Results 1 to 2 of 2

Thread: Solving a congruence modulo p

  1. #1
    Ant
    Ant is offline
    Member
    Joined
    Apr 2008
    Posts
    145
    Thanks
    4

    Solving a congruence modulo p

    $\displaystyle p \equiv 2 (mod 3) $ and $\displaystyle p-1 = 4q$ with p and q distinct primes.

    Show that:

    $\displaystyle 3^{2q} \not\equiv 1 (mod p) $

    I've tried various thing but I can't seem to find a contradiction (after assuming 3^(2q) == 1 mod p).

    Can anyone point me in the right direction please??
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Sylvia104's Avatar
    Joined
    Mar 2012
    From
    London, UK
    Posts
    107
    Thanks
    37

    Re: Solving a congruence modulo p

    Since $\displaystyle p\equiv1\mod4,$ quadratic reciprocity gives $\displaystyle 1 = (-1)^{\left(\frac{p-1}2\right)\left(\frac{3-1}2\right)} = \left(\frac p3\right)\left(\frac3p\right) = (-1)\left(\frac3p\right)$ (where $\displaystyle \left(\frac**\right)$ denotes the Legendre symbol). In other words $\displaystyle 3$ is not a quadratic residue modulo $\displaystyle p;$ by Eulerís criterion, $\displaystyle -1 = \left(\frac3p\right) \equiv 3^{\frac{p-1}2} = 3^{2q}\mod p.$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Modulo and congruence
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: Jun 26th 2010, 05:10 PM
  2. congruence modulo 2^n
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: Nov 3rd 2009, 06:47 PM
  3. Solving for x in a quadratic congruence modulo problem
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: Mar 12th 2009, 08:32 PM
  4. congruence modulo m
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: Feb 27th 2007, 07:59 PM
  5. congruence modulo m
    Posted in the Number Theory Forum
    Replies: 8
    Last Post: Feb 22nd 2007, 03:57 PM

Search Tags


/mathhelpforum @mathhelpforum