Results 1 to 8 of 8

Math Help - two consecutive primes

  1. #1
    Newbie
    Joined
    Jan 2010
    Posts
    4

    two consecutive primes

    Congruence cases.


    (1) Let b be the sum of two consecutive odd numbers that are primes (for example, b = 11 + 13). Prove that b has at least three prime factors (not necessarily different).



    (2) Let b be the sum of two consecutive primes p and p', where p' > p >= 3 (for example, b = 89 + 97). Prove that b has at least three prime factors (not necessarily different).

    Up for the challenge?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Dec 2007
    From
    Anchorage, AK
    Posts
    276
    I don't know about (2), but as for (1), let the smaller odd be 2n+1, for some integer n; we see n\ge{1}, and the other odd number is 2n+3. Thus, their sum is (2n+1)+(2n+3)=4n+4=4(n+1)=2*2*(n+1);
    as n\ge{2}, it has one or more prime factors, and so the sum has at least three (not necessarily different) prime factors.

    --Kevin C.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2010
    From
    Illinois
    Posts
    30

    one of two

    Let's consider #1.
    Consecutive odd integers that are primes implies consecutive primes, neither of which is 2. Hence, both primes are odd.
    Therefore, b is even. It has a factor of 2.
    b=p1 + p2 = 2k.
    Then k = (p1+p2)/2, the average of p1 and p2.
    Assuming p1<p2, k = p1/2 + p2/2 > p2/2 + p2/2 = p2.
    Similarly, k > p1.
    But if p1 < k < p2, then k is the even integer that lies between the two odd primes. Therefore k = 2n for some integer, n. n > 1 since n=1 implies k=2 implies p1=1 and p2=3 which is a contradiction.

    I think you can take a similar approach in #2.
    Good luck.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member Dinkydoe's Avatar
    Joined
    Dec 2009
    Posts
    411
    I don't know about (2), but as for (1), let the smaller odd be 2n+1, for some integer n; we see , and the other odd number is 2n+3. Thus, their sum is (2n+1)+(2n+3)=4n+4=4(n+1)=2*2*(n+1);
    as , it has one or more prime factors, and so the sum has at least three (not necessarily different) prime factors.

    --Kevin C.
    You're assuming that 2 different consecutive primes p,p' can be written as p=2n+1, p'=2n+3 for the same n.
    This implies p'-p= 2 for all consecutive p,p'

    (p=13, p'= 17 are also consecutive primes!)
    Last edited by Dinkydoe; January 26th 2010 at 10:39 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member Dinkydoe's Avatar
    Joined
    Dec 2009
    Posts
    411
    This is how you do it:

    Observe that for any 2 consecutive primes p,p', we have 2|p+p' and

    p<\frac{p+p'}{2} < p' hence \frac{p+p'}{2} is composite since p,p' were consecutive primes.


    Conclusion: p+p' has 3 not necessarily different prime-factors.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jan 2010
    Posts
    2
    Quote Originally Posted by Dinkydoe View Post
    You're assuming that 2 different consecutive primes p,p' can be written as p=2n+1, p'=2n+3 for the same n.
    This implies p'-p= 2 for all consecutive p,p'

    (p=13, p'= 17 are also consecutive primes!)
    The first question asks for two consecutive odd numbers that are prime. Not two consecutive prime numbers that are odd. So I believe he is looking at the problem correctly.

    Your solution does apply to both cases though.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member Dinkydoe's Avatar
    Joined
    Dec 2009
    Posts
    411
    The first question asks for two consecutive odd numbers that are prime. Not two consecutive prime numbers that are odd. So I believe he is looking at the problem correctly
    .

    Ai, you're right, I'm sorry Kevin C. I misinterpreted the question
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Jan 2010
    Posts
    4
    Good job guys
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Consecutive Integers
    Posted in the Algebra Forum
    Replies: 9
    Last Post: September 9th 2011, 04:30 AM
  2. 3 Consecutive Odd Integers
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: December 3rd 2008, 05:41 PM
  3. Four consecutive odd integers...
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: May 5th 2007, 02:26 PM
  4. consecutive #s
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 9th 2006, 04:31 PM
  5. Consecutive Twin Primes Proof
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: December 18th 2005, 05:18 AM

Search Tags


/mathhelpforum @mathhelpforum