Prove that $x^2 \equiv a \pmod{2^e}$ is solvable $\forall e$ iff $a \equiv 1 \pmod{8}$.
Whenever I encounter congruence proof, I'm stuck right away. How can I link $a \equiv 1 \pmod{8}$ with the $a$ in the quadratic congruence equation? The only thing that I know from the top of my head is, to express $a = 8k + 1$, then plug it into the congruence equation to get $x^2 \equiv 8k + 1 \pmod{2^e}$. I still can't see anything from here. I feel like I don't know exactly what do I need to show. Can anyone lecture me on this? Quadratic residue is so confusing, I really can't see its practical purpose.
Thank you