# Proving divisibility of integers [closed]

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 $$k^2$$?

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

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

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!
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 $$k^2 \mid x^2 + y^2$$.