Fibonacci GCD’s, please

Fibonacci numbers exhibit striking patterns. Here's one that may not be so obvious, but is striking when you see it. Recall the Fibonacci numbers:

 n: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,  12,  13,  14, … 
f_n: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, …

Now let's look at some of their greatest common divisors

(gcd's): gcd(f₁₀f₇) = gcd(55, 13) = 1 = f₁
gcd(f₆f₉) = gcd(8, 34) = 2 = f₃
gcd(f₆f₁₂) = gcd(8, 144) = 8 = f₆
gcd(f₇f₁₄) = gcd(13, 377) = 13 = f₇
gcd(f₁₀f₁₂) = gcd(55, 144) = 1 = f₂

Do you see the pattern? The greatest common divisor of any two Fibonacci numbers is also a Fibonacci number! Which one? If you look even closer, you'll see the amazing general result:

gcd(f_mf_nf_{gcd(m, n)}

Presentation Suggestions:
After presenting the general result, go back to the examples to verify that it holds. You may wish to prepare a transparency beforehand with a table of Fibonacci numbers on it, so you can refer to it throughout the presentation.

The Math Behind the Fact:
The proof is based on the following lemmas which are interesting in their own right. All can be proved by induction. 
a) gcd(f_nf_n-1) = 1, for all n 
b) f_m+nf_m+1f_nf_mf_n-1
c) if m divides n, then f_m divides f_n

and the ever important Euclidean Algorithm which states: if n= qm + r, then gcd(n, m) = gcd(m, r). For such n,m we have

gcd(f_mf_ngcd(f_mf_qm+rgcd(f_mf_qm+1f_rf_qmf_r-1gcd(f_mf_qm+1f_rgcd(f_mf_r

where the 2nd equality follows from (b), the 3rd equality from (c) noting that m divides qm, and the 4th equality from noting that f_m divides f_qm which is relatively prime to f_qm+1. Thus

gcd(f_nf_mgcd(f_mf_r

which looks a lot like the Euclidean algorithm but with f's on top! For example since

gcd(100, 80) = gcd(80, 20) = gcd(20, 0) = 20, then
gcd(f₁₀₀f₈₀gcd(f₈₀f₂₀gcd(f₂₀f₀ = 0) = f₂₀

How to Cite this Page: 
Su, Francis E., et al. “Fibonacci GCD's, please.” Math Fun Facts. .

Fun Fact suggested by:
Arthur Benjamin