# 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$$+4=a^3, such that a and b are coprime? I've already found the case where b=11 and a =5, but other than that? And if there do exist other cases, how would I find them? And if not how would I prove so? Thanks in advance. :) -  Well, then you should also remove the part asking for the proof of their non-existence-if-they-don't in your edit. Tricky among other things. – ashley Dec 22 '12 at 9:59 Oh, yikes. I'm sorry. You're right. It was misleading. I think it's okay now, though. – MWarsi Dec 22 '12 at 15:01 ## 4 Answers Using Gaussian integers it is easy to show that the general solution of$$x^2+y^2=z^3$$is$$x=m^3-3m n^2y=3m^2n-n^3z=m^2+n^2$$If x=2 you get m=\pm 1, \pm2 and then you can solve the problem. -  Hey, thanks for that. Now, I've never seen it done that way though. How did you derive the expressions for x, y and z? – MWarsi Dec 23 '12 at 17:07 You factor x^2+y^2=(x+iy)(x-iy). The two factors are relatively prime in Z[i]. Therefore$$x+iy=(m+ni)^3$$. – Pantelis Damianou Dec 23 '12 at 18:23 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$b+2=3k^2+6k+4\implies 3k^2-3k=0\implies k=0,1k=0\implies b=2,a=2$(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. - 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.

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You didn't read the question, did you? – Karolis Juodelė Dec 22 '12 at 9:57
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. – ashley Dec 22 '12 at 10:15
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 – DonAntonio Dec 22 '12 at 10:24
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. – Karolis Juodelė Dec 22 '12 at 14:33
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. – MWarsi Dec 22 '12 at 15:05