Solutions of Diophantine equations in Natural numbers The one of solution of $x^4 - 2y^2 = -1$ is $x = 1$ and $y = 1$. However, the solution $(1, 1)$  of $x^4 - 2y^2 = 1$ is failed. We know $x = 1$ and $y = 1$ is small integers and we can check by trail method. In case more solutions are existing or not how to check? What are the solutions of $x^4 - 2y^2 = 1$?    

The only integer solutions of $x^4-2y^2=1$ are $x=\pm 1$, $y=0$.
For $x$ must be odd. Rewrite our equation as $(x^2-1)(x^2+1)=2y^2$. The greatest common divisor of $x^2-1$ and $x^2+1$ is $2$. Since $x^2+1$ has shape $8k+2$, it follows that $x^2+1=2s^2$, and $x^2-1=t^2$ for some integers $s$ and $t$.
The only solution in integers of $x^2-1=t^2$ is $x=\pm 1$, $t=0$, since $1$ and $0$ are the only perfect squares that differ by $1$.
