find the equation integer solution $$x^3-1=2y^2$$
I have read this proof:proof, have simple methods? Because this problem is a Olympiad problem. So maybe have other methods to solve it since $x>0.y>0$ so $$(x-1)(x^2+x+1)=2y^2$$ since $(x-1,x^2+x+1)=1$ or$3$.so
(1):if $x-1=2u^2,x^2+x+1=v^2,y=uv,u>0,v>0$,then wehave $$(2v)^2=(2x+1)^2+3\Longrightarrow (2v-2x-1)(2v+2x+1)=3$$ then $$\begin{cases} 2v+2x+1=3\\ 2v-2x-1=1 \end{cases},~~\begin{cases} 2v+2x+1=1\\ 2v-2x-1=3 \end{cases}$$ this two case has no solution $x>0$
(2): $$x-1=6u^2,x^2+x+1=3v^2,y=3uv,u,v>0$$
then $$(2v)^2-3(4u^2+1)^2=1 $$ it's pell equation, maybe this following I can't work. Thanks for your help!
As suggested in the proof you link, it is easier to pass to $\Bbb{Z}[\sqrt{-2}]$ right from the start:
If $x$ and $y$ are integers satisfying $x^3-1=2y^2$, then $$x^3=1+2y^2=(1+y\sqrt{-2})(1-y\sqrt{-2}),$$ where the two factors on the right hand side are coprime because their sum is $2$ and their product is odd. Then because $\Bbb{Z}[\sqrt{-2}]$ is a unique factorization domain with trivial unit group it follows that $$1+y\sqrt{-2}=(a+b\sqrt{-2})^3,$$ for some integers $a$ and $b$. Expanding the right hand side and comparing coefficients then shows that $$1=a^3-6ab^2=a(a^2-6b^2),$$ which clearly has no integral solutions apart from $(a,b)=(1,0)$, corresponding to $(x,y)=(1,0)$.