Should I really compute that monster?

Let F n F_n denote the n th n^\text{th} Fibonacci number .

Compute gcd ( F 2000 F 16 + F 1999 F 15 , F 1014 F 1001 + F 1013 F 1000 ) \gcd(F_{2000}F_{16} + F_{1999}F_{15}, F_{1014}F_{1001} + F_{1013}F_{1000}) .

Notation : gcd ( a , b ) \gcd(a,b) denotes greatest common divisor of a a and b b .


The answer is 1.

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1 solution

Brian Riccardi
Feb 14, 2016

It can be proved this property of Fibonacci numbers

F n + m = F n F m + F n 1 F m 1 F_{n+m} = F_{n}F_{m} + F_{n-1}F_{m-1} n , m N 0 \forall n,m \in \mathbb{N_0}

So the problem asks to compute g c d ( F 2016 , F 2015 ) gcd(F_{2016},F_{2015})

A well known fact is that consecutive Fibonacci numbers are coprime and then the answer is 1 \boxed{1}

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