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 $(1,0) \text{ and } (-1,0)$
 A: 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 $y=0$.
A: $$2y^2=x^4-1=(x^2+1)(x^2-1)=(x^2+1)(x+1)(x-1)$$
Suppose that $2y^2$ has an odd prime factor $p$. Then $p^2$ divides $y^2$, and hence, $p^2$ divides $(x^2+1)(x^2-1)$. Since $p$ can not divide both factors, then $p^2$ divides $x^2+1$ or $p^2$ divides $x^2-1=(x+1)(x-1)$. Thas is, $p^2$ divides $x^2+1$, $x+1$ or $x-1$.
This proves that each factor $x^2+1$, $x+1$ and $x-1$ is a perfect square or the double of a perfect square. And there are $1$ or three of them that belong to the second group. The factor $x^2+1$ is a perfect square only if $x=0$, which gives no solution for $y$.
If there is only one double of a perfect square, then we have that $x+1=u^2$, $x-1=v^2$. But then $u^2-v^2=(u+v)(u-v)=2$, and this is impossible since $u+v$ and $u-v$ have the same parity.
If there are three doubles of squares, then $x+1=2u^2$, $x-1=2v^2$ and $2(u^2-v^2)=2$, that is, $u=\pm1$, $v=0$. This gives $x=\pm1$ and $y=0$, and there are no more solutions.
