Consider Diophantine equation $2x^3+1=y^3$
There are two solutions $$x=(0,-1), \, y=(1,-1) $$
Are there anymore integer solutions for $x$ and $y$?
I've tried the following argument but am not satisfied with it.
Identity: $$[(B^3)(x^6)+3(B^2)(x^3)+3(B)]x^3+1=(Bx^3+1)^3$$
Looks like: $$[ 2 ]x^3+1=( y )^3$$
Let $[(B^3)(x^6)+3(B^2)(x^3)+3(B)]=2$ and $(Bx^3+1)=y$
Since $2$ divides $B$ we get possible values for $B$ are $$B=\{-2,-1,1,2\}$$ Testing the $B$ set in equation $$[(B^3)(x^6)+3(B^2)(x^3)+3(B)]=2$$ we can see that the only possible integer solution is $B=2$ and $x=-1$ and so $y=-1$.
The other solution is trivial $x=0,\,y=1$. So from this argument there are no other integer solutions for $x$ and $y$.
My problem with the above argument is that by me saying:
"Let $[(B^3)(x^6)+3(B^2)(x^3)+3(B)]=2$ and $(Bx^3+1)=y$" I am indirectly saying from the start that there and no other solutions.
What I would have to do is prove that:
- all possible solutions for $y$ are of the form $Bx^3+1$ and
- equation $[(B^3)(x^6)+3(B^2)(x^3)+3(B)]=2$ is true for all possible solutions for $x$
But I cannot prove it.
thank you