Does there exist a positive irrational number $\alpha $, such that for any positive integer $n$ the number $\lfloor n\alpha \rfloor$ is not a prime?

Does there exist a positive irrational number $\alpha $, such that for any positive integer $n$ the number $\lfloor n\alpha \rfloor$ is not a prime?
My try if $\alpha=\sqrt{17}$ then $\lfloor n\alpha \rfloor=4n$

Solution 1:

This is called a Beatty sequence. Here is an arxiv paper on the least prime in a Beatty sequence, by Steuding and Technau. There will indeed always be a prime in the sequence (which answers the original question), and the cited paper gives an upper bound for the least such prime for $\alpha>1$. The bound for the OP's sequence (provided $\alpha>1$) is $$p\le L^{35-16\epsilon}\alpha^{2(1-\epsilon)}p^{1+\epsilon}_{m+l}$$

where $L=\log(2\alpha)$, $p_n$ denotes the numerator of the $n^\text{th}$ convergent to the regular continued fraction expansion of $\alpha$, and $m$ is the unique integer such that $p_m\le L^{16}\alpha^2<p_{m+1}$. $\epsilon$ can be chosen arbitrarily small, but $l$ depends on $\epsilon$.