Prove that if $a^p-b^p$ is divisible by $p$, then it is also divisible by $p^2$ $a$ and $b$ are natural numbers and $p$ is a prime number. Prove that if $a^{p}-b^{p}$ is divisible by $p$, then it is also divisible by $p^{2}$.
My attempt: Based on Fermat's theorem $(a^{p}-a)$ and $(b^{p}-b)$ are divisible by p, therefore their difference is divisible by $p$. I.e.
$(a^{p}-a)-(b^{p}-b)=(a^{p}-b^{p})-(a-b)$ is divisible by $p$. Since the left hand side of this equation is divisible by $p$, the right hand side also should be divisible by $p$. Based on the problem's assumption $(a^{p}-b^{p})$ is divisible by $p$ therefore we conclude $(a-b)$ must be divisible by $p$. Now We can factorize $(a^{p}-b^{p})$
$$(a^{p}-b^{p})=(a-b)(a^{p-1}+a^{p-2}b+....+b^{p-1})$$
we have to prove $(a^{p}-b^{p})$ is multiple of $p^2$ which means we have to show the right hand side should be multiple of $p^{2}$. But $(a-b)$ is multiple of p therefore we have to show 
$(a^{p-1}+a^{p-2}b+\cdots+b^{p-1})$ is multiple of $p$ and I am stuck here. Any help would be appreciated.
 A: To start with your beginning.
Knowing that $p\mid a-b$ means that $a\equiv b\pmod p$. So $a^kb^{p-1-k}\equiv a^{p-1}\pmod p$. So:
$$a^{p-1}+a^{p-2}b+\cdots + ab^{p-2}+b^{p-1}\equiv a^{p-1}+a^{p-1}\cdots a^{p-1}=pa^{p-1}\equiv 0\pmod p$$
So both $a-b$ and $a^{p-1}+a^{p-2}b+\cdots + ab^{p-2}+b^{p-1}\equiv pa^{p-1}$ are divisible by $p$, and hence $a^p-b^p$ is divisible by $p^2$.
A: I'm going to assume $p$ is an odd prime. $a^p\equiv  a $ and $b^p\equiv b\bmod p$ therefore $(a-b)$ is a multiple of $p$. If $a$ is a multiple of $p$ then $b$ must also be and it becomes trivial to see both $a^p$ and $b^p$ are multiples of $p^2$. The other case is when neither number is a multiple of $p$, we then apply the Lifting The Exponent Lemma to conclude.
Notice if $p$ is $2$ then it is also true, because if $a$ is odd so is $b$ and odd squares are $1\bmod 4$. If $a$ is even so is $b$ and even squares are $0\bmod 4$.
A: Hint: Once you know that $p$ divides $a-b$, then $a=b+pk$ for some integer $k$.  Try the binomial expansion on
$$a^p-b^p=(b+pk)^p-b^p$$
