I watched this YouTube video that is mainly about primes as factors of the Fibonacci numbers. It notes that every Fibonacci number after F(12) has a new prime factor not previously seen, and this new prime factor will also divide all multiples of that index. Why this works is demonstrated for every third Fibonacci number being even, and extending this argument to other primes is left as an exercise for the viewer. However in the case of 2, the entire cycle repeats every 3 steps. That is to say that mod 2, every Fibonacci number is the same as the one 3 prior.
Whereas for 5, for example, every 5th Fibonacci number is divisible by 5, but the cycle doesn't fully repeat until every 20. I understand why this larger cycle exists, that mod 5, every F(n) is equivalent to F(n-20), but not why the smaller cycle exists within it. I have found many examples where the Fibonacci numbers are cyclical mod some prime only 2 or 4 times the first appearance. For example again, F(11) is 89. For any n divisible by 11, F(n) will have 89 as a factor. And mod 89, F(n) is equivalent to F(n-44). For other primes, the Fibonacci numbers repeat after 2 cycles, such as 47, which first appears as a factor in F(16) and mod 47, F(n)=F(n-32).
Why even with these larger cycles do the primes reappear as factors so periodically along the way?