If $x$ is a square modulo two primes, then it is a square modulo their product > $a, b$ be integers, $p, q$ primes. If $x \equiv a^2 $ (mod $p$) and $x \equiv b^2$ (mod $q$), then $x \equiv c^2$ (mod $pq$) for some interger $c$. I attempted to use Chinese Remainer Theorem, but did not get useful forms, so how to prove it?

By the Chinese Remainder Theorem, there exists an element $c$ such that $$c \equiv a \pmod{p} \\\ c \equiv b \pmod{q}$$

Then $$x \equiv c^2 \pmod{p}\\\ x\equiv c^2 \pmod{q}$$