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How can I prove that $x^2-y^3=1$ has only one solution in the domain of $N^+$?
Namely $x=3, y=2$.
My try:
$x^2-1=y^3 => (x-1)(x+1)-1=y^3$.
$gcd(x-1,x+1)$ is 2 if $x$ is odd and $1$ if $x$ is even.
In the case x-even the only possible solution would be if $x=1$ but $y>0$. Therefore $x$ is odd.
Let $x=2p+1$. We can rewrite the equation as $4p(p+1)=y^3$.
$gcd(p,p+1)=1$, therefore $p(p+1)=2$. Therefore $x=3$.
Is that correct ?

Teodor Dyakov
  • 698
  • Rewrite it as $x^2 - 1 = y^3$. Note that $x^2 - 1 = (x+1)(x-1)$. Continue from there. – quid Aug 03 '16 at 22:29
  • $gcd(x-1,x+1)$ is 2 if $x$ is odd and $1$ if $x$ is even. In the case x-even the only possible solution would be if $x=1$ but $y>0$. Therefore $x$ is odd. Let $x=2p+1$. We can rewrite it as $4p(p+1)=y^3$. $gcd(p,p+1)=0$ therefore $p(p+1)=2$.Therefore $x=3$ Is that orrect ? – Teodor Dyakov Aug 03 '16 at 22:38
  • I'm stuck in the case when $x$ is odd. – Asinomás Aug 03 '16 at 22:43
  • I made a mistake $gcd(p,p+1)=1.$ – Teodor Dyakov Aug 03 '16 at 22:48
  • This is a Mordell's equation with a simple but slightly long solution. See pages 7-8 of this paper: http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/mordelleqn1.pdf – user236182 Aug 03 '16 at 22:57
  • You can find several arguments at http://mathoverflow.net/questions/39561/ which one suits your needs depends a bit on your background. – quid Aug 03 '16 at 23:07

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