<s>[INST] Prove that diophantine equation has only two solutions. I am looking at the following exercise:
$$\text{Prove that the diophantine equation } x^4-2y^2=1 \text{ has only two solutions.}$$
That's what I thought:
We could set $x^2=k$,then we would have $k^2-2y^2=1 \Rightarrow 2y^2=k^2-1 \Rightarrow 2 \mid k^2-1$
Can I use this fact or do I have to do it in an other way?
EDIT: Or could we do it maybe like that:
$k^2-2y^2=1 \Rightarrow (k-\sqrt{2}y)(k+\sqrt{2}y)=1 \Rightarrow k-\sqrt{2}y= \pm1 \text{ and } k+\sqrt{2}y=\pm 1 \Rightarrow  k-\sqrt{2}y=1 \text{ and } k+\sqrt{2}y=1 \text{ OR } k-\sqrt{2}y=-1 \text{ and } k+\sqrt{2}y=-1$
The second case is rejected,because then we get $k<0$
$$k-\sqrt{2}y=1 \text{ and } k+\sqrt{2}y=1 \Rightarrow k=1 \text{ and } y=0 \text{ OR } k=0 \text{ and } y=\frac{1}{\sqrt{2}} \notin \mathbb{Z}$$
So, $k=1 \Rightarrow x^2=1 \Rightarrow x=\pm 1 \text{ and } y=0$
So,the only two solutions are $(1,0) \text{ and } (-1,0)$
 [/INST] You are doing it right. $x^4 - 2 y^2 = 1 \Leftrightarrow x^4 -1 = 2 y^2 \Leftrightarrow (x-1 ) ( x+ 1 ) ( x^2 +1 ) = 2 y^2$.
Now prove that $x$ is odd, then $x = 2 k +1 $, and show that $k ( k+1) ( 2 k^2 + 2 k + 1)$ is a square, and since each factor is relatively prime, $k$ and $k+1$ are squares, conclude that $k=0$ and $x=1$ and $y=0$.
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