Integers that satisfy $a^3= b^2 + 4$ 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. :)

Observe that 
if $a=3k,b^2=a^3-4=(3k)^3-4\equiv-1\pmod 3$ but $b^2\equiv1,0\pmod 3$
if $a=3k+1,b^2=a^3-4=(3k+1)^3-4=9(3k^3+3k^2+k)-3$ which is divisible by $3,$ but not by $9$
So, $a$ must be of the from $3k+2$
Consequently, $b^2-4=a^3-8=(3k+2)^3-8$
$(b+2)(b-2)=9k(3k^2+6k+4)$
Also, as $(a,b)=1,$ both $a,b$ must be odd $\implies (b+2,b-2)=(b+2,b+2-(b-2))=1$ and $k$ is odd
As $k$ is odd, $(k,3k^2+6k+4)=(k,4)=1$
If $b-2=9k,b+2=9k+4$ and $b+2=3k^2+6k+4\implies 3k^2-3k=0\implies k=0,1$
$k=0\implies b=2,a=2$ (but both $a,b$ are odd)
$k=1\implies b=11,a^3=125,a=5$
If $b+2=9k,b-2=9k-4$ and $b-2=3k^2+6k+4\implies 3k^2-3k+8=0$ whose discriminant is negative.
If $b+2=9,a^3=b^2+4=53$
If $b-2=9\implies b=11,a^3=b^2+4=125\implies a=5$
If $b-2=k\implies b=k+2,b+2=k+4$ and $b+2=9(3k^2+6k+4),27k^2+53k+32=0$ whose discriminant is negative.
If $b+2=k\implies b=k-2,b-2=k-4$ and $b-2=9(3k^2+6k+4),27k^2+53k+40=0$ whose discriminant is negative.
