Is this sequence based on the Fibonacci numbers a prime generator?

Question

Consider the Fibonacci sequence $\text{fibo}(n)$ and the fractions

$$ A(n) = \frac{\text{fibo}\left(n^2\right)}{\text{fibo}(n)^2} = \frac{b(n)}{c(n)},$$

where the fractions $\frac{b(n)}{c(n)}$ are in reduced form.

Now, it appears that for all $n > 4$:

1. If $c(n) = 1$, then $b(n)$ is of the form $2^q p$ where $p$ is a prime.

2. If $c(n) > 1$, then either $c(n) $ or $b(n)$ is of the form $2^q p$ where $p$ is a prime.

So this formula generates a prime for every $n$.

A weaker conjecture is that this holds for the cases when $n$ is a prime.

Are these conjectures true? How can we prove them?

Answer 1

For $n=19$, a prime, we have $$\frac{F_{n^2}}{\left(F_n\right)^2}=\frac{297695973435970582594631907579321477163892921001085193295076858332955181}{4181}.$$

The numerator and denominator are odd, the former is divisible by $6567762529$, and the latter is divisible by $37$. So, both of your conjectures are false.