Proving there is no non-abelian finite simple group of order a Fibonacci number "Prove there does not exist a finite simple non-abelian group of order of a Fibonacci number"
I would like to answer the above question, but I currently have few ideas of where to begin.
I understand we will likely only be using results about the prime factorisation of fibonacci numbers. I considered basic results about simple groups, e.g. from the sylow theorems, if a prime $p | |G| $ and $ kp+1$ does not divide $|G|$ for all integers $k$, then the sylow-p subgroup is normal.
However I fail to see how exactly to use this, and other standard techniques.
I have heard before that no Fibonacci number is a perfect number, but again, I cannot see how to use this exactly.
Would someone be able to provide me with hints/ideas?
In particular, is there a specific property of simple groups or Fibonacci numbers that I need consider?
 A: The paper of Florian Luca that proves this result, and was mentioned in the comments, is as follows: Fibonacci numbers, Lucas numbers and orders of finite simple groups, Journal of Algebra, Number Theory and Applications, vol. 4, no. 1, 23--54 (2004).
Unfortunately, the paper is in an obscure journal, no version exists on the arXiv, and the journal website does not appear to be, how shall we say, very good.
I see some ideas with how to prove this myself, but would need to know more about Fibonacci numbers to do so. I satisfied myself with a proof for the alternating groups. (The sporadics can be done by inspection.)
By Carmichael's theorem, every $F_n$ is divisible by some prime not dividing any smaller $F_m$. Since $p$ divides $F_{p\pm 1}$, this primitive divisor must be at least $n-1$. But $|A_n|$ is divisible by exactly those primes at most $n$, and grows faster than $F_n$. In particular, if $|A_n|=F_m$, then $m>n$. Thus you only need to worry about the case $m=n+1$. And there is one: $n=3$, which of course does not yield a simple group.
