Proof about arithmetic progressions

[cragar]

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!.

 

[mfb]

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.

 

[haruspex]

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.

 

[cragar]

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.

 

[mfb]

That is the right answer.