Results 1 to 3 of 3

Math Help - Show that a^2 -1 = 0 mod p has only 2 solutions

  1. #1
    Newbie
    Joined
    Oct 2011
    Posts
    22

    Show that a^2 -1 = 0 mod p has only 2 solutions

    I have an exercise for a course in Cryptography. One of the questions asks us to show that the equation a^2 -1 = 0 mod p has only 2 solutions.

    we are told to consider the group Z sub p whose elements are {1,2,.....,p-1}.

    I am having a hard time even finding a starting point for this problem. Can anyone give me some advice?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,406
    Thanks
    1294

    Re: Show that a^2 -1 = 0 mod p has only 2 solutions

    Quote Originally Posted by restin84 View Post
    I have an exercise for a course in Cryptography. One of the questions asks us to show that the equation a^2 -1 = 0 mod p has only 2 solutions.

    we are told to consider the group Z sub p whose elements are {1,2,.....,p-1}.

    I am having a hard time even finding a starting point for this problem. Can anyone give me some advice?
    \displaystyle \begin{align*} a^2 - 1 &= 0 \\ (a +1)(a-1) &= 0 \\ a = -1 \textrm{ or }a &= 1 \\ a = 1 \textrm{ or } a &= p - 1 \textrm{ mod }p \end{align*}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5

    Re: Show that a^2 -1 = 0 mod p has only 2 solutions

    Quote Originally Posted by restin84 View Post
    I have an exercise for a course in Cryptography. One of the questions asks us to show that the equation a^2 -1 = 0 mod p has only 2 solutions.

    we are told to consider the group Z sub p whose elements are {1,2,.....,p-1}.

    I am having a hard time even finding a starting point for this problem. Can anyone give me some advice?
    No matter which is p>2, a=1 and a=p-1 always satisfy the congruence equation x^{2} \equiv 1\ \text{mod}\ p. In general if we have a congruence equation of the form...

    \sum_{k=0}^{n} a_{n}\ x^{n} \equiv 0\ \text{mod}\ n (1)

    ... then the sum of molteplicities of the roots can't be greater than n...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Is it possible to show this equation has no solutions
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: December 29th 2011, 01:40 AM
  2. Show that x^2*e^|x|=2 has two solutions
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: January 13th 2010, 06:12 AM
  3. Replies: 13
    Last Post: August 1st 2009, 03:02 PM
  4. Using Mean Value Theorem to show solutions ?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 28th 2009, 07:39 AM
  5. Show solutions are global
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: April 4th 2009, 11:06 AM

Search Tags


/mathhelpforum @mathhelpforum