International Zhautykov Olympiad 2005, Problem 6

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tanmoy
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International Zhautykov Olympiad 2005, Problem 6

Unread post by tanmoy » Wed Feb 03, 2016 5:33 pm

Find all prime numbers $p,q<2005$ such that $q \mid p^{2}+8$ and $p \mid q^{2}+8$.
"Questions we can't answer are far better than answers we can't question"

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ahmedittihad
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Re: International Zhautykov Olympiad 2005, Problem 6

Unread post by ahmedittihad » Mon Mar 28, 2016 6:58 pm

$pq$ divides $p^2q^2 +8(p^2 +q^2)+64$
so, $pq$ divides $8(p^2 +q^2 +8)$
so, $p$ divides $q^2$ and $q$ divides $p^2$

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nahin munkar
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Re: International Zhautykov Olympiad 2005, Problem 6

Unread post by nahin munkar » Sat Apr 23, 2016 4:16 pm

ahmedittihad wrote:$pq$ divides $p^2q^2 +8(p^2 +q^2)+64$
so, $pq$ divides $8(p^2 +q^2 +8)$
so, $p$ divides $q^2$ and $q$ divides $p^2$
Ahmed Ittihad,Your approach is wrong.your start was statisfactory indeed; but in your 3rd line, u made a mistake.
look, u told in ur 3rd line that ($p$ divides $q^2$ and $q$ divides $p^2$). But it's not correct. Here, u can check it by counter example where p=3 , q=17. SO, here,your last line is incorrect.
Here,(p,q)=(2,2);(3,17);(17,3);(89,881);(881,89) are the solutions of the problems. :D
# Mathematicians stand on each other's shoulders. ~ Carl Friedrich Gauss

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