# Curves with infinitely many integral points consecutive Fibonacci numbers

I suspect the Granville-Langevin conjecture implies this.

Let $F(x,y)=0$ be a curve with infinitely many integral points $(F_n,F_{n+1})$.

Is it true that either $x^2+x y - y^2 -1$ or $x^2+ x y -y ^2 +1$ divides $F(x,y)$?

Probably this can be generalized to integral points $(F_{f(n)},F_{g(n)})$.

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You don't need any conjectures for this. If two curves have infinitely many points in common then they have a common component (this is a basic fact of the Zariski topology). As $x^2+xy-y^2-1$ and $x^2+xy-y^2+1$ are irreducible, one of them must divide $F$.
Thank you. Why say $(F_{2^n},F_{2^n+1})$ are not on some third curve since there are two known curves? – joro Jan 2 '13 at 12:37
The infinite sequence of points $(F_{2^n},F_{2^n+1})$ is on many curves $F(x,y)=0$. But all these curves have infinitely many points in common with $x^2+xy-y^2+1=0$. So the polynomials $F(x,y)$ and $x^2+xy-y^2+1$ have a non-constant common factor. Thus $x^2+xy-y^2+1$ divides $F$. – Siksek Jan 2 '13 at 12:54