
> 
> Let $x,y \in \mathbb{Z}$ satisfying $3y^2 = 4x^3 - 1$. Does it follow
>  that $x = 1$ and $y = \pm 1$?
> 
> 
> 


Wolfram Alpha says that the answer is positive, but I am not so satisfied with an answer by a computer program since it is (most of the time) not accompanied by a proof, and even if it is, such a proof may be rather long, not illuminating, and difficult to verify.


What I would like to see most is a (reasonably short and complete) proof (or a counterexample).


I have used unique factorization in $\mathbb{Z}[\frac{-1 + \sqrt{-3}}{2}]$to show that it suffices to prove that the only $a,b \in \mathbb{Z}$ satisfying $a^3 + b^3 -6a^2b + 3ab^2 + 1 = 0$ are $(-1,0), \ (0,-1), \ (1,1)$ and verified this using Wolfram Alpha, but I am not sure whether this gets me any closer to a proof.



Are there integer solutions to $3y^2 = 4x^3-1$ other than $(1,1)$ and $(1,-1)$?
The projective form of your curve is $3y^{2} z = 4x^{3} - z^{3}$. This has three obvious points: $(1 : 1 : 1)$, $(1 : -1 : 1)$, and $(0 : 1 : 0)$.


Your curve is isomorphic over $\mathbb{Q}$ to the Fermat cubic, $x^{3} + y^{3} = z^{3}$. This latter curve has only three rational points on it: $(1 : -1 : 0)$, $(1 : 0 : 1)$ and $(0 : 1 : 1)$, and so your curve only has the three rational points named. (If you want to read a proof of the $n = 3$ case of FLT, you can read Euler's proof, which is essentially the same as the one in Hardy and Wright's ``An Introduction to the Theory of Numbers,'' see pages 193-195.)


