Find $q>0$ such that $$[q[qn]]+1=[q^2n],\qquad n=1,2,\dotsc$$ where $[x]$ is the integer-valued function.
In fact, if $q=\frac{\sqrt{5}+1}2$, then $q^2-q=1$, so we can verify $$[q[qn]]+1=[q^2n],\qquad n=1,2,\dotsc$$ and I believe $q=\frac{\sqrt{5}+1}2$ is the only solution of the equation.
Now I have only proved $1.6125\leq q\leq 1.6190$, it still has a long way to go!