<s>[INST] If $x^2\equiv b\pmod{p}$ is solvable, then $x^2\equiv b\pmod{p^2}$ is solvable? If $x^2\equiv b\pmod{p}$ has a solution, does this imply $x^2\equiv b\pmod{p^2}$ has a solution too?
Assume $p$ is an odd prime.
 [/INST] Yes, if $b\not\equiv 0\pmod{p}$. Proof:
$$\left(x^2-b\right)^2=\left(x^2+b\right)^2-b(2x)^2$$
So if $x^2\equiv b\pmod{p}$, then $\left(x^2+b\right)^2\equiv b(2x)^2\pmod{p^2}$. Since $\gcd(2x,p)=1$ (because $p$ is odd and $b\not\equiv 0\pmod{p}$),  $\left(\left(x^2-b\right)\left(2x\right)^{-1}\right)^2\equiv b\pmod{p^2}$.

This way we can also get $x^2\equiv b\pmod{p^n}$ is solvable for all $n\ge 1$.
This also implies: if $p$ is an odd prime and $b\not\equiv 0\pmod{p}$, then if $x^2\equiv b\pmod{p^k}$ has a solution, then $x^2\equiv b\pmod{p^{n+k}}$ has a solution for all $n\ge 0$.
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