How find this sum mod $2017?$

Question

Define Fibonacci Sequence $\{F_{n}\}$ such $$F_{0}=0,F_{1}=1,F_{n+2}=F_{n+1}+F_{n}$$ Find the $$F_{0}+F_{1}+F_{2}+\cdots+F_{2016}\equiv? \pmod {2017}$$

since $2017$ is prime number and $F_{p}\equiv\left(\frac{p}{5}\right) \pmod p$

Answer 1

By Binet's explicit formula$^{(*)}$, if $p>5$ is a prime such that $5$ is a quadratic non-residue $\!\!\pmod{p}$ we have $F_{p-1}\equiv 1\pmod{p}$ and $F_p\equiv -1\pmod{p}$, hence $F_{p+1}\equiv 0\pmod{p}$. Since $$ F_0+F_1+\ldots+F_{p-1} = F_{p+1}-1$$ is straightforward to prove by induction, $$ F_0+F_1+\ldots+F_{2016}\equiv\color{red}{-1}\pmod{2017}$$ readily follows. Do I need to show $(*)$, too?

Answer 2

The sequence $F_0,F_1,F_2,\ldots$ can be written as a sum of two geometric sequences (see the closed form expression on wikipedia). You can then use the sum-formula for geometric series, and at the end you find that the answer is $-1$ mod 2017.

Answer 3

Hint Prove by induction that $$F_0+F_1+..+F_{n}=F_{n+2}-1 $$