-1
$\begingroup$

Given integers $x$ and $y$ and a prime number $k>3$. It turned out that $x + y$ and $x^2 + y^2$ are simultaneously divisible by $k$. Prove that $x^2 + y^2$ is divisible by $k^2$?

$\endgroup$

closed as off-topic by user10354138, José Carlos Santos, AD., Xander Henderson, Jendrik Stelzner May 23 at 15:18

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user10354138, José Carlos Santos, AD., Xander Henderson, Jendrik Stelzner
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Tell us what you have tried. $\endgroup$ – AD. May 23 at 9:37
  • $\begingroup$ Only x^2+y^2=(x+y)^2-2xy $\endgroup$ – Anatoly Karpov May 23 at 9:39
3
$\begingroup$

We write $x^2+y^2=(x+y)^2-2xy$, and notice that since $k$ divides $x^2+y^2$ and $(x+y)$, then $k$ should also divide $2xy$. Since $k>3$, $k$ should divide at least $x$ or $y$. Because $k$ divides $x+y$ and $x$ or $y$, we conclude that $k$ divides $x$ and $k$ divides $y$.

We now again return to the expression $(x+y)^2-2xy$, and notice that $k^2$ obviously divides $(x+y)^2$, and that $k^2$ also divides $2xy$ because $k$ divides $x$ and $k$ also divides $y$. Hope this is clear!

$\endgroup$
  • $\begingroup$ Thanks for your detailed explanations. Figuratively speaking, I got lost between two pines. $\endgroup$ – Anatoly Karpov May 23 at 10:43
1
$\begingroup$

Note that the first statement implies $x \equiv -y \pmod k$, and the second implies $x^2 + (-x)^2 \equiv 0 \mod k$. So we get $k \mid 2x^2$ which implies $k \mid x^2$, in turn implying $k \mid x$ since $k$ is prime. So $k \mid x, k\mid y$ gives $k^2 \mid x^2 + y^2$.

$\endgroup$
  • $\begingroup$ Thanks, perfectly! $\endgroup$ – Anatoly Karpov May 23 at 11:08

Not the answer you're looking for? Browse other questions tagged or ask your own question.