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 $z^2 \equiv a\ (mod\ p^e)$
2
$\begingroup$
elementary-number-theory
-
0I think you mean $b$ is quadratic modulo $p^e$, since the two notions are distinct. – 2012-08-31
-
0first proved that $b$ is quadratic modulo $p^e$ which is true iff $b$ is quadratic modulo $p$ hence the result. – 2012-08-31
-
1Abhishek, 3 is a quadratic residue mod 2 but not mod 8, so I don't understand your comment. By the way, is $p$ supposed to be a prime? If yes, could you please edit this fact into the statement of the question? – 2012-08-31
-
0Sorry Gerry, my bad!! $p$ is supposed to be an odd prime. – 2012-08-31
-
0Abhishek, then 3 is a quadratic residue mod 3 but not mod 9. – 2012-08-31
-
0How is 3 quadratic residue mod 3 as we are talking in $Z_p^*$ and $(Z_p^*)^2$? – 2012-08-31
-
0Per chance this point could be clarified further in the question? Thanks for your attention. – 2012-09-18