Proof about arithmetic progressions

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The discussion revolves around proving the existence of arbitrarily long arithmetic progressions of different positive integers where every two terms are relatively prime. The initial approach involves considering odd integers and their properties, particularly focusing on the differences between terms. A key insight is the use of factorials, specifically the expression n!x + 1, where x is a natural number. The difference between any two terms in this progression is shown to be n!a, which does not share common factors with n!x + 1, ensuring the terms remain relatively prime. The conclusion confirms that this method effectively constructs the desired arithmetic progressions.
cragar
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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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Deleted - sorry, I wasn't thinking straight.
 
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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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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