Let $p$ and $q$ be two different prime numbers and $a∈ \Bbb{Z}$ be an integer number such that $a$ is a quadratic residue modulo both $p$ and $q$. Prove that there exists $x∈ \Bbb{Z}$ such that $a≡x^2$ (mod pq).
I'm envisioning a CRT style argument with letting $a=y^2$ (mod p) and $a=z^2$ (mod q) but can't quite make it work.
Edit:
Using the CRT I have let $1 = pr + qs$ and then shown that $x = ypr + zps$ which satisfies both $x≡y$ (mod p) and $x≡z$ (mod q).
I have then tried to show that $x^2≡a$ (mod pq) but haven't got anywhere.
The Chinese Remainder theorem is the way to go. If $y^2\equiv a\pmod p$ and $z^2\equiv a\pmod q$, then by CRT there is $x$ with $x\equiv y\pmod p$ and $x\equiv z\pmod q$. You should be able to show that $x^2\equiv a\pmod{pq}$.