For an odd prime $p$, if $a = p^fb$ and $p \nmid b, 0 \leq f < e$, how many solutions exist for $z^2 \equiv a\ (mod\ p^e)$.
I have already proved the previous part of this question which was to show that a solution exists if and only if $f$ is even and $b$ is quadratic modulo $p$.
If we let $b \equiv \beta^2\ (mod\ p^e)$, then $z^2 \equiv (p^{f/2}\beta)^2\ (mod\ p^e)$ which gives us $\pm\ p^{f/2}\beta$ as roots modulo $p^e$. How to find number of distinct roots among them?
If $a = p^fb$ and p \nmid b, 0 \leq f < e, how many solutions exist for $z^2 \equiv a\ (mod\ p^e)$
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elementary-number-theory
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0Per chance this point could be clarified further in the question? Thanks for your attention. – 2012-09-18