Consider the set of Fibonacci numbers $F_{i}$ with a prime index $p$, $\mathcal{F}_{\mathbb{P}}$.

The first few numbers in this set are:

$\mathcal{F}_{\mathbb{P}} = \{F_2,\; F_3,\; F_5,\; F_7,\; F_{11},\; F_{13},\; ...\} = \{1,\; 2,\; 5,\; 13,\; 89,\; 233,\; ...\}$

Is every prime factor of these Fibonacci numbers necessarily larger or equal to its index for $p \geq 5$?

Some comments:

Because $\mbox{gcd}(F_{m},F_{n})=F_{\text{gcd}(m,n)}$ for $m,n\ge 1$, a Fibonacci number with a prime index $p$ must necessarily be coprime with every Fibonacci number possessing an index not equal to $1$ or integer multiples of $p$.

Consequently, there are quite a few numbers that cannot be prime factors for each member of $\mathcal{F}_{\mathbb{P}}$, but prime numbers that are not Fibonacci numbers ($7$, for example) can always be potential prime factors based on the above criteria.

A rudimentary search through $p < 50$ shows that no prime factors of $F_p$ are smaller than $p$. Can this be proven/disproven for all $p$?


1 Answer 1


Yes. Consider the matrix $\pmod{p}$

$$ \mathbf{F} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} $$

Note that

$$ \mathbf{F}^n = \begin{bmatrix} F_{n+1} & F_{n} \\ F_{n} & F_{n-1} \end{bmatrix} $$

Now, $\mathbf{F}$ is a member of $\mathrm{GL}_2(\mathbb{F}_q)$, which has order $q(q+1)(q-1)^2$. Thus for prime $q$, if $F_p \equiv 0 \pmod{q}$, then $F_{p-1} \equiv F_{p+1} = f \bmod{q}$, so

$$\mathbf{F}^p = fI$$

which implies(by Fermat's Little Theorem) $$\mathbf{F}^{p(q-1)^2} = (f^{q-1})^{q-1}I = I$$

So by Lagrange's theorem, $p(q-1)^2 \mid q(q+1)(q-1)^2 \implies p \mid q(q+1)$.

You are wondering if prime $q$ can be less than prime $p$. If that is the case, then $q+1 \neq p$, and so $p, q(q+1)$ are coprime, an impossibility.

Note: This is almost magical; it is surprising to me the Fibonacci numbers were that rich in structure...

  • 1
    $\begingroup$ A prime $p>5$ divides $F_{p-1}$ or $F_{p+1}. $ And $F_a$ divides $F_{ab}$ for positive integers $a,b.$ From this it is easy to show that if prime $q$ divides $F_p$ (with prime $p>5$) then $q>p$: The least positive integer $ n$ such that $ q$ divides $F_n$ is $n= p$. But $ n$ is also a divisor of $q-1$ or of $q+1.$ $\endgroup$ May 10, 2018 at 3:32
  • 2
    $\begingroup$ They're rich enough to have their own journal fq.math.ca $\endgroup$
    – qwr
    May 10, 2018 at 3:52

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