Let $f_n$ denote the $n$th Fibonacci number. A positive integer $n$ is called good if $f_{f_n}$ is divisible by $n$ but $f_n$ is not divisible by $n$.

My question is: how many good numbers are there. I think there are infinity many but I can’t prove it. Thanks!

Anyway this is from a russian book called: All about Fibonacci

  • $\begingroup$ What does it mean to say "$f_{f_n}$ is divisible"? $\endgroup$
    – lulu
    Sep 4, 2018 at 12:10
  • $\begingroup$ Also just to clarify : does the Fibonacci series start $1,1,2,3,5,..$ or $0,1,1,2,3,5,...$? Note that this changes the indexing, and the question depends on the indexing. $\endgroup$ Sep 4, 2018 at 12:15
  • $\begingroup$ Please edit your post for clarity. As you believe there are infinitely numbers of the sort you want, perhaps you could write down the first few. As it stands, the definitions aren't clear (at least, not to me). $\endgroup$
    – lulu
    Sep 4, 2018 at 12:21
  • $\begingroup$ FYI : oeis.org/A007570 $\endgroup$
    – mathlove
    Sep 4, 2018 at 12:24
  • $\begingroup$ Voting to close the question as it is unclear what you are asking. If you can, please edit for clarity. $\endgroup$
    – lulu
    Sep 4, 2018 at 12:32

1 Answer 1


Suppose $n$ is good. Since $\gcd(f_m,f_n)=f_{\gcd(m,n)}$, we have $$\gcd(f_{f_{f_n}},f_n)=f_{\gcd(f_{f_n},n)}=f_n,$$ and $$\gcd(f_{f_n},f_n)=f_{\gcd(f_n,n)}<f_n.$$ Therefore, $f_n$ is good. So if there is one good number, there must exist infinitely many good numbers.


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