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Math Help - Show n^13 is congruent to n (mod7)

  1. #1
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    Post Show n^13 is congruent to n (mod7)

    Hi, I have this question,

    Show that for any integer n, n^{13}\equiv{n}(\bmod7)

    I have written this:

    n^{6}\equiv{1}(\bmod7)

    because of Eulers theorem.

    n^{13} = (n^6)^2n

    So (n^6)^2\equiv{1}(\bmod7)
    So n\equiv{n}(\bmod7)
    And 0\equiv{0}(\bmod7) QED

    Is this right? Please help.

    Thanks,

    Katy
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  2. #2
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    Quote Originally Posted by harkapobi View Post
    Hi, I have this question,

    Show that for any integer n, n^{13}\equiv{n}(\bmod7)

    I have written this:

    n^{6}\equiv{1}(\bmod7)

    because of Eulers theorem.

    n^{13} = (n^6)^2n

    So (n^6)^2\equiv{1}\bmod7)
    So n\equiv{n}(\bmod7)
    And 0\equiv{0}(\bmod7) QED

    Is this right? Please help.

    Thanks,

    Katy

    It was going fine but then I didn't understand what you did at the end: so you have that n^6=1\!\!\!\!\pmod 7\Longrightarrow n^{12}=1\!\!\!\!\pmod 7....so now just multiply by n both sides in the last congruence and we're done!

    Tonio
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  3. #3
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    Ahhh yes! Thank you Can't believe I didn't see that.
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