Are any factors of Lucas numbers divisible by a Fibonacci number greater than three?

Question

The congruence relation for Fibonacci and Lucas numbers is stated:
If F_n > 3 is a Fibonacci number then no Lucas number is divisible by F_n.

However, does this apply to the factors as well?

Answer 1

And so it is.

This directly implies that Fibonacci numbers are not the square of any prime, nor the power of a single distinct factor. p>5

First refer to the Fibonacci-Wieferich open problem, and also D. D. Wall's theorem.

Let q, be a prime p minus the Legendre symbol, ie q = p-(p|5) Let F(u(p)), be the smallest Fibonacci number divisible by the prime p. Let L(n), be a Lucas number.

Now since F(q) = F(q/2) * L(q/2), and we know that the gcd < 3, thus p^2 would have to divide either one, or the other.

It cannot divide F(q/2), because that entry point multiplied by p, is coprime to q, ie q does not divide u(p)*p, when u(p) < q, and we know that F(u(p)*p) is always divisible by p^2, but it shares no factors with F(q), F(q/2), etc, thus p^2 does not divide F(q/2).

The entry point must occur at L(q/2) and no earlier, no later, if it is to occur at all. For the same reason respectively, while u(p) < q/2, then q/2 does not divide u(p)*p. This particular application is respective to Lucas numbers, which are similar to Fibonacci numbers.

F(p^2) | F(L(q/2)) iff p^2 | L(q/2)

If there was a Fibonacci number that was a power of only one distinct prime factor, then it would always fail to divide the Lucas number, and so Fibonacci numbers with only one distinct factor are always square free, without multiplicity.

My proof goes on to modify Wall's theorem to include only odd exponents as the only possible true entry points congruent to the primitive quotient.

One result is that: p^2 |L(q/2) iff p^3 | L(q/2)