Congruence question

  • Thread starter Tony11235
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  • #1
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Homework Statement



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.

Homework Equations



Fermat's little theorem.

The Attempt at a Solution



I'm not sure how to show this. Any quick help or examples?
 
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Answers and Replies

  • #2
HallsofIvy
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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.
 
  • #3
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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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