Proof about arithmetic progressions

In summary, the conversation discusses how to prove the existence of arbitrarily long arithmetic progressions formed of different positive integers where every two terms are relatively prime. The solution involves using a progression of numbers that are a certain distance apart and avoiding common divisors.
  • #1
cragar
2,552
3

Homework Statement


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

The Attempt at a Solution


I first thought of looking at odd numbers separated by a powers of 2 but I don't think this forms a progression.
It seems weird to me because if I have a+bx where a and b are fixed constants so I am adding
a multiple of x eventually x will equal a and then 2a so they won't be relatively prime.
unless its like how we can have arbitrarily long composite numbers because of n!+2...n!+n
then I could just maybe add a multiple of the prime between n! and 2n!.
 
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  • #2
It seems weird to me because if I have a+bx where a and b are fixed constants so I am adding
a multiple of x eventually x will equal a and then 2a so they won't be relatively prime.
Right, and the conclusion is that no infinite arithmetic progression with that property can exist.
If you want to avoid a common divisor of 2 for an arbitrary progression length, how can you do that? (You found the answer already)
If you want to avoid a common divisor of 3 for an arbitrary progression length, how can you do that? In particular, what can you say about b?
... generalize to all primes.
 
  • #3
Yes, you have to read the question aright. It's not saying there exists a progression that is arbitrarily long, but that for any given N you can find a progression longer than N.
 
  • #4
ok so I guess I could use this progression 1+n!,1+2n!,1+3n!,...1+(n)n!,
all of these are relatively prime because if i take an two term in this progression and look at their
difference rn!+1-(kn!+1) where k<r<n+1 then i get n!(r-k) and (r-k)<n and none of these terms are divisible
by any of the prime factors of n! or r-k.
 
  • #5
That is the right answer.
 

1. What is an arithmetic progression?

An arithmetic progression is a sequence of numbers in which the difference between any two consecutive terms is constant. For example, 2, 5, 8, 11, 14 is an arithmetic progression with a common difference of 3.

2. What is the formula for calculating the nth term of an arithmetic progression?

The formula for the nth term of an arithmetic progression is a + (n-1)d, where a is the first term and d is the common difference. This formula can be used to find any term in the sequence.

3. How do you prove that a sequence is an arithmetic progression?

To prove that a sequence is an arithmetic progression, you can show that the difference between any two consecutive terms is constant. This can be done by subtracting one term from the next and checking if the result is the same for all pairs of terms.

4. What is the significance of arithmetic progressions in mathematics?

Arithmetic progressions have many applications in mathematics, including in number theory, algebra, and calculus. They are also commonly used in real-world situations, such as calculating interest rates and predicting population growth.

5. Can an arithmetic progression have an infinite number of terms?

Yes, an arithmetic progression can have an infinite number of terms if the common difference is a rational number. However, if the common difference is an irrational number, the progression will have a finite number of terms.

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