Question:
Let $p$ be an odd prime. Prove that if $a\equiv b \pmod p$ then $a^p\equiv b^p \pmod p^2$. Then show the Diophantine equation $x^5+y^5=z^5$ has no integer solutions with $5\not\mid xyz$.
My working so far:
$a\equiv b \pmod p\Longrightarrow a=b+kp$ for some $k\in\mathbb{Z}$ so
$$ a^p = (b+kp)^p = \sum_{i=0}^p {p\choose i}(kp)^i ~b^{p-i} = b^p + k~p^2~b^{p-1} + \sum_{i=2}^p {p\choose i}(kp)^i ~b^{p-i} \equiv b^p\pmod{p^2}.$$
How can I use this to prove the latter statement?
Write the Fermat equation of exponent $5$ in the form $x^5+y^5+z^5=0$. We have $x^5\equiv x \mod 5$, $y^5\equiv y \mod 5$ and $z^5\equiv z \mod 5$, hence $x+y+z\equiv 0\mod 5$. Without loss of generality we may assume, because $5\nmid xyz$, that $x\equiv y\mod 5$, hence $x^5\equiv y^5 \mod 25$, and $−z^5 ≡ x^5 + y^5 ≡ 2 x^5 \mod 25$. However, the equation $x ≡ y \mod 5$ also implies that $−z ≡ x + y ≡ 2 x \mod 5$ and $−z^5 ≡ 2^5 x^5 ≡ 32 x^5 \mod 25$. Combining the two results and dividing both sides by $x^5$ yields a contradiction $2 ≡ 32 \mod 25$.