If the sum of two $p$th powers is divisible by $p$, then it is divisible by $p^2$ 
If $p > 2$ is a prime and $p | (x^p + y^p)$, then show that $p^2 | (x^p + y^p)$

I have been stuck on this problem for a while now. (Though my textbook is prone to mistakes so the original proposition could be wrong).
 A: By Fermat's little theorem, $t^p\equiv t\pmod p$ for any $t$, so if $p\mid x^p+y^p$ then in fact $p\mid x+y$ (or $y\equiv -x\pmod p$). 
Now, since $p$ is odd, $$x^p+y^p=(x+y)(x^{p-1}-x^{p-2}y+x^{p-3}y^2-\cdots+y^{p-1}).$$ The first term is a multiple of $p$, as explained above. The second is $$\sum_{k=0}^{p-1}(-1)^k x^{p-1-k}y^k=\sum_{k=0}^{p-1}x^{p-1-k}(-y)^k\equiv \sum_{k=0}^{p-1}x^{p-1-k}x^k=\sum_{k=0}^{p-1}x^{p-1}=px^{p-1}\equiv0\pmod p.$$
A: By Fermat's Little theorem you have $x^p=x\bmod p$ and $y^p=y\bmod p$. Hence $p$ divides $x+y$. So we can write $y=kp-x$. Hence $y^p=(kp-x)^p$. Expanding by the binomial theorem we get $-x^p$ plus terms in $p^2$.
A: By Fermat's little theorem, if $p$ is prime, $x^p\equiv x\pmod p$.  Therefore, $p\mid x^p+y^p$ iff $x^p+y^p\equiv 0\pmod p$ so that $x+y\equiv 0\pmod p$.
Then $y=kp-x$ for some integer $k$  Therefore, $$x^p+y^p=x^p+\sum_i\binom{p}{i}(kp)^i(-x)^{p-i}.$$  Expanding out a few terms, we can rearrange this as follows:
$$
x^p+\sum_{i=2}^p\binom{p}{i}(kp)^i(-x)^{p-i}+\binom{p}{1}(kp)(-x)^{p-1}+(-x)^p.
$$
Since $p$ is an odd prime, $(-x)^p=-x^p$, which cancels with $x^p$.  All of the terms in the sum are divisible by $p^2$ since $i\geq 2$, it follows that $p^2\mid p^i\mid (kp)^i$.  Finally, $p^2\mid \binom{p}{1}(kp)(-x)^{p-1}=kp^2x^{p-1}$ since $\binom{p}{1}=p$.
A: Hint $\,{\rm mod}\ p\!:\  f(x) = x^p\!+y^p \equiv (x\!+\!y)^2 g(x,y)\,$ since $\,x\equiv -y\,$ is a double root of $f(x)\,$ by $\,f(-y) \equiv 0,$ $\,f'(-y) = p(-y)^{p-1}\!\equiv 0.\ $ So $\,\ p\mid x^p\!+y^p\equiv x\!+\!y\,\Rightarrow\, p^2\mid (x\!+\!y)^2\mid x^p\!+y^p$
