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Math Help - Proof of congruence result.

  1. #1
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    Proof of congruence result.

    How do I prove that

    (5k+1)^5^n is congruent to 1 (mod 5^n+1) ?

    Can I deduce from this result that no number congruent to 1 mod 5 is a primitive root mod 5^m, for any m greater than or equal to 1?
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  2. #2
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    Quote Originally Posted by Cairo View Post
    How do I prove that

    (5k+1)^5^n is congruent to 1 (mod 5^n+1) ?

    Can I deduce from this result that no number congruent to 1 mod 5 is a primitive root mod 5^m, for any m greater than or equal to 1?

    (5k+1)^{5^n}=\sum\limits_{i=0}^{5^n}\binom{5^n}{i}  (5k)^i=1+\sum\limits_{i=1}^{5^n}\binom{5^n}{i}(5k)  ^i

    and as \binom{5^n}{i} is divisible by 5^n\,\,\forall i\geq 1 (proof...?) , we are then done.

    Tonio
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  3. #3
    Member HappyJoe's Avatar
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    \binom{5^n}{i} isn't divisible by 5^n for i=5^n. But anyway, the congruence is mod 5^n+1.
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  4. #4
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    Quote Originally Posted by HappyJoe View Post
    \binom{5^n}{i} isn't divisible by 5^n for i=5^n. But anyway, the congruence is mod 5^n+1.

    Of course: the sum should have been split into three parts: for i=0\,,\,i=5^n\,,\,1\leq i\leq 5^n-1

    Tonio
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  5. #5
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    Sorry, the congruence is modulo 5^(n+1).

    I've tried using induction to prove the result, but got lost.
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  6. #6
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Cairo View Post
    Sorry, the congruence is modulo 5^(n+1).

    I've tried using induction to prove the result, but got lost.
    Tonio's post still applies.
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