An interesting number theory question for math contest training $x^2+x=2y^3$ (finding integer solutions)

Question

This question is in one of my old notebook and I marked it as a solved problem. However, somehow I can't remember the proof or maybe my solution in the past was wrong. The first thing I tried is to muptiply both sides by 4 and then plus to get the equation to the form: $(2x+1)^2=(2y+1)(4y^2-2y+1)$. There are 2 cases, if $2y+1$ and $4y^2-2y+1$ are coprime then the rest is easy. However, if $gcd(2y+1,4y^2-2y+1)=3$, I have tried multiple ways and failed to finish the proof.
Any help would be greatly appreciated.

Answer 1

Welcome. You can simply solve this equation:

$x^2+x-2y^3=0$

$\Delta=1+8y^3$

$y=1$ ⇒ $\Delta=9$ ⇒ $x=1$ and $x=-2$

I think these are only integer solutions. you can work on $\Delta$ for various values of y.

Answer 2

Multiply by $4$ to get: $$(2x+1)^2-1=(2y)^3$$
It is known since Euler that this equation(check page 6) implies $2y=0, 2, -1$. Therefore, $y=0$ and $x=-1$ or $y=1$ and $x=1$.