Results 1 to 3 of 3

Math Help - Proof Perfect Numbers

  1. #1
    Newbie
    Joined
    Feb 2007
    Posts
    23

    Proof Perfect Numbers

    Prove the following theorem.

    Thm: n is an even perfect # iff n = 2^(m-1)*(2m - 1) where m >= 2 and 2^m - 1 is prime.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Apr 2006
    Posts
    401
    Quote Originally Posted by flash101 View Post
    Prove the following theorem.

    Thm: n is an even perfect # iff n = 2^(m-1)*(2m - 1) where m >= 2 and 2^m - 1 is prime.
    Ahh, this will be long. Here goes...

    Proof (<==)

    n = 2^(m - 1)*(2^m - 1) where m >= 2, 2^m - 1 is prime. Show n is an even perfect number.

    Note that since 2^m - 1 is prime implies m is prime.

    Let sigma(n) = sum of divisors.

    sigma(n) = 2n

    m >= 2 implies m - 1 >= 1 implies 2|n implies n is even.

    From 2|n we know 2^{exponent}*(2^m - 1)

    sigma(n) = sigma(2^(m - 1)*(2^m - 1)) = (fill in here after noting the following -------- that something: sigma(2^(m - 1))*sigma(2^m - 1)

    gcd(2^(m - 1), 2^m - 1) = 1, and therefore we can separate sigma over each piece since sigma is multiplicative.

    We know 2^m - 1 is odd; (there are no odd divisors of 2^(m - 1)

    From the red we know (2^m - 1)/2 - 1 * {note sigma(p) = 1 + p} and therefore (2^m -1 + 1) .. since 2^m - 1 is prime, remember.

    = 2^m*(2^m - 1)
    = 2*2^(m - 1)*(2^m - 1) = 2n (do you see how I got this?)

    See next post for (==> part of proof)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Apr 2006
    Posts
    401
    Quote Originally Posted by flash101 View Post
    Prove the following theorem.

    Thm: n is an even perfect # iff n = 2^(m-1)*(2m - 1) where m >= 2 and 2^m - 1 is prime.
    This part is a little harder...at least I think.

    Proof (==>) n is an even perfect number, so sigma(n) = 2n

    Let n = 2^(s)*t where s >= 1 and t is odd.. note we can do this for any number.. do you see why?

    sigma(n) = sigma(2^s*t) = sigma(2^s)*sigma(t) = [2^(s + 1) - 1]/[2 - 1]*sigma(t)

    = (2^(s + 1) - 1)*sigma(t)

    Note: red above that gcd(2^s, t) = 1

    Moving on...

    2n = (2^(s + 1) - 1)*sigma(t)

    2*2^s*t

    2^(s + 1)*t = (2^(s + 1) - 1)*sigma(t)

    Note (2^(s + 1) - 1) is odd

    So, since (2^(s + 1) - 1) is odd, 2^(s + 1)|sigma(t)

    So there exists a positive integer q such that sigma(t) = 2^(s + 1)*q

    2^(s + 1)*t = (2^(s + 1) - 1)*(2^(s + 1)*q)

    Cancel the 2^(s + 1)'s..

    t = (2^(s + 1) - 1)*q

    Recall that we started with n = 2^s*t

    n = 2^s*(2^(s + 1) - 1)*q

    Show q = 1 and that (2^(s + 1) - 1) is prime

    Note that q|t and q < t

    Add q to both sides:

    t + q = (2^(s + 1) - 1)*q + q

    t + q = 2^(s + 1)*q = sigma(t) !!

    If q does not equal 1, q is a divisor of t with 1 < q < t which implies that sigma(t) >= 1 + q + t

    Contradiction!!!

    Therefore, q = 1 and sigma(t) = t + q = t + 1

    Therefore, t is prime and t = 2^(s + 1) - 1

    Hence, n = s^2*(2^(s + 1) - 1) where 2^(s + 1) - 1 is prime and 2 >= 1.

    Q.E.D.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. proof with perfect and prime numbers
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: September 28th 2010, 07:46 PM
  2. Perfect Numbers ?
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: May 21st 2010, 06:27 PM
  3. Perfect Numbers
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: January 24th 2009, 03:47 AM
  4. Perfect Numbers
    Posted in the Discrete Math Forum
    Replies: 9
    Last Post: October 19th 2008, 03:00 PM
  5. [SOLVED] Proof on Perfect Numbers
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: October 20th 2006, 09:07 AM

Search Tags


/mathhelpforum @mathhelpforum