Date: Dec 12, 2012 2:54 PM
Author: AP
Subject: gcd Fibonacci Lucas

Let a, b the zeos of x^2-x-1

and for n>=0
F_n=(1/sqrt(5))(a^n-b^n)

F_0=0 F_1=1
F_(n+1)=F_n+F_(n-1)

L_n=a^n+b^n

L_0=2 L_1=1
L_(n+1)=L_n+L_(n-1)

the following result (and his proof) is well known

gcd(F_n, F_p)=F_gcd(n,p)
(first, one can use F_(n+p)=F_(n+1)F_p+F_nF_(n-1)
to show gcd(F_n,F_p)=gcd(F_n,F_(n+p))

but if this other result is known

gcd(L_n,L_p)

=L_d if n/d and p/d are odd

= 2 if n/d or p/d is even and 3|d

=1 if n/d or p/d is even and 3 do not divide d

(with d= gcd(n,p) )

but I don't find a proof ...

one can read :
Using basic identities Lucas proved this
this result is more subtle than the first