I've been working on this problem for a while, but hit a dead end.
Here's the problem:
Suppose $p$ is an odd prime. Also let $b^2 \equiv a \pmod p$ and $p$ does not divide $a$. Prove there exists some $k \in \mathbb{Z}$ such that $(b+kp)^2 \equiv a \pmod {p^2}$.
Here's what I've tried so far:
$(b+kp)^2 \equiv b^2 + 2bkp + k^2p^2 \equiv a \pmod {p^2}$
Here, I need to find such $k$ that satisfy this congruence. Equivalently, I need to find such $k$ so that $p^2$ divides $(b^2-a) + 2bkp + p^2k^2$ or equivalently show that $p^2$ divides $(b^2-a) + 2bkp$ for some $k \in \mathbb{Z}$.
So far, since $p$ divides $(b^2-a)$, then by definition, there exists some $x \in \mathbb{Z}$ such that $b^2-a = px$. I got stuck here, and I've tried some examples, but I haven't seen any pattern that pertains to this problem.
Any insight would be helpful.