# How can we prove that $F_p \in \mathcal {P} \lor Φ?$

I have created a conjecture which I do not know how to prove. The conjecture is written like:

$F_p \in \mathcal {P} \lor Φ$

$p$ = prime number in correspondence to the appearance of a number in the Fibonacci Sequence.

$\mathcal {P}$ = The set of prime numbers in the Fibonnaci Sequence.

$Φ$ = The set of Fibonacci numbers that have $1+$ divisors which cannot be divided into any earlier Fibonacci numbers.

My conjecture, is that if you have a prime-th Fibonacci number, then that number is either going to be prime or have factors of which are not factors of any earlier Fibonacci numbers. Here is a list of the first $10$ Fibonacci numbers.

1) $1$

2) $1$

3) $2$

4) $3$

5) $5$

6) $8$

7) $13$

8) $21$

9) $34$

10) $55$

As we can see, all prime-th Fibonacci numbers in the list $\in$ $\mathcal {P}$ except for the $2^{nd}$ Fibonacci number because $1$ is not considered a prime, but we will consider it a prime to all numbers. Now as we list the following $20$ Fibonacci numbers this time, we will start to notice that not all prime-th Fibonacci numbers are prime. The $19^{th}$ Fibonacci number is $4181$ which is not a prime number.

$4181 = 37 \times 113$ $\therefore$ $4181 \in Φ$ $\because$ it's factors $(37, 113)$ are not factors of any Fibonacci number before $19$.

All prime-th Fibonacci numbers after $4181$ are prime, but then when we come across the $31^{st}$ Fibonacci number being $1346269$, which is also not a prime number.

$1346269 = 557 \times 2417$ $\therefore$ $134629 \in Φ$ $\because$ it's factors $(557, 2417)$ are not factors of any Fibonacci number before $1346269$.

The most common factor I assume to be divided into fibonnaci numbers is $2^3$ where I assume that $2^3 | F_{6 + 12n} : n \in \mathbb{W}$.

Now at first you would believe that $α(F_p \in \mathcal {P}) > α(F_p \in Φ)$ : $α(χ) =$ amount of times $χ$ appears. But as $P_y \rightarrow \infty$ the gap between $(F_p \in \mathcal {P})_y \land (F_p \in \mathcal {P})_{y + 1} \rightarrow \infty$ at a much faster rate the greater the value of $y$ : $y \in \mathbb{Z^+}$ $\therefore$ $α(F_p \in \mathcal {P}) < α(F_p \in Φ)$ the greater the value of $y$, and when $y > 137$ then $α(F_p \in \mathcal {P}) \ll α(F_p \in Φ)$.

$φ$ can be expressed as a limit:

$$\lim_{k \rightarrow \infty} \frac{F_k + 1}{F_k} = φ : k \in \mathbb{Z^+}$$

$\because$ $F_k = F_{k + 1} - F_{k - 1}$ (you can carry out the mathematics from there). However even if I substitute $k$ as $p$, and then substitute $F_p$ with their unique prime factorisation $\iff$ $F_p \in Φ$, it does not get me anywhere.

Is there a way to prove that this conjecture works either via this approach or another? The largest Fibonnaci prime I know in the sequence so far is $F_{9311}, (9311 = prime)$ so if that is a key fact to finding a proof, please provide it in your answer if you have one. It would be much appreciated.

If you do not know what any of the various symbols in this question mean, I will post an edit to this question tomorrow to explain their definition which will be when I see your answers $-$ I have to go to sleep now. G'night.

Your conjecture is equivalent to stating that for all $n < p$, $\gcd(F_p, F_n) = 1$ if $p$ is prime.
That this is true directly follows from the identity $\gcd(F_m, F_n) = F_{\gcd(m, n)}$.
• So if $n < p, F_{gcd(p,n)} = 1 \therefore φ = \frac{F_{gcd(p,n)} + \sqrt 5}{2F_{gcd(p,n)}}$ – George N. Missailidis Jul 24 '17 at 22:22