4
$\begingroup$

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 $p$?

$\endgroup$

1 Answer 1

8
$\begingroup$

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...

$\endgroup$
2
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.