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

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

Now let's look at some of their s

(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(fmfnfgcd(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(fnfn-1) = 1, for all n
b) fm+nfm+1fnfmfn-1
c) if m divides n, then fm divides fn

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(fmfngcd(fmfqm+rgcd(fmfqm+1frfqmfr-1gcd(fmfqm+1frgcd(fmfr

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

gcd(fnfmgcd(fmfr

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

Su, Francis E., et al. “Fibonacci GCD's, please.” Math Fun Facts. .

Fun Fact suggested by:
Arthur Benjamin

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