Consider the elliptic curve over complex numbers defined by $y^2=x^3+b$, where $re(b)$ and $im(b)$ are both integer-valued.
Is there a way to prove whether or not there exists a point on this curve satisfying the condition $re(x) = 0$, with $im(x)$, $re(y)$, and $im(y)$ all integer-valued?
This is really simple to prove for an elliptic curve over the real numbers- the point exists iff $b$ is a perfect square, and the point does not exist iff $b$ is not a perfect square. But I have no idea how to prove this for an elliptic curve over the complex numbers. I did some arithmetic, to no avail.
${(re(y)+im(y)i)}^2={-im(x)}^3i+re(b)+im(b)i$
${re(y)}^2+2re(y)im(y)i-{im(y)}^2={-im(x)}^3i+re(b)+im(b)i$
$2re(y)im(y)=-{im(x)}^3+im(b)$
${re(y)}^2-{im(y)}^2=re(b)$
I'm probably not aware of the theorem(s) necessary to solve this problem. Any help would be very much appreciated.
For convenience, write $b = a + c i$, $x = u i$, $y = v + w i$, where we want $u, v, w$ to be integers, and $a$ and $c$ are integers. Taking real and imaginary parts, the equation becomes
$$ \eqalign{v^2 - w^2 &= a \cr u^3 + 2 v w &= c\cr} $$ Now the first equation has finitely many solutions: $v^2 - w^2 = (v+w)(v-w)$, so if we factor $a = s t$ where $s$ and $t$ $s \equiv t \bmod 2$, $v = (s+t)/2$ and $w = (s-t)/2$. For each such pair $(v,w)$, we then look at whether $c - 2 v w$ is a cube, and if so take $u = (c - 2 v w)^{1/3}$.