Math 2.0

statement theorem

Gcd of Fibonacci numbers

Statement

Let mm and nn be natural numbers, and let FkF_k be the Fibonacci numbers. Then gcd⁡(Fm,Fn)=Fgcd⁡(m,n).\gcd(F_m, F_n) = F_{\gcd(m, n)}.

Remarks

Example: gcd⁡(F12,F18)=gcd⁡(144,2584)=8=F6\gcd(F_{12}, F_{18}) = \gcd(144, 2584) = 8 = F_6. In particular, FmF_m divides FnF_n when mm divides nn. Consecutive Fibonacci numbers are coprime, because gcd⁡(n,n+1)=1\gcd(n, n+1) = 1 and F1=1F_1 = 1. The proof follows the Euclidean algorithm and uses the addition formula.

Explanations

L1 human

The greatest common divisor of two Fibonacci numbers is again a Fibonacci number. To find which one, take the greatest common divisor of the two indices.

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Sources

  1. Web https://en.wikipedia.org/wiki/Fibonacci_sequence, Divisibility properties

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