On proving quadratic residues have 4 square roots I'm trying to write up a proof of "if a is a quadratic residue modulo N then it has 4 separate square roots". N is the product of two primes and I have to consider the multiplicative group of the integers. 
My reasoning through this is to attempt proof by contradiction (Which I believe is by proving "If a is a quadratic residue and it doesn't have 4 separate square roots") But I seem to be stuck at this primary step. Which method would be best towards proving the primary statement? Any tips on how best to prove it mathematically? 
 A: It suffices to find four square roots of $a=1$ modulo $N=pq$, the product of two primes.  This case is necessary (since $a=1$ is a quadratic residue) and sufficient (since given those four roots we can generate four roots for any quadratic residue by multiplying with them; see below at end of post).
But first a caveat.  It must be assumed that $p,q$ are distinct odd primes.  To see this one can consider counterexamples in $\mathbb{Z}/6\mathbb{Z}$ and $\mathbb{Z}/9\mathbb{Z}$.
Now $p,q$ are coprime, so by the Chinese remainder theorem for any of the four choices of sign in $x=\pm 1 \pmod{p}$ and $x=\pm 1 \pmod{q}$, there exists a unique solution $x$ modulo $pq$.
If we want to be more explicit about this, then suppose $cp + dq = 1$ by their coprimality, where evidently $p \not| d$ and $q \not| c$.
It follows that not only $1$ and $-1$ are square roots of $1$ modulo $pq$, but also $cp-dq$ and $-cp +dq$ are square roots of unity:
$$ (cp-dq)^2 \equiv (cp+dq)^2 \pmod{pq} $$
It remains only to verify these four roots are actually distinct modulo $pq$. By odd parity of $pq$, additive inverses $1,-1$ are distinct, as are $cp-dq$ and $-cp+dq$.
Suppose for contradiction that $cp-dq \equiv 1 \pmod{pq}$.  Then $cp-dq \equiv cp+dq \pmod{pq}$, and thus $2dq \equiv 0 \pmod{pq}$.  It would follow from $pq|2dq$ that $p|d$ and this is a contradiction.  The other possibility, that $cp-dq \equiv -1 \pmod{pq}$, leads to a similar contradiction involving $q|c$.
With this the verification that $\pm 1, \pm cp-dq$ are four distinct square roots of $1$ modulo $pq$ is complete.
Then supposing $a$ is any quadratic residue mod $pq$, and say $x^2 = a$ for nonzero $x$.  It follows that $\pm x, \pm (cp-dq)x$ are four distinct square roots of $a$ modulo $pq$.
A: Start by showing that if $N$ is prime then it has 2 square roots. Then use the chinese remainder theorem for $N$ the product of two primes.
