Hi.

How to prove that GCD(F3k, L3k)=2? and If n dosen't divide with 3 GCD(Fn, Ln)=1

Don't even know how to start that:/....Hope someone can help:)

Ty.

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- Apr 21st 2010, 06:16 AMaurahProof about Fibonacci and Lucas numbers (GCD)
Hi.

How to prove that GCD(F3k, L3k)=2? and If n dosen't divide with 3 GCD(Fn, Ln)=1

Don't even know how to start that:/....Hope someone can help:)

Ty. - Apr 21st 2010, 09:26 AMkompik
Try to show:

a) $\displaystyle L_n=F_{n+1}+F_{n-1}$

b) $\displaystyle \gcd(F_n,F_{n+1})=1$ for every n.

Using these two properties you can get:

$\displaystyle \gcd(F_n,L_n)=\gcd(F_n,F_{n+1}+F_{n-1})=\gcd(F_n,F_n+2F_{n-1})=\gcd(F_n,2F_{n-1}$. This is either 1 or 2, since Fn and F_n-1 are coprime. So you already know that the only possible values are 1 and 2.

To show that they are equal to what you right, just nottice the pattern 0,0,1,0,0,1,.... when looking to the sequence Fn (or Ln) modulo 2.