Proving that $y$ is a square mod $p$ and $-y$ is square mod $q$ Given that $p, q \equiv 3 \pmod 4$, neither $y$ nor $-y$ has a square root mod $pq$, and that $y$ is invertible mod $pq$, how would I prove that $y$ is a square mod one of $p, q$ and $-y$ is a square mod the other?
 A: Although the question does not state explicitly that $p$ and $q$ are prime, the notation suggests that they are. Also, the claimed result need not hold if $p$ and $q$ are not prime. So we assume that $p$ and $q$ sre prime.
The number $y$ is invertible modulo $pq$, so neither $p$ nor $q$ divides $y$. First we show that exactly one of $y$ and $-y$ is a quadratic residue mod $p$, and exactly one of $y$ and $-y$ is a quadratic residue of $q$. 
To prove this, we use a Legendre symbol calculation. We have 
$$(-y/p)=(-1/p)(y/p)=-(y/p).$$ (We have $(-1/p)=-1$ because $p\equiv 3\pmod{4}$.) The same argument works for $q$.
We are told that neither $y$ nor $-y$ is a square modulo $pq$. Suppose first that $y$ is a square modulo $p$. Then $y$ cannot be a square modulo $q$, else $y$ would be a square modulo both $p$ and $q$, and therefore (Chinese Remainder Theorem) modulo $pq$. 
Thus if $y$ is a square modulo $p$, then $-y$ is a square modulo $q$.
Similarly, if $y$ is not a square modulo $p$, then $-y$ is a square modulo $p$. If $-y$ were a square modulo $q$, then $-y$ would be a square modulo $pq$, contradicting the given fact that it is not. So $y$ is a square modulo $q$. This completes the proof.
