Prove that if $p|x^p+y^p$ then $p^2|x^p+y^p$ I can show that $5|x^5+y^5$, by considering $(x+y)^5$ and using binomial expansion. But I am not sure how to show that $25|x^5+y^5$.
More generally, if p is a prime and $p>2$, how do I prove that if $p|x^p+y^p$  then $p^2|x^p+y^p$?

Assuming $x,y\in \mathbb{Z}$, if we have $5\mid x^5 + y^5$ then by binomial expansion it follows that $5\mid (x+y)^5$. But by prime factorization, it is obvious that $5^5\mid (x+y)^5$ and hence $25\mid (x+y)^5$. To see that $25\mid x^5 + y^5$, it is enough to show that $5x^4y + 10x^3y^2 + 10x^2y^3 +5xy^4$ is divisible by $25$ or equivalently $(x^3 +2x^2y + 2xy^2 +y^3)$ is divisible by 5.
Doing this by brute force check for pairs $(x,y)= (1,9) \mod 10$, $(x,y)= (2,8) \mod 10$, $(x,y)= (3,7) \mod 10$, $(x,y)= (4,6) \mod 10$, $(x,y)= (5,0) \mod 10$. [$\mod 10$ these are the only possible solutions]
