Let us say we have the sets of quadratic residues $X = \lbrace x^2 \pmod{p}\rbrace$ and $Y = \lbrace y^2 \pmod{q}\rbrace$.
Is there a way to construct the set of quadratic residues $Z_{\beta} = \lbrace x : x \equiv z^2 \pmod{pq} \land x < \beta \rbrace$ without having to go through $X \times Y$ and applying the Chinese Remainder Theorem $|X \times Y|$ times?
Best Answer
You can compute the two specific numbers $e, d \bmod pq$ with the property that
$$e \equiv 1 \bmod p, e \equiv 0 \bmod q$$ $$d \equiv 0 \bmod p, d \equiv 1 \bmod q.$$
Then to find a number congruent to $x \bmod p, y \bmod q$ you just compute $ex + dy \bmod pq$.