If $m$ and $n$ are positive integers, then $(F_m,F_n)=F_{(m,n)}$.

Question

Edit: The $F$'s are Fibonacci numbers.

I need an idea on how to show the following:

If $m$ and $n$ are positive integers, then $(F_m,F_n)=F_{(m,n)}$.

I believe that using the fact that $F_{m+n}=F_mF_{n+1}+F_nF_{m-1}$ could come in handy. Moreover, Euclid's algorithm may as well be needed. But I am not certain, as there may be better methods to achieve this.

Thanks in advance.

This is most probably a duplicate though I can't find the link right now. — lhf, Apr 25 '12 at 18:23
Here is one answer: http://math.stackexchange.com/questions/60340/fibonacci-modular-results/60353#60353 — sdcvvc, Apr 25 '12 at 18:30
There's a step where he states that $(f_{n-m},f_m)=f_{(n-m,m)}$. Is that something that's well-known? — wjmolina, Apr 25 '12 at 18:36
This is mentioned in Wikipedia article, the Wikipedia article has two references: this link and this book: Paulo Ribenboim, My Numbers, My Friends, Springer-Verlag 2000, p.9. — Martin Sleziak, Apr 25 '12 at 18:39
Josué: The proof is induction on $n+m$, so this is inductive hypothesis you can assume. — sdcvvc, Apr 25 '12 at 18:40
@Josué sdcvvc is correct. The induction is on the sum of the indices. Thus $\rm:(f_{n-m},f_m)=f_{:(n-m,m)}:$ follows by the induction, since it has smaller index sum $\rm:(n-m)+m: <: n+m,:$ by $\rm:m>0,:$ by hypothesis. I just edited the post to make it render better in the latest MathJax release. — Bill Dubuque, Apr 25 '12 at 19:43

score 3 · Accepted Answer · edited Apr 13 '17 at 12:21

3

As noted in the comments by sdcvvc, this answer to an earlier question completely answers this question as well.

edited Apr 13 '17 at 12:21

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answered Jun 07 '13 at 22:55

Brian M. Scott

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If $m$ and $n$ are positive integers, then $(F_m,F_n)=F_{(m,n)}$.

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