Results 1 to 2 of 2

Math Help - The sum of postive divisors Function

  1. #1
    Junior Member
    Joined
    Apr 2008
    From
    Gainesville
    Posts
    68

    The sum of postive divisors Function

    a) Let p and p+2 be twin primes. Prove that sigma(P+2)=sigma(p) + 2
    b)Prove or disprove the converse of part a above.

    For Part a i just can seem to equate the 2.

    I get that sigma(P+2) is [(P+2)^2 - 1]/[(P+2)-1] and rearranging i get
    P^2 + 4p +4 / P+1

    and then i get that sigma(P) +2 is [P^2-1/P-1] + 2 and rearranging i get
    2p^2 +4p -6 / 2(P-1)

    Maybe i am approaching it the wrong way?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,408
    The only two divisors of a prime is 1 and itself and so, the sum of its divisors is p + 1: \sigma (p) = p + 1

    Now apply this to both sides:
    \sigma (p+2) = (p+2) + 1
    \sigma (p) + 2 = (p + 1) + 2
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Sum of divisors function question
    Posted in the Number Theory Forum
    Replies: 5
    Last Post: August 31st 2010, 06:04 AM
  2. Sum of positive divisors function
    Posted in the Number Theory Forum
    Replies: 4
    Last Post: March 17th 2010, 06:42 AM
  3. Postive integrable continuous function problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 16th 2009, 04:00 PM
  4. Sum of Positive Divisors Function
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: October 12th 2008, 03:36 PM
  5. The sum of postive divisors Function
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: August 4th 2008, 11:24 AM

Search Tags


/mathhelpforum @mathhelpforum