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I'm trying to work out following problem: Let $m$ be some Fibonacci number and $n$ some integer. Find all non-trivial pairs $(n,m)$, where $n! = m$. Trivial pair is pair $(1,1)$. I know that for every number $a$ there exists some Fibonacci number that's divisible by $a$. I also tried to bruteforce my way trough, but I didn't succeed. Here is my sub-optimal code:

a = 1
b = 1
mylist = [here are factorials up to 50!]
for i in range(0,100):
c = a+b
a = b
b = c
print(b)
if b in mylist:
    print("found")

I tried many other things including googling it, but I didn't find anything that I can even start with.
I have highschool level of math knowledge.

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    $\begingroup$ Your question is a good one, but if you only have high school level knowledge, then this site is not for you. This is because this site is dedicated to research level mathematics, and so the answers are of that level as well. I recommend that you use your nice question as motivation to learn more mathematics. There are fine books like Hardy-Wright or Erdős-Surányi that you can use as introduction to number theory. Also, for general questions in mathematics see math.stackexchange.com $\endgroup$
    – GH from MO
    Commented Jun 22, 2017 at 20:48
  • $\begingroup$ At a glance, the problem is that the prime factors of $n!$ are all too small and there are too many of them. If $n$ is a prime, then the largest factor of $n!$ is approximately $\log n/(\log\log n)$, and if $n$ isn't prime, they're even larger. Meanwhile, while this is tricky to express precisely, FIbonacci numbers tend to be pretty 'non-smooth' in a technical sense; they don't have a whole lot of prime factors and they tend to have some larger ones. Since $n!=F_k$ implies that $n$ and $k$ are 'close' (very roughly, $k\approx Cn\log n$ for some $C$), these are incompatible characterizations. $\endgroup$
    – Steven Stadnicki
    Commented Jun 22, 2017 at 21:21
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    $\begingroup$ A proof that uses only high school maths is possible. On one hand, $F_n=m!$ implies $v_2(F_n)=v_2(m!)$, but $v_2(F_n) \leq v_2(n)+2$ (this is well-known and fully elementary, no algebraic number theory involved) while $v_2(m!)=m+O(\log m)$ by Legendre-de Polignac. On the other hand, the explicit formula for the Fibonaccis and Stirling's approximation give $m \sim n \log \varphi/\log n$: use this to show that the power of $2$ dividing $n$ has to be too large. Use then the explicit version of Stirling to get precise bounds and check the remaining cases. $\endgroup$
    – user41593
    Commented Jun 22, 2017 at 21:48
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    $\begingroup$ I voted to reopen this question. I encourage others to do the same. $\endgroup$
    – GH from MO
    Commented Jun 23, 2017 at 20:36
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    $\begingroup$ You can find an online copy of the paper On Fibonacci Numbers that are Factorials here: m-a.org.uk/resources/2015_MAG_sample_web.pdf $\endgroup$
    – Sidney Raffer
    Commented Jun 24, 2017 at 3:50

2 Answers 2

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There are only three terms of the Fibonacci sequence which are equal to a factorial:

$$F_{1}= F_{2} = 0!=1! \quad \mbox{and} \quad F_{3}=2!$$

What is more, F. Luca proved in this paper (published in 1999) that the only nontrivial solutions of the diophantine equation $$F_{n} = m_{1}! \cdots m_{t}!$$ are $F_{3}=2!$, $F_{6}=(2!)^{3}$, and $F_{12} = (2!)^{2}(3!)^{2} = 3! 4!$.

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    $\begingroup$ Thanks for the answer, but I don't understand the paper. (I have just highschool level of math). I especially don't understand how to find the solutions from the inequalities in the paper. $\endgroup$
    – JPABA
    Commented Jun 22, 2017 at 19:23
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    $\begingroup$ That's the thing about Number Theory, JPABA; there are so many questions that someone with high school math can ask, that even someone with PhD math will find difficult – or impossible! – to answer. $\endgroup$
    – Gerry Myerson
    Commented Jun 22, 2017 at 23:18
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On Fibonacci numbers that are factorials, Surajit Rajagopal and Martin Griffiths (2014)

This is behind a paywall, but thanks to SJR the paper can be accessed here. The "simple proof" that for any $n\geq 4$ the Fibonacci number $F_n$ is not a factorial starts on page 14.

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