A prime problem in Fibonacci sequence

Question

This problem can be a little too wild but have at it anyways:

Prove that for every positive integer $n$ $\geqslant$ $7$ , $f_{n+1}$ has a prime divisor that doesn't divide $f_{n}-1$ where $f_{n}$ is the n-th number in the fibbonachi sequence.

Answer 1

Starting from Cassini's identity:

$$F_{n+1}F_{n-1}-F_{n}^2=(-1)^n$$

and adding $1$ to both sides we get:

$$F_{n+1}F_{n-1}-(F_{n}+1)(F_{n}-1)=(-1)^n+1$$

When $n$ is even, the RHS is $2$. Since the only fibonacci numbers which are a power of $2$ are $F_3=2$ and $F_6 = 8$ (see e.g. here), then $F_{n+1}$ ($n+1 \gt 6$ for $n \ge 7$) has one or more odd prime factors, but none of them can divide $F_n-1$, otherwise it would divide $2$.

When $n$ is odd, the RHS is $0$. $F_n+1$ and $F_n-1$ cannot have any common factor other than $2$. $F_{n+1}$ and $F_{n-1}$ are coprime. Then, if all prime factors of $F_{n+1}$ divide $F_{n}-1$ we have:

$$F_{n+1}=2^{k}(F_n-1)$$

for some $k \gt 0$. If $k = 1$ we have $F_{n+1}=F_n+F_{n-1}=2F_n-2$ and then $F_{n-1}=F_n-2$, impossible, otherwise still an absurd:

$$F_n = \frac{F_{n-1}+2^k}{2^k-1} \lt F_{n-1}$$

because:

$$F_{n-1} \gt \frac{2^k}{2^k-2}$$