I'm blanking on this seemingly simple problem. Let $a \in \mathbb{Z}$ be an integer, $p$ prime. When is it true that $a$ is a square in the ring of p-adic integers $\mathbb{Z}_p$?
I understand that the case $p = 2$ will be different, and that I'll likely have to use Hensel's Lemma, but don't know how to approach this. Thank you.
If $p \not = 2$, then any root of $x^2-a$ that you find modulo $p$ (i.e. in $\Bbb Z/p\Bbb Z$) can be uniquely lifted to a root modulo $p^2$, then $p^3$, and so on, since the derivative is nonzero modulo $p$. In contrast, if $\xi^2 \not \equiv a\!\! \mod p$, then a $p$-adic integer that is equivalent to $\xi\!\! \mod p$ can't be a root. This solves the task for $a$ that is coprime to $p$. If $a = m p^k$, $(p,m)=1$ and $k > 0$, then either $k$ is even and you can reduce the equation to $y^2 - m = 0$, or $k$ is odd and there is no solution by multiplicativity of $|\cdot|_p$.
Let $p = 2$. Theorem 4.1 here implies the following variant of Hensel's lemma (which reduces to the usual if $N = 0$): if $f \in \Bbb Z_p[x]$, $\alpha_0 \in \Bbb Z_p$, $|f'(\alpha_0)|_p = p^{-N}$, $N \ge 0$, and $\alpha_0$ is a root of $f$ modulo $p^{2N+1}$, then there exists a unique $\alpha$ that is equivalent to $\alpha_0$ modulo $p^{N+1}$ and is a root of $f$.
Here we have $N = 1$, and if $a$ is odd, then you can easily see that all odd residues modulo $8$ square to $a \equiv 1\!\! \mod 8$. If $a \equiv 2\!\! \mod 4$, then there are no roots, and if $a \equiv 0\!\! \mod 4$, you can solve the equation for $a/4$.
