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Math Help - modulo on an expression

  1. #1
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    modulo on an expression

    Hi everybody,

    Is there any theorem that shows:

    n^s-1\equiv 0 (mod (s+1))

    when s+1 is prime?

    (this is valid for every n>1)

    or need to be prooven?
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Yes, that is Fermat's little theorem. It states that for any prime p and any integer a, a^p \equiv a \mod p.
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  3. #3
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by gdmath View Post
    n^s-1\equiv 0 (mod (s+1))

    when s+1 is prime?

    (this is valid for every n>1)
    This (in the form you have written) is not valid for every integer n>1, only when n and s+1 are coprime (in which case it is also valid for n=1). However, if you re-write it as n^{s+1}-n\equiv0\,(\bmod\,s+1) then it is valid for every integer n (including 1).

    It is more usual to write the prime s+1 as p. Thus Fermatís little theorem is usually presented as n^{p-1}\equiv1\,(\bmod\,p) (which is valid if \gcd(n,p)=1) or n^p\equiv n\,(\bmod\,p) (which is valid for any integer n).
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