I want to show that for $d= m^2+2$ the equation

$$x^2-dy^2 = -2$$

has infinitetly many integer solutions. By trying out one can see that $x=\pm m, y=\pm 1$ are solutions, but how do we know that there exist infinitely many solutions?

I have already studied the similar equation $x^2 -dy^2=1$ and shown that for that a fundamental solution is given by $(m^2+1, m)$.


1 Answer 1


If $d\in\Bbb N$ is not a square, and $$x^2-dy^2=k\tag1$$ has some integer solution for $k\ne 0$ then $(1)$ has infinitely many integer solutions. This is because Pell's equation $$x^2-dy^2=1\tag2$$ has infinitely many integer solutions, and if $(x_1,y_1)$ is a solution to $(1)$ and $(x_2,y_2)$ is a solution to $(2)$ then $(x_1x_2+dy_1y_2,x_1y_2+y_1x_2)$ is also a solution to $(1)$.

  • 1
    $\begingroup$ Thank you! Do you mean "is also a solution to $(1)$" in the end? $\endgroup$
    – mathcourse
    Feb 11, 2018 at 11:42
  • 2
    $\begingroup$ How did you come up with looking at $(x_1x_2 + dy_2y_2, x_1y_2 + y_1x_2)$? $\endgroup$ Feb 11, 2018 at 14:15
  • $\begingroup$ There is a typo; please fix it.. On the last line, it should say $dy_1 y_2$ instead of $d y_2 y_2$. $\endgroup$
    – evaristegd
    Jun 24, 2019 at 23:40

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