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. :)
 A: Update: This is a Mordell equation and from the ref E_-00004 from this table all the known solutions were provided here :
E_-00004: r = 1   t = 1   #III =  1
          E(Q) = <(2, 2)>
          R =   0.4503206856
           4 integral points
            1. (2, 2) = 1 * (2, 2)
            2. (2, -2) = -(2, 2)
            3. (5, 11) = -2 * (2, 2)
            4. (5, -11) = -(5, 11)

Fine references about this kind of problems are :


*

*de Jonquières' 1878 paper (french)

*Conrad's paper for simple impossibilities proofs but not only since the theorem $3.3$ is the proof that no other solutions in $\mathbb{Z}$ exists for your equation.


In Jonquières' paper one finds "D'autres fois, mais rarement, on démontre qu'il n'existe qu'une seule solution. C'est ce qui a été fait par Fermat, Euler et Legendre pour les équations $x^3-2=y^2$, $x^3-4=y^2$...¨. 
This means that no other solution exist and that this was proved by one or more between Fermat, Euler and Legendre (I'll search references).
A: $a=5, b=11$ is one satisfying it. I don't think this is the only pair. 
A: 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.
A: 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.
