# 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 $x^4 - 2y^2 = 1$?

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Well, $x = 1$, $y=0$ is a solution. Examining mod 4 shows that $x$ has to be odd and $y$ has to be even. I am sure other residues can produce more limitations. –  Arthur Dec 22 '12 at 13:13

## 1 Answer

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 $1$.

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