0
$\begingroup$

Solve this: $y^2+2=x^3$ or

Prove that $y^2+2=x^3 => (x,y)=(3,\pm 5)$

I know that it could be obvious to some of you by trial and error but I need a methodical approach.

Thanks in advance!

$\endgroup$
6
  • $\begingroup$ One observation is that $y$ is even if and only if $x$ is even, but if $y=2m,x=2n$ we have $4m^2+2 = 8n^3$, which happens if and only if $2m^2+1 = 4n^3$, which cannot be since LHS is odd while RHS is even. Hence, neither $x$ nor $y$ can be even... $\endgroup$
    – gt6989b
    Commented Dec 8, 2014 at 22:53
  • $\begingroup$ @gt6989b Yeah thats a nice observation thanks $\endgroup$
    – L887
    Commented Dec 8, 2014 at 22:55
  • $\begingroup$ Another one is just as easy: since $y^2 \geq 0$, we must have $x>0$, and the problem is symmetric in $y$, i.e. if $(x,y)$ is a solution, then $(x,-y)$ is a solution as well. Hence, let's look for $y \geq 0$ and it suffices to show that $(3,5)$ is a unique solution. $\endgroup$
    – gt6989b
    Commented Dec 8, 2014 at 22:58
  • $\begingroup$ @gt6989b Nice one again but how can you say that this suffices to show that (3,5) is a unique solution? $\endgroup$
    – L887
    Commented Dec 8, 2014 at 23:00
  • $\begingroup$ i am not saying this suffices to show that (3,5) is unique; i said, showing (3.5) is unique suffices to solve the problem, since we now can assume $y>0$... $\endgroup$
    – gt6989b
    Commented Dec 8, 2014 at 23:01

0

You must log in to answer this question.

Browse other questions tagged .