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 $p^2|x^p+y^p$?
 A: 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]
A: Notice for all prime $p$ and integer $n$, $p | n^p - n$.
In particular, $p | x^p - x$ and $p | y^p - y$. This means
$$p|x^p + y^p\quad\implies\quad p|x+y$$
Write $x+y$ as $mp$ for some integer $m$, we have
$$\begin{align}
x^p + y^p 
&= x^p + (mp-x)^p\\
&= x^p + (-x)^p + \binom{p}{1}(mp)(-x)^{p-1} + (mp)^2\left(\sum_{k=2}^p\binom{p}{k}(mp)^{k-2}(-x)^{p-k}\right)
\end{align}$$
When $p$ is a prime $> 2$, $p$ will be odd and the first two terms cancel out. 
The third term is a multiple of $p^2$ because $\binom{p}{1}(mp) = mp^2$. 
For the rest, it is a multiple of $p^2$ because of the overall $(mp)^2$ factor.
