Solving a Congruence Question: Fermat's Little Theorem

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SUMMARY

The discussion centers on demonstrating that \( a^{1728} \equiv 1 \mod p \) for primes \( p = 7, 13, \) and \( 19 \) using Fermat's Little Theorem. The theorem states that if \( p \) is a prime and \( p \nmid a \), then \( a^{p-1} \equiv 1 \mod p \). By calculating \( 1728 \mod 7, 1728 \mod 13, \) and \( 1728 \mod 19 \), the results are \( 6, 12, \) and \( 18 \) respectively, leading to the conclusion that \( a^{6} \equiv 1 \mod 7 \), \( a^{12} \equiv 1 \mod 13 \), and \( a^{18} \equiv 1 \mod 19 \), thereby confirming the theorem's application.

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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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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 into 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.
 

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