Brazilian Mathematical Olympiad

We have $(x - k)^2 + x^2 = 1$, so $2x^2 - 2k x + k^2 - 1 = 0$. This has 0, 1 or 2 real solutions according as $k^2 > 2$, $k^2 = 2$ or $k^2 < 2$.

$k = \sqrt{2}$ gives $x = \frac{1}{\sqrt{2}}$, $y = \frac{1}{\sqrt{2}}$ or $y = -\frac{1}{\sqrt{2}}$, so there are two solutions to the original set. Similarly for $k = -\sqrt{2}$.

If $|k| < \sqrt{2}$, then $x = \frac{k \pm \sqrt{2 - k^2}}{2}$, $y = \pm x$. That gives 4 solutions unless one of the values of $x$ is 0, in which case we get 3 solutions. So $k = 1$ gives $(x, y) = (0, 0)$ or $(1, 1)$ or $(1, -1)$. $k = -1$ gives $(x, y) = (0, 0)$ or $(-1, 1)$ or $(-1, -1)$.

If $|k| > \sqrt{2}$, there are no solutions.