Exercise 5.3.17* (Solutions of $x^4 - 2y^2 = 1$)

Proof. Suppose that the integers $x,y$ satisfy $x^4-2y^2=1$. Then $x^4=2y^2+1$ is odd, so $x$ is odd. Moreover $2y^2=x^4-1=(x^2-1)(x^2+1)$. Since $2\mid x^2-1$ and $2\mid x^2+1$, then $4\mid 2y^2$, thus $2\mid y^2$, so $y$ is even. So we can write the equation $x^4-2y^2=1$ as

We put $z=\frac{y}{2},s=\frac{x^2-1}{2},t=\frac{x^2+1}{2}$. Then $z,u,v$ are integers, and $2z^2=\mathit{st}$, where $t=(x^2+1)∕2>0$ and $s\ge 0$, because $\mathit{st}=2z^2$.

If $d\in \mathbb{Z}$ satisfies $d\mid s$ and $d\mid t$, then $d\mid t-s=\frac{x^2+1}{2}-\frac{x^2-1}{2}=1$. This proves that $s\wedge t=1$. By Problem 5, the equality $2z^2=\mathit{st}$ implies the existence of integers $u,v$ such that $s=u^2$ and $t=2v^2$, or $s=2v^2$ and $t=u^2$, so

• In the first case,

  $$1=4v^2-x^2=(2v-x)(2v+x),$$

  therefore

  $$\begin{array}{lll}\hfill \{\begin{array}{cc}2v-x\hfill & =1,\hfill \\ 2v+x\hfill & =1,\hfill \end{array}\text{or}\{\begin{array}{cc}2v-x\hfill & =-1,\hfill \\ 2v+x\hfill & =-1,\hfill \end{array}& & \hfill \end{array}$$

  which gives $x=0$. This is impossible because $x$ is odd.

• In the second case,

  $$1=x^2-4v^2=(x-2v)(x+2v),$$

  therefore

  $$\begin{array}{lll}\hfill \{\begin{array}{cc}x-2v\hfill & =1,\hfill \\ x+2v\hfill & =1,\hfill \end{array}\text{or}\{\begin{array}{cc}x-2v\hfill & =-1,\hfill \\ x+2v\hfill & =-1,\hfill \end{array}& & \hfill \end{array}$$

  which gives $x=1$ or $x=-1$, and $2y^2=x^4-1=0$, thus $y=0$.

If $x$ and $y$ are integers for which $x^4-2y^2=1$, then $x=\pm 1,y=0$. □