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International Zhautykov Olympiad 2005, Problem 6
Posted: Wed Feb 03, 2016 5:33 pm
by tanmoy
Find all prime numbers $p,q<2005$ such that $q \mid p^{2}+8$ and $p \mid q^{2}+8$.
Re: International Zhautykov Olympiad 2005, Problem 6
Posted: Mon Mar 28, 2016 6:58 pm
by ahmedittihad
$pq$ 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 $p^2$
Re: International Zhautykov Olympiad 2005, Problem 6
Posted: Sat Apr 23, 2016 4:16 pm
by nahin munkar
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.
