# Congruence question

1. Feb 12, 2007

### Tony11235

1. The problem statement, all variables and given/known data

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. 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. Feb 13, 2007

### HallsofIvy

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. Feb 13, 2007

### Tony11235

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.