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Congruence question

  1. Feb 12, 2007 #1
    1. The problem statement, all variables and given/known data

    Let p = 7, 13, or 19. Show that [tex] a^{1728} \equiv 1 (mod p) [/tex] for all a such that p does not divide a.
    2. Relevant equations

    Fermat's little theorem.

    3. The attempt at a solution

    I'm not sure how to show this. Any quick help or examples?
    Last edited: Feb 12, 2007
  2. jcsd
  3. Feb 13, 2007 #2


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    Well, since you state "Fermat's little theorem" as a "relevant equation" it might be a good idea to write that out and see what happens. (There are two common forms of it- one is "more relevant" than the other.)

    You might also want to calculate what 1728 (mod 7), 1728 (mod 13) and 1728 (mod 19) are.
  4. Feb 13, 2007 #3
    Yeah so if you take 1728 (mod 7), 1728 (mod 13) and 1728 (mod 19), you get 6,12,18. And if you plug them in to Fermat's theorem, you get [tex] a^{6} \equiv 1 (mod \ 7) [/tex] and so on. This certainly follows Fermat's theorem of the form [tex] a^{p-1} \equiv 1 (mod \ p) [/tex] but so what? What have I really shown? I don't think I've shown much.
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