Find natural number solutions to $x^2-2y^2=2$

Question

One solution I found is $x=2, y=1$. But my question is:

Is this the only natural solution?

Answer 1

Suppose $(a,b)$ is a solution of the negative Pell equation $x^2-2y^2=-1$ (many $x$ values of solutions are listed at OEIS A02315). Then a solution of $x^2-2y^2=2$ can be obtained as below:

$$a^2-2b^2=-1$$

$$-2a^2+4b^2=2$$

$$(2b)^2-2a^2=2$$

Answer 2

Hint: if $(x,y)$ is a solution, then so is $(3x+4y, 2x+3y)$.

Answer 3

With an equation of the form $x^2-Dy^2=N$, whenever $(x, y)$ is a solution, $(ax+by, cx+dy)$ is a solution for some $a, b, c, d$ which depend on the coefficients of the original equation but not on $x$ or $y$. Here's one way to find those coefficients $a, b, c, d$ in the OP's case:

\begin{align*} (ax+by)^2-2(cx+dy)^2 &= 2\\ \implies (a^2-2c^2)x^2+2(ab-2cd)xy+(b^2-2d^2)y^2 &= 2\\ \implies a^2-2c^2 &= 1;\\ ab &= 2cd;\\ b^2-2d^2 &= -2. \end{align*} So if you can find a solution to $a^2-2c^2 = 1$, you may take $d=a$ and $b=2c$. $a=3, b=2$ works, so we get the recurrence which Robert Israel pointed out.

More generally, if the original equation is $x^2-Dy^2=N$, \begin{align*} (ax+by)^2-D(cx+dy)^2 &= N\\ \implies (a^2-Dc^2)x^2+2(ab-Dcd)xy+(b^2-Dd^2)y^2 &= N\\ \implies a^2-Dc^2 &= 1;\\ ab &= Dcd;\\ b^2-Dd^2 &= -D\\ \implies d^2-\frac{b^2}{D} &= 1\\ \implies d^2-D\Big(\frac{b}{D}\Big)^2&=1 \end{align*} So again, if you can find a solution to $a^2-2c^2 = 1$, you may take $d=a$ and $b=Dc$.