How can I conclude from the given relation that consecutive Fibonacci numbers are relatively prime?

Question

I've proved the relation $F_nF_{n-1}=F_n^2-F_{n-1}^2+(-1)^n,n\ge2$ but I'm not getting how can I conclude that consecutive Fibonacci numbers are relatively prime from this relation?

Answer 1

Let $d>1$ be a common divisor of $F_n$ and $F_{n-1}$ then $d^2$ divides the left-hand side of $$F_nF_{n-1}-F_n^2+F_{n-1}^2=(-1)^n$$ which is a contradiction because the right-hand side is $(-1)^n$.

Answer 2

Say $d$ is their $\gcd$, then $F_n = da$ and $F_{n-1}=db$. Now put this into equation and you get $$d^2(ab-a^2+b^2)= (-1)^n$$ Now what can you say for $d$?