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Let $m$ and $n$ be relatively prime and $b \in (\mathbb Z/ mn \mathbb Z)^\times$. Then $b$ is a quadratic residue modulo $mn$ if and only if $b$ is a quadratic residue modulo $m$ and modulo $n$.

I am struggling with this proof. It seems like I should be able to do it but every time I get stuck.

Thank you for you help.

Annie
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For the somewhat harder direction, suppose that $s^2\equiv b\pmod{m}$ and $t^2\equiv b\pmod{n}$. By the Chinese Remainder Theorem the system of congruences $x\equiv s\pmod{m}$ and $x\equiv t\pmod{n}$ has a solution.

If $x$ is a solution, then $x^2\equiv b\pmod{m}$ and $x^2\equiv b\pmod{n}$. It follows that $x^2\equiv b\pmod{mn}$, and therefore $b$ is a quadratic residue modulo $mn$.

We leave the other direction to you.

André Nicolas
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  • Thank you. This is what I was doing, but was stuck on the fact that s and t have to be relatively prime, is that true? – Annie Nov 13 '14 at 23:14
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    No, they do not have to be relatively prime. The moduli $m$ and $n$ have to be relatively prime for the Chinese Remainder Theorem, but we are told that they are. – André Nicolas Nov 13 '14 at 23:17