• Support PF! Buy your school textbooks, materials and every day products Here!

Congruence question

  • Thread starter Tony11235
  • Start date
254
0
1. 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.
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?
 
Last edited:

Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
41,734
893
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.
 
254
0
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.
 

Related Threads for: Congruence question

Replies
10
Views
1K
Replies
1
Views
800
Replies
4
Views
903
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
9
Views
1K
  • Last Post
Replies
4
Views
5K
  • Last Post
Replies
6
Views
3K
Top