Are Fibonacci numbers with a square prime index always divisible by $F_p$?

Question

I am doing some research on sequences and I need some help. The sequence of $F_{p^2}$ seems sort of different. It seems that because the index only has one distinct prime factor, as a result the only repeated factor is $F_p$. Otherwise all the other factors are primitive factors.

I don't know this is true or not. If so, is there proof?

Answer 1

It is well known (see here ) that $\gcd(F_m,F_n)=F_{\gcd(m,n)}$ and so $\gcd(F_p,F_{p^2})=F_p.$