If $p$ is a prime number and $p\mid a^p-b^p$. Then $p^2\mid a^p -b^p$. If $p$ is a prime number and $p\mid a^p-b^p$. Then $p^2\mid a^p -b^p$. 
 A: The case $a=b$ is going to be true anyways.
WLOG let's assume $a=b+x,x>0$
$$(b+x)^p-b^p=x^p+\sum_{r=1}^{p-1}\binom{p}{r}x^rb^{p-r}$$
Since it is divisible by p, $p|x$.
As $p>1,p^2|x^p$.Also $\binom{p}{r}x^rb^{p-r}$ is divisible by $p^r*p=p^{r+1}$
Since $r>0,r+1\geq 2$
Therefore, $$p^2|x^p+\sum_{r=1}^{p-1}\binom{p}{r}x^rb^{p-r}$$
A: You have $$a^p\equiv a\pmod p\\b^p\equiv b\pmod p .$$
$($*follows from Fermat's ittle theorem*$)$
So, $$a^p-b^p\equiv a-b \pmod p.$$
Now, you have $$a^p-b^p\equiv 0 \pmod p.$$ So, $$a-b\equiv 0 \pmod p\\\implies a\equiv b \pmod p\\\implies p\mid a-b\tag 1\\$$
Now, note that $$a^p-b^p=(a-b)(a^{p-1}+a^{p-2}b+a^{p-3}b^2+\dots+b^{p-1})$$
Now, $$\begin{align}
a^{p-1}&\equiv a^{p-1}\pmod p\\
a^{p-2}b&\equiv a^{p-2}a=a^{p-1}\pmod p[\text{since} \;a\equiv b\pmod p]\\
\vdots\\
b^{p-1}&\equiv a^{p-1} \pmod p
\end{align}$$
So, summing them up, $$(a^{p-1}+a^{p-2}b+a^{p-3}b^2+\dots+b^{p-1})\equiv pa^{p-1}\equiv 0\pmod p$$
So, $$p\mid a-b\\p\mid (a^{p-1}+a^{p-2}b+a^{p-3}b^2+\dots+b^{p-1})$$
So, $$ \bbox[yellow]{p^2\mid a^p-b^p}$$
