# Congruence question

1. 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.
2. Homework Equations

Fermat's little theorem.

3. The Attempt at a Solution

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

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HallsofIvy
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.