If $a, d$ are relatively primes and $n$ is arbitrary given number, then does there exist a number in $\{a+d, a+2d, ..., a+nd\}$ which is relatively prime to any other number of $\{a+d, a+2d, ..., a+nd\}$?
I calculated a few cases with my bare hands, it seems to me true.
For sufficiently large $n$, by Dirichlet’s theorem, there must be a prime among $a+[\frac{n+1}{2}]d, ..., a+nd$, so this is true.
But I’d like to see whether it is true for any given $n$.
Could you tell me this is true or not?
Thank you in advance.
I took the question to ask that none of them is coprime to all the rest - that there is at least one partner for each.
This is a list of $26$ numbers from $a+d$ to $a+26d$, which all share factors.
Suppose $a+d$ is a multiple of $2,5,11$ and $17$.
Suppose $a+2d$ is a multiple of $3$.
Suppose $a+3d$ is a multiple of $7$ and $19$.
Suppose $a+4d$ is a multiple of $13$.
Then $a+kd$ are all even for all odd $k$.
$a+2d,a+8d,a+14d,a+20d,a+26d$ are multiples of $3$.
$a+4d,a+17d$ are multiples of $13$.
$a+6d,a+16d$ are multiples of $5$.
$a+10d,a+24d$ are multiples of $7$.
$a+12d,a+d$ are multiples of $11$.
$a+18d,a+d$ are multiples of $17$.
$a+3d,a+22d$ are multiples of $19$.
Use the Chinese Remainder Theorem to find $a,d$ coprime to each other and to the primes less than twenty to set this up.
$$713+6767*[1:26]$$