Results 1 to 2 of 2

Math Help - [SOLVED] Problem involving Fermat's Little Theorem

  1. #1
    Newbie
    Joined
    Apr 2009
    Posts
    2

    [SOLVED] Problem involving Fermat's Little Theorem

    The solution to this problem is probably painfully obvious to most people, but I'm not making the leap for some reason. The problem is:

    Show that 1^p+2^p+3^p+...(p-1)^p \cong 0(mod\; p) when p is an odd prime.
    So, by Fermat's Little Theorem, we know that:

    a^p \cong a(mod\; p) and a^{p-1} \cong 1(mod\; p).

    And thus, 1^p+2^p+3^p+...(p-1)^p \cong 1+2+3+... (p-1)\;(mod\; p).

    So, by my thinking, I need to show that 1+2+3+... (p-1) \cong 0\;(mod\; p), but I'm not sure how to proceed.

    Any help at all would be greatly appreciated.
    Last edited by ikarous; April 19th 2009 at 12:14 PM. Reason: Fixed typos.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Apr 2009
    Posts
    2
    Solved. I'm dumb.

    Since (p-1) is even,

    1+2+...(p-1) \cong 1+2...+ (p-2) + (p-1) \cong 1+2 ... + -2 + -1 \cong 0 \; (mod\; p).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: January 10th 2011, 08:51 AM
  2. [SOLVED] Fermatís Little Theorem
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: March 3rd 2009, 04:29 PM
  3. Replies: 3
    Last Post: February 14th 2009, 07:35 PM
  4. Replies: 8
    Last Post: January 24th 2008, 02:49 AM
  5. Fermatís Little Theorem Problem
    Posted in the Number Theory Forum
    Replies: 4
    Last Post: March 14th 2007, 08:11 AM

Search Tags


/mathhelpforum @mathhelpforum