, 2017

Solution:

First, note that $5$ is not a quadratic residue modulo $127$. We are looking for the period of the Fibonacci numbers $\bmod\ 127$. Let $p=127$. We work in $\mathbb{F}_{p^{2}}$ for the remainder of this proof. Let $\alpha$ and $\beta$ be the roots of $x^{2}-x-1$. Then we know that $F_{n}=\frac{\alpha^{n}-\beta^{n}}{\alpha-\beta}$. Note that since $x \rightarrow x^{p}$ is an automorphism and since automorphisms cycle the roots of a polynomial we have that $\alpha^{p}=\beta$ and $\beta^{p}=\alpha$. Then $F_{p}=\frac{\alpha^{p}-\beta^{p}}{\alpha-\beta}=-1$ and $F_{p+1}=\frac{\alpha \beta-\beta \alpha}{\alpha-\beta}=0$ and similarly we obtain $F_{2p+1}=1$ and $F_{2p+2}=0$. Thus since $2p+2$ is a power of $2$ and since the period does not divide $p+1$, we must have the answer is $2p+2=256$.