Prove if there exists $[b]\in{Z}_{pq}$ such that $[b]^2=[a]$ in ${Z}_{pq}$, then there are exactly four distinct $[x]\in{Z}_{pq}$

Question

Let $p$ and $q$ be distinct odd primes. Let $a\in{Z}$ with $gcd(a,pq)=1$. Prove that if there exists $[b]\in{Z}_{pq}$ such that $[b]^2=[a]$ in ${Z}_{pq}$, then there are exactly four distinct $[x]\in{Z}_{pq}$ such that $[x]^2=[a]$ in ${Z}_{pq}$.

Any help is appreciated, thanks!

Answer 1

$pq|x^2-b^2=(x+b)(x-b)$, so there are $4$ possibilities: $pq|x+b$, $pq|x-b$, $p|x+b$ and $q|x-b$, $p|x-b$ and $q|x-b$. Since $(2b, pq)=1$ we see these $4$ cases are mutually exclusive. The first two cases give two solutions: $x\equiv -b$ and $x\equiv b$ mod pq; each of the last two cases also gives one solution by the Chinese remainder theorem.