Skip to main content
1 of 16
Jim White
  • 225
  • 1
  • 7

Oh, I think the answer is definitely yes!


Let $\{k \to x,y\}$ be any solution of (), and let $K$ be the set of $k$ for which () has any solution with $0<k<y-1$. In a paper recently recently submitted to Glasnik Matematicki we call these solutions exceptional solutions to ().

One interesting result we obtain is that, if $k \in K$, then $y < (2 - \sqrt{3}k$.

A particular feature of (
) is its symmetry wrt $k$ and $y$. So for any solution $\{k \to x,y\}$ there is a corresponding solution $\{y \to x,k\}$.

It follows that if $k \neq y \pm{1}$, then either $k \in K$ or $y \in K$.

Jim White
  • 225
  • 1
  • 7