Does $x^2 \equiv 0 \mod p$ have a solution other than $x = 0$? I think it doesn't since if it did, by Euclid's lemma $p$ divides $x \in \{1,\dots, p-1\}$.  Is this correct?
 A: Here's a cute proof using little Fermat. If $\rm\ P = 1\!+\!2N\ $ is prime then
$$\rm\ \color{#C00}{X^2\!\equiv 0}\ \Rightarrow\ X \equiv X^P \equiv X^{1+2N}\!\equiv\, X\, (\color{#C00}{X^2})^N\!\equiv\, X\cdot  \color{#C00}0^N\! \equiv\, 0$$
Remark $\ $ This proof is meant only for pedagogical variety. As you mention, it can be deduced by Euclid's lemma, which implies the fundamental prime divisor property $\rm\ p\mid ab\:\Rightarrow\: p\mid a\ \ or\ \ p\mid b.$
A: If $x^2\equiv 0\mod p$, then $x^2 = np$ for some $n\in\Bbb{Z}$. Assuming $x\in\Bbb{Z}\setminus p\Bbb{Z}$, this implies that $np$ is a perfect square. If $p$ is a prime, then $x^2 \equiv 0\mod p\implies x = np$ for some  $n\in\Bbb{Z}$, as the only way $np$ can be a square with $p$ prime is if $n = b^2 p$.
A: If $x^2 \equiv 0 \pmod p$ and $x \neq 0 \pmod{p}$ then $x$ has an inverse mod $p$. Multiplying by its inverse you get that $x \equiv 0 \pmod p$ contradiction....
A: $x^2\equiv 0 \pmod{p} \iff p|x^2 \Rightarrow p|x \Rightarrow x\equiv 0 \pmod{p}$
A: If $x^2 \equiv 0 \pmod{p}$ then the prime $p$ divides $x^2$, so it has to divide one of its factors, i.e. $p \mid x$, and $x \equiv 0 \pmod{p}$.
