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Proof about arithmetic progressions

  1. Jun 21, 2013 #1
    1. The problem statement, all variables and given/known data
    Prove that there exist arbitrarily long arithmetic progressions formed of different
    positive integers such that every two terms of these progressions are relatively prime.
    3. The attempt at a solution
    I first thought of looking at odd numbers separated by a powers of 2 but I don't think this forms a progression.
    It seems weird to me because if I have a+bx where a and b are fixed constants so I am adding
    a multiple of x eventually x will equal a and then 2a so they wont be relatively prime.
    unless its like how we can have arbitrarily long composite numbers because of n!+2.....n!+n
    then I could just maybe add a multiple of the prime between n! and 2n!.
     
  2. jcsd
  3. Jun 21, 2013 #2

    mfb

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    Right, and the conclusion is that no infinite arithmetic progression with that property can exist.
    If you want to avoid a common divisor of 2 for an arbitrary progression length, how can you do that? (You found the answer already)
    If you want to avoid a common divisor of 3 for an arbitrary progression length, how can you do that? In particular, what can you say about b?
    ... generalize to all primes.
     
  4. Jun 21, 2013 #3

    haruspex

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    Yes, you have to read the question aright. It's not saying there exists a progression that is arbitrarily long, but that for any given N you can find a progression longer than N.
     
  5. Jun 24, 2013 #4
    ok so I guess I could use this progression 1+n!,1+2n!,1+3n!,.........1+(n)n!,
    all of these are relatively prime because if i take an two term in this progression and look at their
    difference rn!+1-(kn!+1) where k<r<n+1 then i get n!(r-k) and (r-k)<n and none of these terms are divisible
    by any of the prime factors of n! or r-k.
     
  6. Jun 24, 2013 #5

    mfb

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    That is the right answer.
     
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