Prove that for a fixed integer $n >1$, all the solvable congruences $x^2 \equiv a \pmod n$ with $ \gcd(a, n) = 1$ have the same number of solutions.
My try: Let a solution be $x_0$ and let $x$ be any other solution. Then $$x^2\equiv x_0^2\pmod{n}\implies(x-x_0)(x+x_0)\equiv 0 \pmod{n}\implies n \mid (x-x_0)(x+x_0)$$
Now how to proceed? Am I going right?
If you know a bit of group theory, then this is simple.
The map $x \mapsto x^2$ is a homomorphism of $U(n)$, the group of units mod $n$.
The non-empty fibers (that is, the inverse image of elements in the image) are all cosets of the kernel of this map and so have the same number of elements.
If you don't know this bit of group theory, then you can do it explicitly.
If $x^2 \equiv y^2$, then $(xz)^2 \equiv x^2 z^2\equiv 1$, where $yz\equiv 1$. This means that $x \equiv yt$ with $t^2 \equiv 1$.
Conversely, if $x \equiv yt$ with $t^2 \equiv 1$, then $x^2 \equiv y^2$.
This means that set of elements $x$ with $x^2 \equiv a$ has the same number of elements as the set of elements $t$ with $t^2 \equiv 1$.