Quite straightforward, but it's proving somewhat frustrating for me.
Show that for all integers r,s with 2(r^2)+1= s, one of r and s must be divisble by 3.
If $\displaystyle r$ has form $\displaystyle 3k+1$ then $\displaystyle 2(r^2)+1$ is divisible by $\displaystyle 3$, so $\displaystyle s$ is divisible by $\displaystyle 3$. If $\displaystyle r$ has form $\displaystyle 3k+2$ the same idea applies. Otherwise $\displaystyle r$ has form $\displaystyle 3k$ but then there is nothing to show.