How to prove or disprove following statement :
There are infinitely many primes of the form : $\lfloor \sqrt {3} \cdot n \rfloor $
Note: This is a problem I made myself.
There is a theorem that states :
$\lfloor nx \rfloor = \begin{cases} n\lfloor x \rfloor, & \text{if } 0 \leq \{x\} < \frac{1}{n} \\ n\lfloor x \rfloor +1, & \text{if } \frac{1}{n} \leq \{x\} < \frac{2}{n} \\ n\lfloor x \rfloor +2, & \text{if } \frac{2}{n} \leq \{x\} < \frac{3}{n} \\ \vdots \\ n\lfloor x \rfloor +n-1, & \text{if } \frac{n-1}{n} \leq \{x\} < 1 \\ \end{cases}$
where $\{x\}$ is a non-integer part of $x$ .
Hence :
$\lfloor \sqrt{3}\cdot n \rfloor = \begin{cases} n, & \text{if } 0 \leq \sqrt{3}-1 < \frac{1}{n} \\ n +1, & \text{if } \frac{1}{n} \leq \sqrt{3}-1 < \frac{2}{n} \\ n +2, & \text{if } \frac{2}{n} \leq \sqrt{3}-1 < \frac{3}{n} \\ \vdots \\ 2n-1, & \text{if } \frac{n-1}{n} \leq \sqrt{3}-1 < 1 \\ \end{cases}$
How can I proceed from here ?