Proof about arithmetic progressions

In summary, the conversation discusses proving the existence of arithmetic progressions with different positive integers where every two terms are relatively prime. The attempt at a solution involves considering odd integers, powers of 2, and factorials. Ultimately, it is determined that the progression n!x+1 will work, where x is a natural number and n can be as large as desired. The reasoning behind this is that any two terms in this progression will have a difference of a multiple of n!, making it impossible for them to have any common factors.
  • #1
cragar
2,552
3

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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  • #2
Deleted - sorry, I wasn't thinking straight.
 
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  • #3
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
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
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.
 

Related to Proof about arithmetic progressions

What are arithmetic progressions?

Arithmetic progressions are sequences of numbers where the difference between each consecutive term is constant.

Why are arithmetic progressions important in mathematics?

Arithmetic progressions are important because they have many real-world applications, such as in calculating interest rates and predicting future values. They also provide a basis for more complex mathematical concepts and proofs.

What is the formula for finding the nth term in an arithmetic progression?

The formula for finding the nth term in an arithmetic progression is an = a1 + (n-1)d, where an is the nth term, a1 is the first term, and d is the common difference between terms.

How do you prove the existence of an arithmetic progression with a given set of numbers?

To prove the existence of an arithmetic progression, you can use the difference between consecutive terms to show that it is constant. You can also use mathematical induction to show that the progression continues infinitely.

What are some common applications of arithmetic progressions in science and technology?

Arithmetic progressions are commonly used in fields such as computer science, physics, and engineering. They are used to model growth rates, predict future trends, and analyze data in various applications.

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