Proof about arithmetic progressions

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Homework Statement


Prove that there exist arbitrarily long arithmetic progressions formed of different positive integers such that every 2 terms of these progressions are relatively prime.

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 I am 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.
 
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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?
 
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.
 
cragar said:
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.
Glad that worked - seemed like it should but I hadn't checked the details.