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Using mathematical induction, I was trying to prove that consecutive Fibonacci series numbers are coprime(or relatively prime). It is a ubiquitous topic and there are many elegant proofs regarding this. Still, I tried to prove it because I am still new to the topic of mathematical induction and am trying to get better at it. So, here's how I went about it

Let $f(n)$ be the nth term of the Fibonacci series. I am taking $f(3)$ as the base case.
$f(3) =2$ which is coprime with $f(2)$. Hence, our base case is true.
Now, let us assume that it is true for $f(k)$ where $k ∈ [3,n]$
$f(n+1) = f(n) + f(n-1)$
Now $f(n)$ and $f(n-1)$ are coprime(Our assumption).
Hence $f(n+1)$ is a coprime and hence it is true that consecutive numbers in the Fibonacci series are coprimes

Now, this led to an interesting observation: at least three consecutive numbers in the Fibonacci series are always coprime. So, I manually wrote out some terms of the Fibonacci series and it got me wondering: What is the maximum number of consecutive numbers in the Fibonacci series which are coprime?

Since I couldn't think of a way how to go about it, I tried to go manually and get a feel for it. I haven't done many terms but for now, the upper limit seems to be 5(the first $5$ numbers: 1,1,2,3,5 and another one thanks to lulu is 233, 377, 610,987 and 1597).

So, is $5$ the limit or can we do better than that?

Bill Dubuque
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Random Math Enthusiast
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  • Every third $F_n$ is even, Every fourth $F_n$ is divisible by $3$. That should let you produce an upper bound. Then look at small $n$ to get a long coprime streak to establish a lower bound. – lulu Mar 03 '25 at 16:26
  • @lulu Since every fourth $Fn$ is divisible by 3, it seems to me that the best that we can do is 4 consecutive coprime other than the first five. So, looks to me that 5 should be the best that we can do and there is no chance of doing any better than 4 in the future series. – Random Math Enthusiast Mar 03 '25 at 16:59
  • So now you need to exhibit a coprime string of length $5$, but this is not difficult. – lulu Mar 03 '25 at 17:01
  • @lulu So can we consider it established that the length of the coprime string lies between 3 and 5 or do I need to prove anything else to be confident in my statement? – Random Math Enthusiast Mar 03 '25 at 17:03
  • @lulu Yes I did write in the question statement that the first five numbers are the only apparent one of length 5 – Random Math Enthusiast Mar 03 '25 at 17:42
  • Oh, I wouldn't think that. Take the $5$ which follow $144$ for instance. – lulu Mar 03 '25 at 17:44
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    No need to chat. I agree that you did list the first $5$. If you want to research further, I would suggest looking at the strings of length $5$ which follow multiples of $6$ (just as the first $5$ follow $0$). Perhaps you can prove that infinitely many of those length $5$ strings are coprime (if that's true). – lulu Mar 03 '25 at 17:46
  • the chat message was an accident. I clicked the "automatically move this discussion to chat" suggestion by Stack which sent that message. Let me see if I can arrive at any proof – Random Math Enthusiast Mar 03 '25 at 17:50
  • To stress: it may not be true, or if true it may not be provable. But if you are interested, good luck! – lulu Mar 03 '25 at 17:52
  • 5 coprime $F_n,\ldots F_{n+4}$ start at $n=1,7,10,13,19,22,25,31,34,37,43,46,49,55,58,\ldots$ For related problems see e.g. https://doi.org/10.1016/j.jnt.2012.05.022 . – R. J. Mathar Mar 04 '25 at 17:10
  • Hint $\ \gcd(F_n,F_m) = F_{\large \gcd(n,m)}\ \ $ – Bill Dubuque Mar 10 '25 at 10:00
  • First differences of numbers in comment by @R.J.Mathar are $6,3,3,6,3,3,6,3,3,6,3,3\dots$. – Gerry Myerson Mar 10 '25 at 12:18
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    I added an equivalent conjecture as a comment to https://oeis.org/A004772 . – R. J. Mathar Mar 11 '25 at 13:20

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