Proof about arithmetic progressions

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    Arithmetic Proof
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Homework Help Overview

The discussion revolves around proving the existence of arbitrarily long arithmetic progressions of different positive integers where every two terms are relatively prime. The original poster explores various approaches, including the properties of odd integers and factorials, to establish a valid progression.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster considers using odd integers and their properties, questioning the implications of using consecutive odd integers. They also explore the idea of using factorials in the progression. Other participants raise questions about the relationship between common factors and the step size in the progression.

Discussion Status

Participants are actively engaging with the problem, providing hints and exploring different formulations of the arithmetic progression. Some guidance has been offered regarding the use of factorials, and there is a sense of productive direction as participants refine their approaches.

Contextual Notes

There is an ongoing discussion about the assumptions regarding the nature of the integers in the progression and the implications of using different step sizes. The original poster's attempts indicate a need for clarity on the conditions under which the integers remain relatively prime.

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.
 
Last edited:
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.
 

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