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Proof about arithmetic progressions

  1. Mar 26, 2014 #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 2 terms of these progressions are relatively prime.
    3. The attempt at a solution
    First i started to think about the odd integers like 2x+1 and how consecutive odd integers are relatively prime because their difference is 2 , but im pretty sure the question is asking about any 2 terms in the progression. Then i started to think about progressions of odd numbers that were separated by powers of 2 but then this would be a progression because the difference between consecutive numbers would not be the same. Then I was trying to think of something i could do with a factorial.
     
  2. jcsd
  3. Mar 26, 2014 #2

    AlephZero

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    Deleted - sorry, I wasn't thinking straight.
     
    Last edited: Mar 26, 2014
  4. Mar 27, 2014 #3

    haruspex

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    If two terms have a common factor, p, what can you say about their difference? If the step size is n!, what can you say about the relationship between p and n?
     
  5. Mar 27, 2014 #4
    ok thanks for the hint haruspex, I think n!x+1 will work where x=1,2,3....... and n is a natural number as large as you like. if we look at the difference between any 2 of the terms in the progression.
    like n!(x+a)+1-(n!x+1) =n!a where n>x+a , if n!(x+a)+1 and n!x+1 had common factors then it should divide their difference but their difference is n!a, and none of those factors divide n!x+1 because their would be a remainder because n! contains all the factors 1 up to n.
     
  6. Mar 28, 2014 #5

    haruspex

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    Glad that worked - seemed like it should but I hadn't checked the details.
     
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