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. :)

  • $\begingroup$ Well, then you should also remove the part asking for the proof of their non-existence-if-they-don't in your edit. Tricky among other things. $\endgroup$
    – ashley
    Dec 22, 2012 at 9:59
  • $\begingroup$ Oh, yikes. I'm sorry. You're right. It was misleading. I think it's okay now, though. $\endgroup$
    – AlpArslan
    Dec 22, 2012 at 15:01

4 Answers 4


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).


$a=5, b=11$ is one satisfying it. I don't think this is the only pair.

  • $\begingroup$ You didn't read the question, did you? $\endgroup$ Dec 22, 2012 at 9:57
  • 1
    $\begingroup$ Oh, i very well did. but you seem to have missed the first version of his Q. he added the line starting "Ah well.. " after my answer appeared here-- see my comment up there. he is getting a flag. he is a cheat. $\endgroup$
    – ashley
    Dec 22, 2012 at 10:15
  • 1
    $\begingroup$ I think it's important not to write offensive or even "tough" commentaries towards other people since, unfortunately, it is a very common practice over here to change the original question, either by the OP or by someone else. Ashley's answer correctly addressed the OP. +1 $\endgroup$
    – DonAntonio
    Dec 22, 2012 at 10:24
  • $\begingroup$ I apologize. I wrongly interpreted the time of your answer. It's odd that a question can be edited for a while without leaving any mark. $\endgroup$ Dec 22, 2012 at 14:33
  • $\begingroup$ Oh, I never added the line mentioning 5 and 11. That was always a part of the original question. But I concede, my original question was misleadling-ly phrased. $\endgroup$
    – AlpArslan
    Dec 22, 2012 at 15:05

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.

  • $\begingroup$ Hey, thanks for that. Now, I've never seen it done that way though. How did you derive the expressions for x, y and z? $\endgroup$
    – AlpArslan
    Dec 23, 2012 at 17:07
  • $\begingroup$ You factor $x^2+y^2=(x+iy)(x-iy)$. The two factors are relatively prime in $Z[i]$. Therefore $$x+iy=(m+ni)^3$$. $\endgroup$
    – PAD
    Dec 23, 2012 at 18:23
  • 1
    $\begingroup$ The above derivation is incorrect, for example if $x=y=2$ then $x+iy$ and $x-iy$ are not relatively prime. $\endgroup$ Jul 21, 2014 at 14:54

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, $b^2-4=a^3-8=(3k+2)^3-8$


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.

  • 1
    $\begingroup$ But what if each factor on the left-hand side shares a factor with two or more factors on the right-hand side? I don't believe your proof considers that possibility. $\endgroup$ Nov 13, 2014 at 22:14

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