So this is a question in my discrete math's Textbook and according to our teacher our schools math's department has not been able to solve this question in 4 years of trying. So I bring this challenge to you. Either try and prove it or prove that it is not possible and that there is a mistake in the question.
This was under the Fermat's little theorem section of discrete mathematics.
$p$ is a prime number bigger than two. $p\mid(x^p + y^p)$ prove that $p^2\mid x^p + y^p$
By FLT $p|x+y$. Thus $\sum_{k=0}^{p-1}{x^ky^{p-1-k}(-1)^{p-1-k}} = \frac{x^p+y^p}{x+y}$ is congruent to $px^{p-1}$ mod $p$, so is divisible by $p$. So $p^2|x^p+y^p$.
First, by Fermat's little theorem, $x^p-x\equiv_p 0$, hence $x+y\equiv_p0$. Or $x\equiv_p (-y)$
Second, $$\frac{x^p+y^p}{x+y} = \sum_{k=1}^px^{p-k}(-y)^{k-1}\equiv_p p x^{p-1} \equiv_p 0.$$ Now $x^p+y^p = \frac{x^p+y^p}{x+y} \cdot (x+y)$, or, in other words, a product of two terms, and both of these terms are divisible by $p$, hence we conclude that $x^p+y^p\equiv_{p^2}0$.
Using Fermat's Little Theorem
$x^p+y^p\equiv x+y\pmod p$
If $p$ divides $x^p+y^p,x$ must divide $x+y,y=kp-x$(say)
$$x^p+y^p$$
$$=x^p+(kp-x)^p$$
$$=x^p-(x-kp)^p$$
$$=x^p-\left(x^p-\binom p1(kp)^1 x^{p-1}+\text{terms containing } p^2 \right)$$
There are more possibilities than proving something or proving that is not possible. E.g. some mathematical problems are such that you can prove that you cannot prove that you can prove that you cannot prove something.
Anyway, let $x=0, y=p^{1/p}$ we thus have $p|x^p+y^p$ but $p^2 \not | x^p + y^p \forall$ prime $p$
Perhaps there was some more constraints not mentioned?
