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Find $\gcd(f_{n+1}, f_{n+2})$ by using Euclidean algorithm for the Fibonacci numbers whenever $n>1$. How many division algorithms are needed? (Recall that the Fibonacci sequence $(f_n)$ is defined by setting $f_1=f_2=1$ and $f_{n+2}=f_{n+1}+f_n$ for all $n \in \mathbb N^*$, and look here to get information about Euclidean algorithm)

J. M. ain't a mathematician
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    $\gcd(F_{n+1},F_{n+2})=\gcd(F_{n+1},F_{n+2}-F_{n+1})=\gcd(F_{n+1},F_n)$, and then use induction... – anon Sep 09 '11 at 09:54
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    @anon: You could consider fleshing that out to a full answer? Given that the OP doesn't seem to be in the business of accepting answers it may not be worth your while? – Jyrki Lahtonen Sep 09 '11 at 10:56
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    Are Fibonacci numbers the "worst case" as far as efficiency of Euclid's algorithm is concerned? – Michael Hardy Sep 09 '11 at 13:46
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    @Michael: Yes. At each step, you can only subtract $F_i$ once from $F_{i+1}$, so the number of iterations needed is maximal, given the size of the two initial numbers. – TMM Sep 09 '11 at 14:24
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    @Michael: I think that the proof of that (sometime in the early 1800s, I believe) was one of the first analyses of an algorithm as well as one of the first practical applications of the Fibonacci numbers. – Mike Spivey Sep 09 '11 at 15:26

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anon's answer: $$ \gcd(F_{n+1},F_{n+2}) = \gcd(F_{n+1},F_{n+2}-F_{n+1}) = \gcd(F_{n+1},F_n). $$ Therefore $$ \gcd(F_{n+1},F_n) = \gcd(F_2,F_1) = \gcd(1,1) = 1. $$ In other words, any two adjacent Fibonacci numbers are relatively prime.

Since $$\gcd(F_n,F_{n+2}) = \gcd(F_n,F_{n+1}+F_n) = \gcd(F_n,F_{n+1}), $$ this is also true for any two Fibonacci numbers of distance $2$. Since $(F_3,F_6) = (2,8)=2$, the pattern ends here - or so you might think...

It is not difficult to prove that $$F_{n+k+1} = F_{k+1}F_{n+1} + F_kF_n. $$ Therefore $$ \gcd(F_{n+k+1},F_{n+1}) = \gcd(F_kF_n,F_{n+1}) = \gcd(F_k,F_{n+1}). $$ Considering what happened, we deduce $$ (F_a,F_b) = F_{(a,b)}. $$

Yuval Filmus
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    Here's the complete proof and then some. Note that you don't need to first prove $(F_{n+1},F_n) = 1$ since it is a special case. – Bill Dubuque Sep 09 '11 at 19:51
  • @anon: How do you get that gcd($F_{n+1},F_{n+2}$)=gcd($F_{n+1},F_{n+2}−F_{n+1}$)=gcd($F_n,F_{n+1}$)? I know that $F_{n+2}=F_{n+1}+F_n$, but how you get that $F_{n+2}=F_{n+2}−F_{n+1}=F_{n+1}$ –  Sep 10 '11 at 15:05
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    @alvoutilla: I'm not saying $F_{n+2}=F_n$, I'm using the well-known fact that $\gcd(a,b)=\gcd(a,b-a)$. If you're having trouble realizing why this works, consider thinking of $a$ and $b$ in terms of their prime factorizations (and use $\gcd(cn,cm)=c\cdot\gcd(n,m)$). – anon Sep 10 '11 at 22:48
  • @anon: But a and b are primes in this case(because they have not prime factorization except trivial one( a and b itself)? Where I use gcd(cn, cm) = c*gcd(n,m) and how I use it? like this gcd(ca,cb)=c gcd(a,b)= ...? –  Sep 12 '11 at 15:12
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    @alvoutila: You really shouldn't analyze the Euclidean algorithm without being fully familiar with basic gcd properties first. :) The identity I gave is independent of whether or not $a,b$ are primes. First let's prove it holds when $a,b$ are coprime. Suppose that $\gcd(a,b)=1$ so that they are coprime and let $d=\gcd(a,b-a)$. Since $d|a$ and $d|(b-a)$, $d$ must also divide their sum $a+(b-a)=b$ (the sum of two multiples of a number $d$ must also be a multiple of the number). But $d|a,d|b\implies d=1$ because they're coprime! (continued) – anon Sep 12 '11 at 19:10
  • (...) Now we prove the identity in full generality: suppose $\gcd(a,b)=d$. Then $$1=(a/d,b/d)=(a/d,b/d-a/d)$$ (by our previous reasoning), so multiplying both sides by $d$ gives $\gcd(a,b)=d=(a,b-a)$, QED. If there's any part you didn't follow please let me know. – anon Sep 12 '11 at 19:11
  • @anon: But although they are coprime( what does it mean?) they don't have prime factorization? –  Sep 13 '11 at 07:50
  • @alvoutila: Every whole number has a prime factorization! Two numbers are coprime if $\gcd(a,b)=1$, or in other words they share no common factor other than $1$. So $6=2\cdot3$ and $35=5\cdot7$ are coprime and they both have prime factorizations. Beyond this I'm not sure how to help you learn the relevant introductory number theory here. Were you able to follow the proof I gave? If not, you could post it as a distinct question and I think someone will make a more thorough treatment of the issue. – anon Sep 13 '11 at 07:58