When is $x^2 - 75 y^2 = 0$ in $\mathbb{Z}_p$ solvable? 
Exercise: For which prime numbers does the equation $x^2 - 75 y^2 = 0$ have non-trivial solution in the $p$-adic integers $\mathbb{Z}_p$?

For $p\neq 5$, the non-trivial solvability of the equation $x^2 - 75 y^2 = 0$ in $\mathbb{Z}_p$ is equivalent to the solvability of the equation $f(z):=z^2 - 75=0$ in the $p$-adic numbers.
But if $z = u \cdot p^{-n}$ with $u \in \mathbb{Z}^\times_p$ and $n\in\{1, 2, 3, \ldots\}$, then
$$u^2 \cdot p^{-2n} - 75 = 0 \Longleftrightarrow u^2 - 75 p^{2n} = 0$$
and we would have to conclude that $u^2 \equiv 0\bmod{p}$, which is not possible, so $z\in\mathbb{Z}_p$.
Now let $p > 5$. Find the $p$ prime for which the Legendre-symbol $$\left(\frac{75}{p}\right) = \left(\frac{3}{p}\right) = (-1)^{\frac{p-1}{2}} \left(\frac{p}{3}\right) = +1\, .$$
$\left(\frac{p}{3}\right) = +1$ exactly if $p\equiv 1 \bmod{3}$ and $(-1)^{\frac{p-1}{2}} \equiv +1 \bmod{p}$ exactly if $p = 1 \bmod{4}$.
So for $\left( \frac{75}{p} \right) = +1$, either $p \equiv 1 \bmod{3}$ and $p\equiv 1\bmod{4}$ or $p \equiv 2 \bmod{3}$ and $p\equiv 3\bmod{4}$. By the Chinese remainder theorem this is equivalent to $p \equiv \pm 1 \bmod{12}$, e.g.:
$$\{11, 13, 23, 37, 47, 59, 61, 71, 73, 83, 97, 107, \ldots \}\, .$$
By Hensel's lemma in these $\mathbb{Z}_p[X]$ the solutions are also liftable because $f'(z) =2 z \not \equiv 0 \bmod{p}$.
Now the remaining cases:
$p = 2$: $z^2 - 75 \equiv 0 \bmod{4} \Longleftrightarrow z^2 \equiv 3 \bmod{4}$, which is not possible.
$p = 3$: $z^2 - 75 \equiv 0 \bmod{9} \Longleftrightarrow  z^2 \equiv 3 \bmod{9}$, which is not possible.
$p = 5$: $z^2 - 75 = 0$ in $\mathbb{Q}_5$, which means $z^2 - 75 \equiv 0 \bmod{125}$ must be solvable in $\mathbb{Z}_p$. So $\bigl(\frac{z}{5}\bigr)^2 \equiv 3 \bmod{5}$ must be solvable, but since $\left( \frac{3}{5} \right) = -1$ this is not possible.
To sum up the results: $$x^2 - 75 y^2 = 0 \text{ has  non-trivial solutions in $\mathbb{Z}_p$} \Longleftrightarrow p \equiv \pm 1 \bmod{12}\, .$$

Is the proof ok? How could one make it cleaner? Thanks!
 A: Proof looks OK but I would streamline it quite a bit.
$x^2-75y^2=0$ is just the same as $x^2=3(5y)^2$ so for $x,y <>0$ and $p<>5$ (which makes sure that 5 has an inverse) we get $3=[x(5y)^{-1}]^2$ which means that 3 is a quadratic residue mod p or, in Legendre symbol notation,
$$(\frac{3}{p})=+1$$
Using the quadratic reciprocity theorem, hence
$$(\frac{p}{3})(\frac{3}{p})=(-1)^{\frac{p-1}{2}\frac{3-1}{2}}$$, we derive that
$$(\frac{p}{3})=(-1)^{\frac{p-1}{2}}$$
Which excludes $p=2$ and $p=3$, and else indeed leads to $p\equiv \pm 1 \bmod{12}$, as you describe; quite simply because either $p\equiv 1 \bmod{4}$ and then we need $(\frac{p}{3})=+1$ thus $p\equiv 1 \bmod{3}$, or $p\equiv 3 \bmod{4}$ and then we need $(\frac{p}{3})=-1$ thus $p\equiv 2 \bmod{3}$.
The first leads to $p\equiv +1 \bmod{12}$; the second to $p\equiv -1 \bmod{12}$.
$p=2$ means $x^2=y^2$ which gives only $x=y=1$ as a near-trivial solution;
$p=3$ or $p=5$ means $x^2=0$ which gives only trivial solutions $x=0$, $y=$any.
