Congruence question

Homework Statement

Let p = 7, 13, or 19. Show that $$a^{1728} \equiv 1 (mod p)$$ 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

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HallsofIvy
Science Advisor
Homework Helper
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.

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 $$a^{6} \equiv 1 (mod \ 7)$$ and so on. This certainly follows Fermat's theorem of the form $$a^{p-1} \equiv 1 (mod \ p)$$ but so what? What have I really shown? I don't think I've shown much.