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AP
Posts:
137
Registered:
3/4/09


Re: gcd Fibonacci Lucas
Posted:
Dec 14, 2012 4:52 PM


On Wed, 12 Dec 2012 20:54:33 +0100, AP <marc.pichereau@wanadoo.fr.invalid> wrote:
>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 ... I have find a proof ...I think >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


