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My second observation is the following. Let $p$ be a prime not equal to $5$. Then $5$ is a quadratic residue modulo $p$ if and only if $p\equiv\pm1\pmod5$. And $5$ is not a quadratic residue modulo $p$ if and only if $p\equiv\pm2\pmod5$.

If $p$ is a prime and $m$ the period of $F_n\pmod{p}$, then $p\equiv\pm1\pmod5$ implies $m|(p-1)$.

I am looking for a generalization of the above cited statment.

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Hello! thank you so much for editing my question, which I faild to type. –  pavani.neta Sep 13 '11 at 6:32

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You might e.g. look at the following discussion: http://groups.google.com/group/sci.math/browse_thread/thread/ab71277480f8b345/d511366e2337beb9

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Very useful like I have seen now. Thank you very much for your post. –  pavani.neta Sep 13 '11 at 9:08
    
Thank you so much for your post –  pavani.neta Sep 13 '11 at 9:10

The cycle modulo p, of the fibonacci series, divides m, where m=p+1 if p=2,3 mod 5, and m=p-1 if p=0, 1, 4. mod 5. In the series, you might have to double the number m to get the full cycle.

Fermats little theorm applies. For composite numbers, it is the lcm of the individual primes, eg 77 = 7*11, so the period is lcm(16, 10) =80.

powers of primes give the value 2m *p^{n-1}.

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