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Proof about sum of integers

  1. Jun 20, 2013 #1
    1. The problem statement, all variables and given/known data
    Prove that every integer bigger than 6 can be written as a sum of 2 integers
    bigger than 1 which are relatively prime.
    3. The attempt at a solution
    Ill first look at the case where our number is odd.
    Let x be an odd integer. I will just add (x-2)+2=x since x is odd so is x-2 and 2 is even
    so x-2 and 2 are relatively prime.

    Now Lets look at the case where our number 2y is even.
    and y is even. 2y=y+y=(y+1)+(y-1) now since y is even y+1 and y-1 are odd. and y-1 and y+1 are odd numbers separated by a factor of 2.
    Lemma 1: Let n be an odd number. Lets assume for contradiction that n and[itex] n+2^x [/itex] have a common factor so it should divide their difference but [itex]n+2^x-n=2^x[/itex] but n and [itex]n+2^x [/itex] do not have a factor of 2 because they are odd.
    so y+1 and y-1 are relatively prime by lemma 1.


    Now lets look at the case where 2z=z+z where z is odd.
    we will just look at 2z=z+z=(z+2)+(z-2) since z is odd z-2 and z+2 are odd and they are odd numbers separated by a power of 2 so they are relatively prime.
     
  2. jcsd
  3. Jun 20, 2013 #2

    lurflurf

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    Homework Helper

    Consider 3 cases
    1)2n+1=(n)+(n+1)
    any n
    2)2n=(n+1)+(n-1)
    n odd
    3)4n=(2k+1)+(2k-1)
    any n
     
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