Results 1 to 2 of 2

Math Help - Factor Theorem Troubles

  1. #1
    Newbie
    Joined
    Oct 2009
    Posts
    8

    Factor Theorem Troubles

    Hey Guys, I have two problems that have to do with the factor theorem. I've been stumped on them for sometime now and can no longer put them off as my assignment is due soon.

    1) Use the factor theorem to show that if 2^p - 1, where p does not equal 3, is a prime number, then p is neither divisible by 4 or divisible by 3. {Alternatively, prove that if p is divisible by 4 or 3, the 2^p - 1 is divisible by some number other than +/- itself or +/-1.}

    Known, factor theorem: p(x) = q(x)(x-a) +r(x)

    I've tried setting 2^p - 1 = 0, but I get P log 2 = 0, I guess if someone could give me a hint or push me in the right direction I'd greatly appreciate it. thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello cndman
    Quote Originally Posted by cndman View Post
    Hey Guys, I have two problems that have to do with the factor theorem. I've been stumped on them for sometime now and can no longer put them off as my assignment is due soon.

    1) Use the factor theorem to show that if 2^p - 1, where p does not equal 3, is a prime number, then p is neither divisible by 4 or divisible by 3. {Alternatively, prove that if p is divisible by 4 or 3, the 2^p - 1 is divisible by some number other than +/- itself or +/-1.}

    Known, factor theorem: p(x) = q(x)(x-a) +r(x)

    I've tried setting 2^p - 1 = 0, but I get P log 2 = 0, I guess if someone could give me a hint or push me in the right direction I'd greatly appreciate it. thanks
    We use two results here:
    x^2-1=(x-1)(x+1) and x^3-1=(x-1)(x^2+x+1)
    First, suppose that p =4m, m \in \mathbb{N}, m\ge1. Then
    2^p-1 = 2^{4m}-1
    =(2^{2m})^2-1

     =(2^{2m}-1)(2^{2m}+1)
    \Rightarrow 2^p-1 has a factor 2^{2m}-1\ge 3 and a factor 2^{2m}+1\ge 5.
    Now suppose that p = 3m, m \in \mathbb{N}, m\ge2. Then
    2^p-1 = 2^{3m}-1
    =(2^{m})^3-1

     =(2^{m}-1)((2^{m})^2+2^m+1)
    \Rightarrow 2^p-1 has a factor 2^{m}-1\ge 3 and a factor (2^{m})^2+2^m+1\ge 21
    Grandad
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Need help on factor theorem.
    Posted in the Algebra Forum
    Replies: 2
    Last Post: February 9th 2010, 05:07 PM
  2. Factor Theorem
    Posted in the Algebra Forum
    Replies: 6
    Last Post: November 22nd 2009, 10:55 AM
  3. Factor theorem
    Posted in the Algebra Forum
    Replies: 8
    Last Post: November 2nd 2009, 02:10 PM
  4. Residue theorem troubles
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 16th 2009, 06:17 AM
  5. Factor Theorem and Remainder Theorem
    Posted in the Algebra Forum
    Replies: 2
    Last Post: September 8th 2007, 10:50 AM

Search Tags


/mathhelpforum @mathhelpforum