p divides Fibonacci(A*p^2+B)-F(A+B) ?

benoit abcloitre at wanadoo.fr
Wed Sep 10 18:22:26 CEST 2003


Let F(n) denotes the n-th Fibonacci number. It seems that if p is a 
prime different from 5 :

p divides F(A*p^2+B)-F(A+B) for any (A,B) in N^2.

Someone told me about a simple proof for  A=1 et B=0 ,

Since 2^(n-1)*F(n)=sum(2k+1<=n, C(n,2k+1)*5^k). putting n=p ^2 and 
since p divides C(p^2, 2k+1), except for k=(p^2-1)/2 we have :

2^(p^2-1)*F(p^2)== 5^(p^2-1)/2 (mod p)

Since  p^2-1=(p-1)(p+1) and 2^(p-1)==5^(p-1)=0 (modp)

F(p^2)==1 (mod p)

I didn't find a similar general proof for any (A,B).

Is this property true?  If  p divides F(A*p^2+B)-F(A+B) for any (A,B) 
in N^2, is p prime? Is there any composite q with this property?

Benoit Cloitre.
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