TITLE: Integers that satisfy $a^3= b^2 + 4$
QUESTION [5 upvotes]: Well, here's my question:
Are there any integers, $a$ and $b$ that satisfy the equation $b^2$$+4$=$a^3$, such that $a$ and $b$ are coprime?
I've already found the case where $b=11$ and $a =5$, but other than that?
And if there do exist other cases, how would I find them? And if not how would I prove so?
Thanks in advance. :)

REPLY [3 votes]: Using Gaussian integers it is easy to show that the general solution of 
$$x^2+y^2=z^3$$
is 
$$x=m^3-3m n^2$$
$$y=3m^2n-n^3$$
$$z=m^2+n^2$$
If $x=2$ you get $m=\pm 1, \pm2$ and then you can solve the problem.