# How find all positive $a^3=b^2+2000000$

Find all positive integer $a$ and $b$,such $$a^3=b^2+2000000$$ This problem is from china Math competition(2013,7.10)

So I think this problm have nice methods,because is from competition. Thank you everyone.

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and I know this problem is call:en.wikipedia.org/wiki/Mordell_curve,and inf.unideb.hu/~pethoe/cikkek/67_MORDELL.pdf –  math110 Jul 24 '13 at 2:19
and I think this prblem have nice methods,because is from competition,Thank you –  math110 Jul 24 '13 at 2:19
The original problem is not to find all solutions, but ask whether it has solution and if it has, is the solution unique? Since $(300,5000)(129,383)$ are two different solutions, we have answered the original problem. To find all solutions is more difficult than it, I doubt there exist a nice and elementary methods. –  Next Jul 24 '13 at 5:57
Here is a way to get it: Let $c=1000=2^3\times5^3,t=\sqrt{-2},$ then $a^3=b^2+2c^2=(b+ct)(b-ct).$ Note that $\mathbb Q(t)$ is a UFD. Denote $d=(b+ct,b-ct)=(b+ct,2ct),$ hence $d\mid 2ct.$ Denote $b+ct=dx,b-ct=dy,(x,y)=1,a^3=d^2xy,$ balabala.. –  Next Jul 24 '13 at 5:59

We are looking for the positive integer solutions of the Mordell-Bachet Diophantine equation $$y^2=x^3-d,$$ with $d=2000000$. This has positive integer solutions (one coming from $y^2=x^3-2$, which is $(3,5)$, yielding $(300,5000)$). To see this, one can proceed as shown in Theorem $3.4$ of K. Conrad's article http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/mordelleqn1.pdf. Note that $x^3=(y+\sqrt{-d})(y-\sqrt{-d})$. The factors on the RHS are relatively prime. If we have unique factorisation, this can be used for a nice proof.