GCD of Three Fibonacci Numbers

Question

I'm trying to prove that the $\gcd$ of Fibonacci numbers is $1$, i.e. $\gcd(F_{i}, F_{i+3}, F_{i+k}) = 1$, where $i \neq 0 \pmod 3$.

I've tried to combine the formulae of $\gcd(F_{m}, F_{n}) = F_{\gcd(m,n)}$ with the usual way to compute the $\gcd$ of three numbers: $\gcd(a, b, c) = \gcd(a, \gcd(b, c))$. However, this usually produces issues due to the restriction of $i \neq 0 \pmod 3$.

Is there any way to circumvent this restriction?

Answer 1

$\gcd(F_i,F_{i+3}) = F_{\gcd(i,i+3)} = F_{\gcd(i,3)}=F_1=1$

(The term $F_{i+k}$ is irrelevant)