Use induction to show that $F_{2n} \equiv n(-1)^{n + 1} \pmod 5 \forall n \ge 1$

Question

With $F_n$ being $n^{\textrm{th}}$ Fibonacci number, use induction to show that $F_{2n} \equiv n(-1)^{n + 1} \pmod 5$ for all $n \ge 1$.

I tried hard to prove this identity, but each time I'm failing to adjust the exponent of "$-1$". It would be nice if someone could come up with proof.

Thank you

Answer 1

Hint: Sometimes it is easier to prove something stronger. Prove simultaneously that

• $F_n \equiv 2n \bmod 5$ when $n \equiv 0 \bmod 4$

• $F_n \equiv 1n \bmod 5$ when $n \equiv 1 \bmod 4$

• $F_n \equiv 3n \bmod 5$ when $n \equiv 2 \bmod 4$

• $F_n \equiv 4n \bmod 5$ when $n \equiv 3 \bmod 4$

Answer 2

Hint: Prove that $F_n \equiv 2n \cdot 3^n \bmod 5$.

(This comes from Binet's formula mod $5$, because $3$ is a double root of $x^2-x-1$.)