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.
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.