Diophantine equation exercise

Question

Prove that the diophantine equation $x^4-2(y^2)=1$ has only 2 solutions. Any hint on how to start and what to do .. I do not have a lot of experience on non linear diophantine equations and do not know how to approximate them. So far I can see that for $y=0$, $x=1$ is one solution.

Answer 1

There is the following result of J. H. E. Cohn concerning the Diophantine equation $$ x^4-Dy^2=1. $$ Theorem (Cohn 1997): Let the fundamental solution of the equation $v^2 − Du^2 = 1$ be $a + b\sqrt{D}$. Then the only possible solutions of the above equation are given by $x^2 = a$ and $x^2 = 2a^2 − 1$; both solutions occur in only one case, $D = 1785$.

For $D=2$ we have $a=1$ so that $x^2=1$ and $y=0$. For the proof of Cohen's result, see here.
For an easy proof for the case of $D=2$ see Solutions of Diophantine equations in Natural numbers.