.
Moreover, if $ m$ is odd, then $ G[m]\cong \mathbb Z_m$;
otherwise $ G[m]\cong\mathbb Z_m\times\mathbb Z_2$.

My question is: why the forms for $ G[m]$ are different when $ m$ takes values from odd numbers and from even numbers?

This is because of the second factor of $ G$. Note that for odd $ m$ and every $ x\in\mathbb Z_2$, $ mx\equal{}0$ iff $ x\equal{}0$.


Prove that in any triangle $ ABC$ the following inequality holds true:

\[ \frac{2(a^3\plus{}b^3\plus{}c^3)\plus{}3abc}{9abc}\geq\frac{R}{2r}.\]

[quote="Cezar Lupu"]Prove that in any triangle $ ABC$ the following inequality holds true:
\[ \frac {2(a^3 \plus{} b^3 \plus{} c^3) \plus{} 3abc}{9abc}\geq\frac {R}{2r}.\]
[/quote]
It's obviously wrong. Try $ a\equal{}b\equal{}1$ and $ c\rightarrow2.$ :wink:


In triangle $ ABC$ lies a point $ M$. Lines $ AM,BM,CM$ intersect opposite sides $ BC,CA,AB$ at $ A_1,B_1,C_1$ respectively. Let $ K$ be a point on $ A_1C_1$ such that $ B_1K$ is perpendicular to $ A_1C_1$. Prove that $ KB_1$ bisects the angle $ AKC$.

[quote="mr.danh"]A1C1 meets AC at E. We have (ACB1E) is harmonic, KB1 is perp to KE, hence result.[/quote]

That's what I had in mind as well  :)


How many even integers between 4000 and 7000 have four different digits?

[hide]Let's split this into two cases: Where the first digit is even, and where the first digit is odd.

If the first digit is even, we have $2$ possibilities for the first digit, $4$ for the last, and $8\cdot 7$ for the second and third digits, for $2\cdot8\cdot7\cdot4=448$ possible orders.

If the first digit is odd, it must start with $5$. $1$ possibility for the first digit, $5$ for the last digit, and $8\cdot7$ for the second and third digits. In total, there are $1\cdot8\cdot7\cdot5=280$ possibilities for this case.

Therefore, our answer is $280+448=\boxed{728}$ such integers. [/hide]

Our cases can be quickly written as follows:

4_ _0
4_ _2
4_ _6
4_ _8
5_ _0
5_ _2
5_ _4
5_ _6
5_ _8
6_ _0
6_ _2
6_ _4
6_ _8

For each of the cases, there are $8 \cdot 7$ ways to obtain a desired result. Since there are $13$ cases, our end answer is simply $8 \cdot 7 \cdot 13 =\boxed{728}$.

[quote=ilovepizza2020]Is this alcumus?[/quote]

Did Alcumus even exist in 2006?

No. I was just looking through my Combo 1 (MC with proofs) problem sets and this was on it so I looked at the collections and posted my solution :)

I just realized this was bumped 14 years..
[hide=solution]There are $3$ possible cases for the first digit (4,5, or 6), $9$ for the second, etc. If the first digit is even, there are $2 \times 4 \times 8 \times 7 = 448$ possibilities. If it is odd, there are $1 \times 5 \times 8 \times 7 = 280$ possibilities. This gives a total of $448+280 = \boxed{728}$ possibilities. [/hide]

[quote=ilovepizza2020][quote=Possible]Dang, I forgot about the 4000-7000 restriction and I got the problem wrong :([/quote]

How did you get it wrong?[/quote]

I forgot to consider the 4000-7000 restriction


inequalities
LaTeX
number theory proposed
number theory

prove or disprove:

$ 3$ is the only prime of the form $ (\alpha^2+\beta^2)^2-\alpha^2\beta^2$ where $ \alpha,\beta$ - positive int.

$ (\alpha^2\plus{}\beta^2)^2\minus{}\alpha^2\beta^2$

If $ \alpha$ is also restricted to being a positive integer, this expression is a difference of two perfect squares. The only way a difference of two perfect squares can be prime is if the squares are consecutive. In this case, that would mean $ \alpha^2 \plus{} \beta^2 \equal{} \alpha\beta \plus{} 1$.

If $ \alpha > \beta$ then $ \alpha^2 \plus{} \beta^2 < \alpha^2 \plus{} 1$, which would make $ \beta^2 < 1$.
Because $ \beta^2$ is a positive integer, that inequality is impossible.

If $ \beta > \alpha$ then $ \alpha^2 \plus{} \beta^2 < \beta^2 \plus{} 1$, which would make $ \alpha^2 < 1$.
Because $ \alpha^2$ is a positive integer, that inequality is impossible.

If $ \alpha \equal{} \beta$ then $ 2\alpha^2 \equal{} \alpha^2 \plus{} 1$. That is only true when both $ \alpha$ and $ \beta$ are equal to 1 (or trivially -1), which gives the unique result of 3.

If, however, $ \alpha$ is not restricted to being a positive integer...

Set $ \alpha \equal{} \sqrt(2)$ and $ \beta \equal{} 1$. The result gives 7 which is also prime.

(a^2+b^2-ab)(a^2+b^2+ab) is prime and a^2+b^2-ab < a^2+b^2+ab
=> a^2+b^2-ab=1 =>(a-b)^2=(1-ab)
min value of (a-b)^2=0
=>max value of ab=1 => a=1,b=1
which results in that 3 is the only prime integer of the given form.

[quote="praneeth"]$ (a^2\plus{}b^2\minus{}ab)(a^2\plus{}b^2\plus{}ab)$ is prime and $ a^2\plus{}b^2\minus{}ab < a^2\plus{}b^2\plus{}ab$
=> $ a^2\plus{}b^2\minus{}ab\equal{}1 \equal{}>(a\minus{}b)^2\equal{}(1\minus{}ab)$
min value of $ (a\minus{}b)^2\equal{}0$
=>max value of $ ab\equal{}1$=>$ a\equal{}1,b\equal{}1$
which results in that $ 3$is the only prime integer of the given form.[/quote]

how did you know that $ a^2\plus{}b^2\minus{}ab\equal{}1$  !!!

because $ a$ is real number !!

If you review LaTex code , there is written "alpga" instead of "alpha"
So it must be:
[quote]
prove or disprove:
3 is the only prime of the form $ (\alpha^2\plus{}\beta^2)^2\minus{}\alpha^2\beta^2$ where $ \alpha,\beta$ - positive ints.
[/quote]


I'm curious as to what different strategies teams find useful for the team portion of the MATHCOUNTS competition?  Do you split up the problems, or work on them all together, or some combination of the above?

We have a secret, but I'm not giving it away. Our team has won the PA states 2 of the last 3 years and the chapters 5 of the last 6 years, and we're in the states again this year. So what we do works. :D :D :D :D :D :D

our team had our "superstar", two middle-level skilled people, and one weaker person. on team competition, we split into two pairs with our "superstar" with out weaker teammate and our two middle-level people and divided up the questions in half. it worked fairly well.

[quote="mathmom"]Ok, what strategy would you recommend for a weaker team than your own?  Say one with two medium-strong members, and two weaker members.  ;)[/quote]

by experience, i would put one medium-strong member with a weaker member and have the other pair the same. our team did better when we were balanced as opposed to having two different level sub-teams within one team.

Last year, on our team, we simply separated into two subsets of two. We had three stronger members, Nate, Todd and I. And a weaker member, Nick. Nick, however was good at catching stupid mistakes and such, so we paired him with Nate who often made the dumb mistakes. Todd and I just seemed to work really well together and had different thought processes so we could easily combine our efforts to solve the problems. Since Todd and I worked better together, we started from the back where the more difficult problems are, and Nate and Nick started at the front. Once we had answers for most of them, we'd combine answers and debate answers and then work together on teh ones we didn't have answers for.

[i][NOTE: names were changed and added for clarification][/i]

[quote="mathmom"]That was my instinct too.  How did you divide the problems up?  Did you just take 1-5 and 6-10, say, or did you choose them based on areas of strength, or something else?[/quote]

categorizing the problems into areas of strength could be time consuming (don't know, never tried it) but our team just had 1-5 for one subteam and 6-10 for the other like you mentioned.

Here are some ideas:

--Have a sheet near the center of the table with your answers.  Mark your initials next to each answer you verify/answer.  Disputes can be resolved efficiently.  Worked very well last two years at nats.
--2 groups of 2, each works on predesignated 5 together.
--Each person does a predesignated 2 or 3.
--Have your strongest member work on problems from the back and your second strongest member help the remaining two if needed (and work on problems if not).
--Your team is good enough to have everyone go over all the problems 100 times. :D

A good idea that I have found is to have your best 2 kids do odds or evens.  If the other 2 aren't too strong you could take your top kid and do 10,8,6.... and #2 kid do 9,7,5.......

Your lower two kids could do 1,2,3... and 5,4,3.....

I like this because the top 2 kids can get through their half at about the same pace and then fill in any blanks left by the other two.

My kids are all pretty good, so for chapter level I actually do it reverse and have the top 2 start at the top so we can be more sure of getting the first 5(usually easier) problems correct and then all 4 work for a bit on the harder ones.

Good luck.  We might get snowed out of school tomorrow, but chapter is saturday, so snow, shmow!!!

Weakest kid does 1 to 5
Second-weakest kid does 5 to 1
Second-strongest kid does 6 to 10
Strongest kid does 10 to 6

I think that's your secret ;) . Don't worry, our team has a better method...

I abbreviated the names of my team members.

N. ties S. to the chair and staples his mouth shut
J. works on 6 to 10
I work on 10 to 6
N. works on 5 to 1

:)
Lets just keep the stapling to that team only
 :)

Our team had a similer stratergy to all of yours but now we have 2 superstars and 2 medium up people so basicly the 2 superstars to all 10. and the others to any that they know.   :lol:  :lol:  :lol:  :lol:  :lol:  :lol:  :lol:  :lol:  :lol:  :lol:  :lol:
WE took an old state competion and got a medium sprint score, failed the target, didn't ace the team and still came only .5 points behind the winer that year. That shows our closeness in scores :D  :D  :D  :D  :D  :D  :D

[quote="Treething"]Weakest kid does 1 to 5
Second-weakest kid does 5 to 1
Second-strongest kid does 6 to 10
Strongest kid does 10 to 6
[/quote]
That's what we do, except that we don't discriminate between weak and strong. I'm definitely not the strongest on my team, but I do 10 to 6 anyway, because I want to, and our strongest member usually does 1 to 5, because he wants to. But I'm giving an overly orderly impression of my team. The real truth is that we start off with the idea of dividing the labor evenly, but disintegrate into doing whatever we feel like doing regardless of strategic considerations. We got first in our chapter anyway, though, so obviously it works.

the people on this side of the table do 1-5 and the people on the other side of the table do 6-10

here's what we do:
we split off into groups of two: two middle level people together, strongest" and "weakest" together.
the "weakest" is always the person with the official sheet and is responsible for solving easier problems as well as checking to see if answers are reasonable and/or in the right form
the strongest person does the last two or three problems and after that, starts doing the entire tests... (that's my job).
the two middle level people solve problems and check the strongest persons work.
also: ask for extra scratch paper: at least 10 sheets... one per problem... take the corresponding sheet, do the work on it, circle your answer, initial it, and pass it off... therefore, person checking does not have to re-work problem and instead can just read your work.
hope this helps and I hope you are not from Texas... or else i just gave away our secrets.

what works the best is that the people who work together the best should be in a pair.  that way everything will go faster, since your allowed to talk during team... but if someone talks really loud we just assign them problems to check or do blah blah blah

YEA... OUR TEAM'S FIRST IN SAN DIEGO!!!!!!!!!!! goin to state

My team has a very good strategy, IMO:

Weakest does 1, 2, 3
2nd weakest does 4, 5, 6
They check each other.

Strongest does 9, 10
2nd strongest does 7, 8
They check each other.

If the two people agree on the same answer, then it's put aside. If they disagree, a third person checks it and/or the first two people discuss how they each got their answer.

If someone can't solve a problem, they switch with someone else.

It works very well for us.

One might even say this method is  :sqrt: (yuk yuk yuk).

Our team came in first in the West Windsor area. Goin' to states =D !

The entire top four was my school, and the entire top 12 (aside from places 7 and 9) was the West Windsor-Plainsboro school district. Sheer ownage.

[quote="Ars"]Our team had a similer stratergy to all of yours but now we have 2 superstars and 2 medium up people so basicly the 2 superstars to all 10. and the others to any that they know.
WE took an old state competion and got a medium sprint score, failed the target, didn't ace the team and still came only .5 points behind the winer that year. That shows our closeness in scores       [/quote]

Stop [i]lying[/i] Ars!!!  In practice before the chapters, I some of them.  You and dean didn't do all 10. And our strategy isn't even close to any of their's.

2 of our people are cheching and 2 are doing the problems

[quote="AznXtreme418"]My team had 1 super smart guy and three strong people. But I was paired off with this annoying little 7 who slacks off but is good at checking. So you shouldn't only split according to strength. You should let them point their teammates to some extent and only intervene if it's really not working.[/quote]
i completely agree... i was paired with a 7th grade slacker, but the super smart kid was paired with the other really annoying super smart kid, because they work well together, yet because the 7th grade slacker understands what we're here for, he tries and helps me out :D

U NEED PPL WHO WORK WELL TOGETHER...GET TO KNOW THEM...OR IT DOESN'T WORK OUT

My team has a super genius the defending state champ, one other really good guy,
one average guy, and another not so good guy.
the state champ does 10 and 8
the other good guy does 7 and 9
the average guy does 1 3 and 5
and the bad guy does 2 4 and 6
we also trade problems if they are to hard for some of us even though sometimes when we do that the defending state champ ends up doing all the problems
<snip by moderator rcv>

We split ours up into two pairs doing five each also, but another team I know of has an interesting strategy:

Weakest: 1-4
Second-Weakest: 3-6,9
Second-Strongest: 5-8,10
Strongest: 1,2,7-10

This way, every problem is checked once automatically.

[quote="dasherm"]There were well more than one seventh grader to make nats last year, maybe you are talking your state.[/quote]
Ya.. I'm talking about CT./..

[quote="dasherm"]... every problem is checked once automatically.[/quote]Why stop at "once automatically" if there is time for "more than once automatically"? Having said that, in most cases if two members come up with the same answer the problem is probably solved correctly.

How about this real story: during a practice 2 of our team members gave "0.6" as the answer to a particular problem and  the other 2 members did not bother to use their extra time to check it.  The problem asked the answer to be in fraction!  :D

By automatically, I meant without having to regroup and find new problems to do.

You are right: "At least once" would be better than "once automatically".

My coach, bob fischer, has a strategy that works for us (3 miedium-strongs and a medium), and apparently for his other team (2 superstars, 2 strongs)

One person starts at one and goes down, one person starts at 5 and goes up, one person starts at 6 and goes down, one person starts at 8/9/10 and goes up. It works pretty well or us (weaker team). Then we go and see which problems we can't do and stuff.

My team doesn't really have a strategy sorta deal
we have 2 strong people, and 2 medium people.
We usually grab random problems =P and we all check each others and stuff...
like (not real names)
joe does #2
bob does #9
sue does #7
billy does #10

Then after they're done with those problems, one person checks the other people's answers while the other's work on another problem, then other random stuff happens.
 :P just random stuff, no strategy, and we still got 1st  :D

We were going to have the strongest/weakest and second strongest/weakest work together, but we changed it to strongest/2nd weakest and 2nd strongest/weakest because two of our members cooperated better. We were going to have one person do the evens going up and the person with him do the evens going down and the other team would have one person doing odds going up and the other odds going down. But we changed after the first 2 minutes or so and just did any problem we could, then had everyone check each other's answers at the end (we had 3 minutes...).

[quote="theaznkungfu"]

YEA... OUR TEAM'S FIRST IN SAN DIEGO!!!!!!!!!!! goin to state[/quote]

We came in second...


number theory
prime numbers
number theory unsolved

i  found this problem very ineresting. it appeared  in the hungary-israel competution 2005.

 Does there exist a sequence of 2005 consecutive positive integers that contains 25 prime numbers?

I believe that the answer is yes, there is such a sequence.

We know that there are more than 25 primes from 1 to 2005.  We also know that there is a sequence of 2005 consecutive integers with no primes.  (Example: Consider the sequence $2006!+2$, $2006!+3$, ..., $2006!+2006$.)

Imagine a window that covers 1..2005, and moves to the right one step at a time.  At each step, the number of primes will change by $\pm 1$.  Thus, at some point before $2006!+2$, the number will be exactly 25.

I know this was a very old topic but it was from a recent link :wink:

The other day I was working on some problems from the contest this was from and thought this problem was a joke!
My proof was exactly Ravi B's, except I also noted that 109 is the 26th prime. Also, his is slightly incorrect - the number of primes can chance by 0. It is more accurate to say the the change in the number of primes is an integer between -1,1  :P
I also considered the sequence starting at $2006!+2$. But I defined the sequence $\{N\}$ to represent $N,N+1,...,N+2004$ which helped some notation.


real analysis
algebra
partial fractions
calculus
calculus computations

Can anyone show me how to take the Inverse laplace transform of $ Y(s) \equal{} (e^{ \minus{} s})[\frac {1}{(s \plus{} 1)^2}]$? Convolutions or partial fractions?

Combining two famous formulas of Laplace transform, we have

$ \mathcal{L}^{\minus{}1}\{e^{\minus{}as}F(s\minus{}b)\} \equal{} e^{b(t\minus{}a)}f(t\minus{}a)u(t\minus{}a)$

where $ F(s) \equal{} \mathcal{L}\{f(t)\}$ and $ u$ is the unit step function. Thus we have

$ \mathcal{L}^{\minus{}1}\left\{\frac{e^{\minus{}s}}{(1\plus{}s)^2}\right\} \equal{} e^{\minus{}(t\minus{}1)}(t\minus{}1)u(t\minus{}1)$


[quote="Brent Woodhouse"]There may be more than one solution, but by just using common sense I found $950095$
This yields $10001$ when divided by $95$.[/quote]
I don't think $950095$ is a palindrome...

Sorry, sometimes things seem so easy you forget to check them :oops: .

Here's a real answer:
The 6-digit palindrome we're looking for must be a multiple of 95, so the unit's digit must be 5 or 0.  0 doesn't work because this number is a palindrome and you can't have zero for the first digit.  That leaves a sure 5 for the units digit which makes the hundred thousands digit 5.
That is limited enough, so you can multiply four digit palindromes in the range $5000 - 7000$ by 95 to check for answers.

The only one that works is $(5555)(95)=527725$, where $527725$ and $5555$ are the two palindromes.

[quote="frt"]Find a 6-digit palindrome such that when divided by $95$, the remainder is $0$ and the quotient is another palindrome.[/quote]

Is there a way to do this problem without trial and error? I mean, I know that the last digit must be $5$, but still, you have to check through $100$ possible six digit palindromes :o

[quote="Iversonfan2005"]Is there a way to do this problem without trial and error? I mean, I know that the last digit must be $5$, but still, you have to check through $100$ possible six digit palindromes :o[/quote]
Yeah, you can restrict the range of the new palindrome.

Let $a$ be the original 6-digit palindrome and $b$ be the new palindrome (i.e., $a=95b$)

Since the last digit of $a$ is 5, the leftmost digit of $a$ is also 5, which means $500005\le a\le 599995$.

Therefore, $\frac{500005}{95}\le b\le \frac{599995}{95} \Longleftrightarrow 5264\le b\le 6315$ and there are only $7$ possibilities for $b$.


[quote="shobber"][quote="cezar lupu"]Determine the nature of a triangle such that $h_a=l_b=m_c$ :D  ;)[/quote]
$h$ is the altitude, $m$ is the median, what about $l$?[/quote]

l is bisector

$l_b$- length of the bisector of the angle $\widehat {ABC}$.


A skateboarder of mass $M$ stands on one end of a skateboard of mass $m$ and length $\ell$, initially situated on a horizontal surface, motionless. At a given moment, the skateboarder jumps under an angle $\alpha$ (with the horizontal) and lands on the skateboard at the other end. We neglect friction, the movement of the skateboard on the vertical and the length of the skateboarder's shoes.
[url=http://www.google.com/search?hl=en&q=i+may+rise+but+i+refuse+to+shine&btnG=Google+Search][img]http://i75.imagethrust.com/i/642504/skater.jpg[/img][/url]
Determine:
a. the initial speed of the skateboarder;
b. the kinetic energy of the skateboard during the skateboarder's jump;
c. the numerical value of the impulse when the skaterboarder returned to his first love, the skateboard.

PS: The official solution confuses me. Let's see your ideas.

[hide]
If the magnitude of the rider's horizontal velocity is $v_{R}$, and that of the board is $v_{B}$ during the jump, then by conservation of momentum $M v_{R}= m v_{B}$. Also if $t$ is the duration of the jump, then $0 = (v_{R}\tan{\alpha}) t-\frac{1}{2}g t^{2}$, so $t = \frac{2 v_{R}\tan{\alpha}}{2}$. And since $l = (v_{R}+v_{B}) t$, $v_{R}= \sqrt{\frac{m g l \cot{\alpha}}{2 (M+m)}}$. Finally, $v = v_{R}\sec{\alpha}= \sqrt{\frac{m g l \csc{2 \alpha}}{(M+m)}}$.

Part B is easy. $KE = \frac{1}{2}m v_{B}^{2}$. Then you can substitute some stuff.

Part C is also easy. $|I| = M v$.
[/hide]


email
AwesomeMath
summer program
induction
algorithm
parameterization
number theory

May we start discussing Admission Test A, since the deadline already passed?

Probably not unless you email and ask Dr. A about it,
He'll probably say yes, but i would not post anything about it,

eh, I didn't get around to sending it in.

I'd say not, since solutions are not on AwesomeMath yet, and there are people with regular and late registration can use Test A.

I'd say we wait until we see solutions posted on AwesomeMath.

Actually, I believe you cannot apply for regular/late registration with Test A...

[quote="hamt30"] there are people with regular and late registration can use Test A.
[/quote]
yes, but they still should have sent in it by the test a deadline

Uhh...there are new tests for the regular and late registrations.

Let's just wait for the official reply of one AwesomeMath staff, just to be sure we are not cheating anybody, although I don't see how we can do that since the deadline already passed

Alright, I emailed Titu so you guys can start discussing test A now.

I guess a few comments:

6. is nice. The equation x^2/a^2+y^2/b^2-z^2/c^2=K is a hyperboloid of one sheet. Look up the graph of this by searching google. Anyway, it is an interesting thing to note that this surface is curved but it has lines that lie on the surface.

The other solution is just to find the intersection of the two planes in terms of a parameter.

9. Is a nice problem. It is very closely tied with linear recurrences is general. The two methods that I thought of using was either induction or the division algorithm and plugging in the roots into both sides. Then noting that the solutions for ax+b are the fibbonacci sequence by doing a binet like argument.

I thought that 6 was rather easy. Just like you said, I set a parameter, and solved it in 3-4 lines.
If somebody wants to compare answers, I got:
[hide]
1. is a picture
2. none
3. circle- no, line-yes
4. 83160
5. 1,9
6. 1
7. n/a
8. n/a
9. n/a
10. 9
[/hide]
 I was wondering how people solved number  4 and 10.
For 4, I incorporated a few lemmas about number of divisors, and how LCM of 12 smallest positive integers is less or equal the LCM of any other 12 consecutive positive integers.
For 10, my whole proof was based on careful observation. I counted total number of squares, and how many are eliminated once you put domino on the board.

Although I just take a look with this test, I think your answer for 4 is wrong because of the factorization is: 2^3.3^3.5.7.11, which obviously can't include 12 consecutive integers! I think this problem is not hard if you care about its smallest value will be obtained whenever 12 divisors of it are the smallest, which means 12!

Another solution to number 6 is to rewrite the equations as follows:
\begin{align*} 6x + 7z & = 9y + 2 \\
7x - 6z & = - 2y + 9. \end{align*}
Then square both sides and add the results.

Another solution is to use complex numbers:
\begin{align*} (6 - 7i)(x + zi) & = 6x + 7z + (6z - 7x)i \\
& = 9y + 2 + (2y - 9)i \\
& = (9 + 2i)(y - i). \end{align*}
Now take the absolute value of both sides.

A very similar problem appeared in 2002 AMC 10 B, Problem 20.

lena_siz's answer to number 4 is correct, but not the answer to number 2.

Oopps, you're right Ravi :oops: . I forget that it can be divisible from 1 to 12 without including all of them! Sorry for my bad explanation!

Ravi B, can you please explain my mistake?
I factored $ 32p\plus{}1\equal{}k^3$ as a difference of cubes, and then, found that there do not exist primes such that:
$ 32p\equal{}(k\minus{}1)(k^2\plus{}k\plus{}1)$.
 This is because all factors of 32 are even, and thus, there are many restrictions on what (k-1) and (k^2+k+1) can be equal to.

Your factorization is the right approach.  But what happens when $ k \equal{} 33$?

It's easy to see that $ k$ must be odd as $ 32p$ is even and $ 1$ is odd.  Thus, $ k \minus{} 1$ must be even and $ k^2 \plus{} k \plus{} 1$ must be odd.
This implies that $ k \minus{} 1 \equal{} 32\rightarrow k \equal{} 33$.
Then just compute it.

Number 6 was almost the same as that AMC 10 problem as you said ravi.  Luckily, I did that test a few days before for practice =).

Number 9: I used induction... It's easy to notice the answer at the beginning, and if you use induction to assume it's true for n=k, then for n=k+1, the second to last remainder will simply be the remainder from n=k times x.

Number 10: I got 9 xD

Number 2: You use the fact that the only square doubled is the middle one.  Let that be x, then since the column and row have the same sum $ \frac {45 \plus{} x}{2}$, you can see that $ x$ must be odd.  I forget the numbers, but then it's simple to check the only two cases.

Number 1: Try dissecting the 6x6 into the 4x9.

Number 3: for the circle, since 16 can only add to one other number less than 16 to form a perfect square, it's obviously impossible. It also doesn't work for 8.  Since you know those two, you simply put them on the ends of the line.

There are others, but I don't have time...

[quote="lena_siz"]Ravi B, can you please explain my mistake?
I factored $ 32p \plus{} 1 \equal{} k^3$ as a difference of cubes, and then, found that there do not exist primes such that:
$ 32p \equal{} (k \minus{} 1)(k^2 \plus{} k \plus{} 1)$.
 This is because all factors of 32 are even, and thus, there are many restrictions on what (k-1) and (k^2+k+1) can be equal to.[/quote]

Since $ k^2\plus{}k\plus{}1$ is odd, all of $ 32$ is in $ k\minus{}1$. And since $ k^2\plus{}k\plus{}1$ is always greater than 1, $ k^2\plus{}k\plus{}1\equal{}p$ and $ k\minus{}1\equal{}32$, so $ 32p\plus{}1\equal{}33^3$. After that, it's computation.  :read:

For #9, it is very easy to use induction.

#1 I knew how to dissect, I just needed to show how I got there.

[quote="1=2"]I just needed to show how I got there.[/quote]

No you don't, you can just show the dissection.

Alright, I guess I just made a calculation error, b/c I was doing the same thing with number 2. Thanks for explaining it though!!  :)

I loved the SFFT problem - I've been waiting to use that!

Oh number 7? I've been waiting also xD. But it was more of diophantine equations later on in the problem. I think I forgot to include on my proof for number 10 that $ 8*6<49$ -_-... Each $ 1*2$ takes at most $ 6$ of the $ 2*2$ squares out.  There are $ 49$ of them, so you must have at least $ 9$. Great... Hope I remembered to include that ~_-.

Thanks Carl. You make us feel so  good about ourselves T.T

[quote="CatalystOfNostalgia"]#7 can be done in 3 lines using inequalities.[/quote]

.....

Could you show us how?

You can't just write those words on a #7 sheet and hand that in and get a perfect...

Rearranging gives $ n\equal{}\dfrac{3m}{5m\minus{}4}$, we need $ m,n\in\mathbb{Z}$. For $ m>2$, $ 5m\minus{}4>3m>0$, so $ n$ can't be an integer. Also, for $ m<0$, $ 0<\dfrac{3m}{5m\minus{}4}<\dfrac{3}{5}$. Thus, we only need to check $ m\equal{}0,1,2$, which give $ (0,0),(1,3),(2,1)$.

Once you find the answers on scrap paper, you can simply write that down as a proof. However, I used the solution sheets as scrap paper =) xD...


$100$ fractions are written on a board, their numerators are numbers from $1$ to $100$ (each once) and denominators are also numbers from $1$ to $100$ (also each once). It appears that the sum of these fractions equals to $a/2$ for some odd $a$. Prove that it is possible to interchange numerators of two fractions so that sum becomes a fraction with odd denominator.
[i]N. Agakhanov, I. Bogdanov [/i]

[quote="Rust"]$ \sum_{k\not \equal{} b,2b}\frac {k}{k} \plus{} \frac {2b}{b} \plus{} \frac {b}{2b} \equal{} 100 \plus{} \frac {1}{2} \equal{} \frac {201}{2}$.[/quote]

Please provide more details...


As most people know, North Carolina got a new education lottery this year.  It has been a while and I was wondering how much money different school systems are getting from it.

I know WSFCS (Winston-Salem/Forsyth County Schools) got 7 million to build stuff (of 250 million needed), but other than that, I don't know.

Also, do you think we should have gotten a lottery?

dude since when do we discuss like actual issues on here

You know, I have no idea (in response to both of the above posts)...

You know when you guys got a lottery, it hurt our lottery, because people on the NC side of the NC/SC border stopped coming to SC to buy Powerball tickets.  It actually hurt businesses as well.

I think a lottery is good though, especially since it helps education (although we need the money much more than NC!)

Are you sure about it helping education  :?:

So far, the only changes I have seen was $\$7,000,000$ donated to help our school system build/rebuild some schools.  Note that $\$250,000,000$ was needed.

I imagine that hurt Virginia as well...

Yeah, it has helped, but most of it goes to college scholarships.  For example, I could get $\$$10,000 (pretty much a full ride) to USC from the LIFE scholarship thing, which is funded by lottery money.


Let $ a,b,c$ be real numbers such that $ (a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2) \equal{} 1.$ What is the maximal value of the following expression
$ ab(a \plus{} b) \plus{} bc(b \plus{} c) \plus{} ca(c \plus{} a) \minus{} 2abc$?

$ [ab(a \plus{} b) \plus{} bc(b \plus{} c) \plus{} ca(c \plus{} a) \minus{} 2abc ]_{max} \equal{} \sqrt {2},$
$ [ab(a \plus{} b) \plus{} bc(b \plus{} c) \plus{} ca(c \plus{} a) \minus{} 2abc ]_{min} \equal{} 1.$


function
search
algebra unsolved
algebra

Can we construct a bijective function $ f: \mathbb{N}\rightarrow\mathbb{R}$?

No.
If function is bijective, it has to be defined for $ f(n)$ for every $ n \in N$, and it have to have a solution $ f(x) \equal{} r$ for every $ r \in R$

Since $ m(N) < m(R)$  then $ f(n)$ would have  more then $ 1$ value, therefore $ f$ is not a function.

EDIT: such function can be injective, but not surjective, therefore it is not bijective.

There is a bijection $ \mathbb{R}\to\mathcal{P}(\mathbb{N})$ the [url=http://en.wikipedia.org/wiki/Power_set]power set[/url] of $ \mathbb{N}$!
(Try to construct one using the [url=http://en.wikipedia.org/wiki/Cantor%E2%80%93Bernstein%E2%80%93Schroeder_theorem]Schröder-Bernstein theorem[/url])

Then any surjection $ f: \mathbb{N}\to\mathbb{R}$ gives a surjection $ g: \mathbb{N}\to\mathcal{P}(\mathbb{N})$.
And then the set $ M: \equal{}\{n\in\mathbb{N}: n\not\in g(n)\}$ has the form $ M\equal{}g(m)$ for some $ m\in\mathbb{N}$.
But then $ m\not\in g(m)\Leftrightarrow m\in M\equal{}g(m)$. Contradiction!
This is of course [url=http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument]Cantor's diagonal argument[/url].

[quote="nguyentrang"][quote="NapoleonXIV"]

Since $ m(N) < m(R)$  then $ f(n)$ would have  more then $ 1$ value, therefore $ f$ is not a function.

[/quote]

What is $ m$?  :maybe:[/quote]

$ m(A)\equal{}|A|$ is how many elements are in set $ A$.

[quote="nguyentrang"]Well, I appreciate your solution. But I wonder if there is an elementary one...[/quote]
It's an one-line contradiction. What could be more elementary? :huh:
Here is a surjection
$ \mathbb{R}\to\mathcal{P}(\mathbb{N})$
$ x\mapsto \{n\in\mathbb{N}: [3^nx]\equal{}3[3^{n\minus{}1}x]\}$
where $ [\cdot]$ is the floor function.

[quote="NapoleonXIV"][quote="nguyentrang"][quote="NapoleonXIV"]

Since $ m(N) < m(R)$  then $ f(n)$ would have  more then $ 1$ value, therefore $ f$ is not a function.

[/quote]

What is $ m$?  :maybe:[/quote]

$ m(A) \equal{} |A|$ is how many elements are in set $ A$.[/quote]

But both sets are infinite

[quote="nguyentrang"][/quote][quote="NapoleonXIV"][quote="nguyentrang"][quote="NapoleonXIV"]

Since $ m(N) < m(R)$  then $ f(n)$ would have  more then $ 1$ value, therefore $ f$ is not a function.

[/quote]

What is $ m$?  :maybe:[/quote]

$ m(A) = |A|$ is how many elements are in set $ A$.[/quote][quote="nguyentrang"]

But both sets are infinite[/quote]

Yes, they are, but set of Real numbers is still greater then set of Natural numbers.

Search for cardinality and you'll see that.
Here is quote from wikipedia:http://en.wikipedia.org/wiki/Cardinality
"For example, the set R of all real numbers has cardinality strictly greater than the cardinality of the set N of all natural numbers,"...


Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that:
\[ (a\plus{}b\plus{}c)\left(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{b}\right)\geq7\plus{}\sqrt{(a^2\plus{}b^2\plus{}c^2)\left(\frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2}\right)\minus{}5}\]

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that:
\[ (a \plus{} b \plus{} c)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{b}\right)\geq7 \plus{} \sqrt {(a^2 \plus{} b^2 \plus{} c^2)\left(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}\right) \minus{} 5}\]
[/quote]

$ \Longleftrightarrow  2(a^2b\plus{}ac^2\minus{}3abc\plus{}b^2c)(a^2c\minus{}3abc\plus{}bc^2\plus{}ab^2)\geq 0$.

 :lol:

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that:
\[ (a \plus{} b \plus{} c)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{b}\right)\geq7 \plus{} \sqrt {(a^2 \plus{} b^2 \plus{} c^2)\left(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}\right) \minus{} 5}\]
[/quote]
The following Inequality is stronger and wellknow
\[ (a \plus{} b \plus{} c)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{b}\right)\geq\left(1 \plus{} \sqrt {1 \plus{} \sqrt{(a^2 \plus{} b^2 \plus{} c^2)\left(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}\right)}}\right)^2\]

[quote="dduclam"][quote="arqady"]Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that:
\[ (a \plus{} b \plus{} c)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{b}\right)\geq7 \plus{} \sqrt {(a^2 \plus{} b^2 \plus{} c^2)\left(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}\right) \minus{} 5}\]
[/quote]
The following Inequality is stronger and wellknow
\[ (a \plus{} b \plus{} c)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{b}\right)\geq\left(1 \plus{} \sqrt {1 \plus{} \sqrt {(a^2 \plus{} b^2 \plus{} c^2)\left(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}\right)}}\right)^2\]
[/quote]

Its Vasc's inequality posted in 2004 GM-B(5)

Try also the following:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1750073#1750073

Another AM-GM:
Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that:
\[ (a \plus{} b \plus{} c)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{b}\right)\geq1 \plus{}2 \sqrt[3] {6(a^2 \plus{} b^2 \plus{} c^2)\left(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}\right) \plus{}10}\]


A 1.05 kg block is attached to a spring with spring constant 16 N/m. While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 41cm/s.
What is the block's speed at the point where x= 0.45 A?

A stands for amplitude.
A 221 g ball is tied to a string. It is pulled to an angle of 5.60 degrees and released to swing as a pendulum. A student with a stopwatch finds that 14 oscillations take 14.5 s.  How long is the string?

5.60 degrees is pretty small so we can use the approximation $ \sin\theta\approx\theta$.  The period of a pendulum with length $ \ell$ with gravity constant $ g$ is given by $ T\equal{}2\pi\sqrt{\frac{\ell}{g}}$  Note that the period of a simple pendulum like your example does not depend on the mass (assuming it is a point mass).

[quote="JRav"]5.60 degrees is pretty small so we can use the approximation $ \sin\theta\approx\theta$.  The period of a pendulum with length $ \ell$ with gravity constant $ g$ is given by $ T \equal{} 2\pi\sqrt {\frac {\ell}{g}}$  Note that the period of a simple pendulum like your example does not depend on the mass (assuming it is a point mass).[/quote]

But i don't hink 0.23cm is the correct answer...

I actually got an answer of .27 but anyway, the answer is in meters, not cm.  So that would be 27 cm, which certainly seems reasonable.


Anyway, here's yet another Mock AMC. I tried really hard, so if it's bad, don't say anything or I'll feel really bad and go cry. In any case, rules are the same as usual; the test has been checked over, so any ambiguities should be resolved. Otherwise, post it in this thread (general questions) or PM me (specific questions, related to answers). Difficulty should be fairly consistent with AMC problems, not that I would know, but that's ok. Enjoy!

[url=http://wangsblog.com/jeffrey/mockAMC.pdf]The Test[/url]

Oh, and regarding submitting answers, I guess I'll do all that scoring, ranking, etc. just PM me your answers. You'll have a week or so, so let's say due by next Wednesday 1/17.

[quote="mysmartmouth"]OK, I won't post my test until yours ends.  The letters should have been switched...but whatever.  I'm gonna take this right now!

#6 - Is this missing something?[/quote]

As far as I know, it's solvable. It might seem a little ambiguous, but just don't think too hard about it.

I suppose the lack of "never" as an answer choice makes it answerable.

I just wanted to add that I HATE geometry, but I really liked every geometry problem you had on the test.  Nice job with those.  But why must there be trig... :moose:

[quote="paladin8"][quote="mysmartmouth"]I just wanted to add that I HATE geometry, but I really liked every geometry problem you had on the test.[/quote]

Perhaps we have a common interest here...[/quote]
That means the writer is mighty talented.

nicely done...the problems feel like those of an actual amc [probably 12]...20,21,24 are nice problems

I liked 21 and 24 a lot too.  24 is probably too easy for a 24, but I'm not complaining  :wink:\

EDIT: wtf purple plume...or should i say now a ranger... change your avatar.

EDIT: Haha we'll see = Sean did something stupid and thinks he solved the problem.

[quote="mysmartmouth"]I liked 21 and 24 a lot too.  24 is probably too easy for a 24, but I'm not complaining  :wink:[/quote]

Haha, we'll see.

[quote="arjunlandes"]Hmm...can you check 17? Maybe I'm being stupid, but I don't see the right answer there.[/quote]

I'm pretty sure it's there. If you want, I can PM you the solution (unless you are taking the test and wish to submit answers, of course).

[quote="arjunlandes"]I'm taking the test, so I'll look at it again.  Just to be sure, loglog|x| means log of the log of |x|, right?[/quote]

Yup.

The message contains only quote and code tags. Please add some text of your own.

I'm sorry but I don't understand number 6. I don't see what it means.

Yes. the test is very beastly.

Paladin8, I added this Mock AMC to the wiki.

Number 6 might be worded incorrectly, I'm not sure.  But just use common sense and you'll see what the writer meant.

[quote="mysmartmouth"]Haha, 25 was a beast.  Nice problem.  Did you make that up?  If not, I wanna know the source (after the competition ends).[/quote]

I did make it up. All the problems are original :).

[quote="bpms"]When is this due?[/quote]

Wednesday, 1/17. At night, I guess. Whatever.

[quote="arjunlandes"]Hmm...can you check 17? Maybe I'm being stupid, but I don't see the right answer there.[/quote]

I apologize to everyone who has taken the test. arjunlandes is corret; the right answer choice was not there for question 17, the point will be given to everyone. The correct answer should be $4 \cdot (10^{\frac{1}{10}}-1)^{2}$.

EDIT: I won't be sending the new scores (you can calculate that), but they will affect the rankings/stats at the end. Also, please use the LINK from now on; it has the newest version (reworded question 6 and changed the answer choices for question 17).

Thanks for the test. The author must be really brilliant... :)

Please sign me up to take this test also.

Today's the last day to submit answers. I should have results/statistics up by tomorrow.

[quote="Ubemaya"]Can I turn it in today?

I didn't hear about it until today...

and if I can't, can I just submit it to see what score I got?[/quote]

Sure, but I won't update the statistics. You can just check your answers in the other thread anyway.

Sorry for my reviving this old thread!But can anyone post the answer key of this test since I need to see where my score is?? I need to see the solution of 1 or 2 problems in this test(23 and another).

Could you give the answers to the test.
I am just practicing for the AMC 12.
My email is
Aggarwal.nitish@gmail.com

Thanks

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=129159

Now please stop bumping this topic.


trigonometry
trig identities
Law of Cosines
Olimpiada de matematicas

[img]http://img409.imageshack.us/img409/1228/bisecea3.png[/img]

Sea un $\Delta ABC$ con $\measuredangle \ BAC = 120^\circ$. Se trazan las respectivas bisectrices en cada uno de los vértices como se muestra en la figura. Pruebe que $\overline{ED}\perp \overline{FD}$

Saludos.

A hint[sorry, but I don`t speak spanish :) ]

F is the exincenter of the triangle $ACD$ and $E$ is the exincenter of the triangle $ABD$.

I think it works with Law Of Cosines and with the bisectrix's formula: $l_{a}=2bc/(b+c) * cos{A/2}$

Sorry because i don't speak in spanish but i understand a little.

En realidad no necesito Hints porque estos problemas ya los he resuelto.

Los expongo acá para compartirlos.

Saludos.

P.D.: en lo posible, evitar el uso de la trigonometría.

Pues como dijo anyciup, obervas facilmente que E y F son excentros de ABD y ACD respectivamente, y con ves que DF y DE son bisectrices de ADB y ADC respectivamente. Ya con eso concluimos que EDF=180/2=90.

Bueno, ahí hay una forma de sacarlo.

Quisiera que expusiesen otra forma de sacarlo sin hacer uso de excentro.

Saludos.

[hide="Unos pequeños Hints"]

Sea $I$ el incentro del $\Delta ABC$. Podría ser de utilidad que el $\measuredangle \ CIB = 150^\circ$ y también buscar dos triángulos isósceles por ahí..., o hallar unos cuadriláteros cíclicos...[/hide]

[color=blue]Como el suplemento de $\begin{equation*}\begin{aligned}\measuredangle CAB \end{aligned}\end{equation*}$ mide $\begin{equation*}\begin{aligned}60^\circ \end{aligned}\end{equation*}$, es inmediato que $\begin{equation*}\begin{aligned}\overline{AE}\end{aligned}\end{equation*}$ y $\begin{equation*}\begin{aligned}\overline{AF}\end{aligned}\end{equation*}$ son bisectrices externas de los triángunlos $\begin{equation*}\begin{aligned}ABD \end{aligned}\end{equation*}$ y $\begin{equation*}\begin{aligned}ACD \end{aligned}\end{equation*}$, respectivamente. Como también $\begin{equation*}\begin{aligned}E \end{aligned}\end{equation*}$ y $\begin{equation*}\begin{aligned}F \end{aligned}\end{equation*}$ conforman bisectrices internas de los triángulos anteriormente nombrados, y respectivamente. Es inmediato que $\begin{equation*}\begin{aligned}\overline{ED}\end{aligned}\end{equation*}$ y $\begin{equation*}\begin{aligned}\overline{FD}\end{aligned}\end{equation*}$ son bisectrices de $\begin{equation*}\begin{aligned}\measuredangle ADC \end{aligned}\end{equation*}$ y $\begin{equation*}\begin{aligned}\measuredangle ADB \end{aligned}\end{equation*}$, una vez más respectivamente, debido a que sobre los puntos $\begin{equation*}\begin{aligned}E \end{aligned}\end{equation*}$ y $\begin{equation*}\begin{aligned}F \end{aligned}\end{equation*}$ se encuentran los excentros de los triángulos antes mencionados. Luego $\begin{equation*}\begin{aligned}\measuredangle EDF &= \frac{{\measuredangle ADC+\measuredangle ADB}}{2}&= \frac{{180^\circ }}{2}&= 90^\circ \end{aligned}\end{equation*}$, y queda demostrado el enunciado.[/color]

$conejita$ ya habia posteado este problema antes: http://www.mathlinks.ro/Forum/viewtopic.php?t=108306