We have two numbers that are both a product of two distinct primes, $N<M$ and some message $1 \leq m \leq N-1$.
I have been asked to show that it is possible to find $m$ given $c,d$ where
$m^2 \equiv c \pmod N$ and $m^2 \equiv d \pmod M$
However, currently the only ways I can think of are effectively as difficult as just factorising $N$ or $M$, which I think is not the point of this exercise.
I am aware that there are 4 solutions modulo $N$ and 4 modulo $M$ to this quadratic, but as we can't easily find these 4 roots, I'm not sure that this is actually that helpful. I don't really know what else I might be able to use to solve this.
Using the Chinese Remainder Theorem, we can find the value of $m^2 \pmod{NM}$. But, $m < N < M$, so $m^2 < NM$. In particular, the number $m^2$ is just the same as the value $m^2 \pmod{NM}$, so we can simply take the square root and find $m$.