1. Mar 26, 2014

### cragar

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. Mar 26, 2014

### AlephZero

Deleted - sorry, I wasn't thinking straight.

Last edited: Mar 26, 2014
3. Mar 27, 2014

### haruspex

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?

4. Mar 27, 2014

### cragar

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.

5. Mar 28, 2014

### haruspex

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