Special privilege enjoyed by Elliptic Curves with Complex Multiplication

I have never really known the answer to this question either. There has to have been a motivation for this. All I can think is that in having CM, the lattice associated the the curve (if working over $\mathbb{C}$) has more symmetry, so that the geometry dictates much more towards the arithmetic.

I am only just studying this stuff myself so I am no expert. The main use of CM that I know of (but have never read much into) is that given a number field $K$, you can use an elliptic curve with CM by $\mathfrak{O}_K$ to get the Hilbert class field of $K$, by simply adjoining the $j$-invariant of the curve to $K$. Knowing the Hilbert class field is a very powerful thing...for example it tells you great things about representations of primes by binary quadratic forms.

For example for certain $n$ (namely those that are square-free and not $3$ mod $4$) we may study when a prime $p$ can be written as $x^2+ny^2$ for integers $x,y$. Now algebraically this is the same as $p$ splitting, into principal prime ideals, in the ring of integers $\mathbb{Z}[\sqrt{−n}]$ of the number field $\mathbb{Q}(\sqrt{−n})$. Now the principal part is important here because when we involve the Hilbert class field, this is the same as saying that the ideal splits completely in this field. So $p$ can be written in this form iff $p$ splits completely in the Hilbert class field of $Q(\sqrt{n})$

But as I just said, you can get this Hilbert class field by finding an elliptic curve with CM by $\mathbb{Z}[\sqrt{−n}]$ and adjoining its $j$-invariant to $\mathbb{Q}(\sqrt{-n})$. So really, once this has been accomplished, the minimal polynomial $f(x)$ of the $j$-invariant over $\mathbb{Q}$ gives a criterion for the possible $p$'s that can be written as $x^2+ny^2$ (for those $n$'s we restricted to).

Basically (for $p$ not dividing $n$ or the discriminant of $f(x)$):

$p = x^2 + ny^2$ if and only if $\left(\frac{-n}{p}\right) = 1$ and $f(x) \equiv 0$ mod $p$ has a solution.