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AP
Posts:
137
Registered:
3/4/09


gcd Fibonacci Lucas
Posted:
Dec 12, 2012 2:54 PM


Let a, b the zeos of x^2x1
and for n>=0 F_n=(1/sqrt(5))(a^nb^n)
F_0=0 F_1=1 F_(n+1)=F_n+F_(n1)
L_n=a^n+b^n
L_0=2 L_1=1 L_(n+1)=L_n+L_(n1)
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_(n1) 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 3d
=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


Date

Subject

Author

12/12/12


AP

12/14/12


AP


