Find all $n!$ that are products of two Fibonacci numbers

Question

Find all positive integers $n$ such that $n!$ is the product of two Fibonacci numbers

I found this discussion, but it doesn't quite help my problem. Check if the number is a result of multiplying two fibonacci numbers

Answer 1

With the convention that $F_0=0$ and $F_1=1$, for $k>12$ the $k$-th Fibonacci number $F_k$ has a prime factor $p$ satisfying $p\geq k-1$, see here. It follows that if $n!=F_aF_b$ for some positive integers $a$ and $b$, then $a,b\leq n+1$. It is not hard to verify that $F_k\leq\varphi^k$ for all $k\geq0$, and so $a,b\leq n+1$ implies that $$n!=F_aF_b\leq F_{n+1}^2\leq(\varphi^{n+1})^2=\varphi^{2n+2}.$$ Of course the left hand side increases more rapidly than the right hand side as soon as $n\geq\varphi^2$, i.e. $n\geq3$. Then a quick check shows that $7!>\varphi^{16}$, so any solution to $n!=F_aF_b$ must have $n\leq6$. From here it is a routine verification to find that $$0!=F_1F_2,\quad 1!=F_1F_2,\quad 2!=F_2F_3,\quad 3!=F_3F_4,\quad 4!=F_4F_6,\quad 6!=F_5F_{12},$$ and that $5!=120$ is not a product of two Fibonacci numbers.

Answer 2

I might be overly complicating the stuff but I'll give it a try:

A first step might be to determine all Fibonacci numbers that are equal to some factorial.

from https://mathoverflow.net/questions/114850/is-every-fibonacci-number-fibonacci-prime

Carmichael's theorem says that for all $n \ne 1,2,6,12$, there exists a prime divisor $p$ of $F_{n}$ such that $p \nmid F_k$ for (positive integer) $k \lt n$

the second statement is equivalent to a sort of counterpart going in the other direction: for all $n \ne 1,2,6,12$, there exists a prime divisor $p$ of $F_{n}$ such that $p \nmid F_k$ for $k \gt n$ save only for the obvious exceptions where $k$ is a multiple of $n$ (since $F_{n}\mid F_{k}$ when $n \mid k$).

• As suggested in Doug M's comment, let us prove that for $i>12$, $F_i$ has a prime factor greater or equal than $i-1$.

Assume that $F_i$ has all its prime factors smaller that $i-1$, then since there exists a prime divisor of $i$ which doesn't divide any $F_k$ with $k<i$, and $f(p)\leq p+1$, the only possibility is that this factor is $i-1$ or $i$ (if one of them is prime)[this actually happens, take for instance $F_{23}$].

• We want to know when some prime $p$ divides a Fibonacci number, its done here, https://www.fq.math.ca/Scanned/1-2/vinson.pdf, where it is proven that, if $f(m)$ is the first time that $F_n$ is divisible by $m$, then $F_i \equiv 0 \:(m) \Longleftrightarrow f(m) | i$.

Notice that $f(p)\leq p+1$ ; $f(2^\ell) = 3.2^{\ell-2}$ ; $f(5)=5$ ; and if $m$ and $n$ are coprime, then $f(m.n)=f(m)f(n)$.

• Now take $n$ such that $2^\ell\le n < 2^{\ell+1}$ and let $p$ be the largest prime in $n$'s decomposition. We're looking for $i$ such that $F_i = n!\:$. Now that means that $i\leq p+1$ (otherwise $F_i$ would have a prime factor greater than $p$). Meanwhile, if $n\geq 5$, then $2^\ell$ and $5$ divide $n!$, hence $f(2^\ell)f(5) = 3.2^{\ell-2} .5 |i$. But $3.2^{\ell-2} .5> 2^{\ell+1} >n $, this is absurd.

This means that the only possible solutions to $F_i=n!$ are for $n=0,1$ or $2$ (see also Peter comment)

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Step 2 is to do the same with $n!=F_iF_j$. A different, probably easier, approach is to say that $n! \geq \sqrt{2\pi n}\left( {n \over e}\right)^n$ while $F_iF_j \leq F_{\max i,j}^2 \leq F_{n+1}^2$ (using the fact that $F_n$ has prime factor of size a least $n-1$ as we proved before).

Hence we want $n! \geq \sqrt{2\pi n}\left( {n \over e}\right)^n \geq 2^{2(n+1)} > F_{n+1}^2 \geq F_iF_j $ this is true for $n \geq 12$ (and again, we conclude using Peter's computations)

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disclamer: this might need a bit of cleaning and checking at some point, sorry if it's a bit messy