TITLE: An interesting number theory question for math contest training $x^2+x=2y^3$ (finding integer solutions)
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: 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 $x=1$.