# How to make the arithmetic progression contain infinitely many primes by adding a rational number to common difference?

Let $\alpha>0 \in \mathbb{R}$, $x_n=x_0+dn$, where $x_0=\frac{p_0}{q_0}$ and $d=\frac{p_1}{q_1}$ are rational numbers (e. g. $x_0=\frac 1 2$, $d=\frac 1 3$: $x_2= \frac 5 6\ldots$ or $x_0=\frac 1 2$, $d=\frac 3 4$: $x_0=\frac 1 2\$,$x_1=\frac 5 4\$, $x_2=2\ldots$)

Dirichlet proved that if $b$ and $r$ are relatively prime, there are always infinitely many primes of the form $bk+r$. But can I always add a rational number $s=\frac{p_2}{q_2} \in (0,\alpha)$ to common difference $d$ so that there would be infinitely many primes in progression $x_n=x_0+(d+s)n$?

• Consider the subsequence having integer values. Is it again of the form $kn+l$ with coprime integers $k,l$? – Dietrich Burde Feb 11 '18 at 19:28
• @DietrichBurde I can impose additional conditions to s, but can I always find arbitrarily small rational number that could "fix" my sequence? – Tyrone Ward Feb 11 '18 at 20:28