I have the following question:
Prove that for every $ K \in \mathbb{N} , K\gt 1$ there exists an arithmetic progression with length K of co prime numbers. the numbers themselves aren't necessarily prime. but they are co prime.
define how to choose the common difference and why it works ?
I am having a struggle proving this, I had a feeling its related to induction and or euclid
Actually there are infinitely many AP's exist $\forall K$.
Let all elements of the sequence be odd, therefore the increment is even.
Let $r ≥ max(K-1,2)$
and $d = r\#$, where $\#$ is the primorial function, factorial could be used instead.
Let $A = \{1, 2, ..., K - 1\}$.
Let the starting term be $a_1 = d + 1$ (it could be parameterized by multiplying $d$ with an integer, so that even $a_1 = 1$),
and let the increment be $d$ (again, it could be parameterized by multiplying with an integer),
so any other term is $a_i = d + 1 + \alpha d$, $\alpha \in A$.
If the elements have a common prime factor, then it divides their difference, and vice versa, if it divides their difference, then it must divide the elements.
Any two elements' difference is $\beta d$, $\beta \in A$, and no prime factor of this divides any $a_i$ or $a_1$.