# Infinitely many primes of the form $\lfloor \sqrt {3} \cdot n \rfloor$?

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 ?

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I don't know, if this helps... It is easy to see (Beatty's Theorem) that all the positive integers belong to exactly one of the sets $$\{[n\sqrt3]\mid n\in\mathbf{Z}_+\}$$ and $$\{[n\frac{\sqrt3+3}2]\mid n\in\mathbf{Z}_+\}.$$ This follows because $a=\sqrt3$ and $b=(3+\sqrt3)/2$ are both irrational and satisfy the equation $1/a + 1/b =1$. So it would suffice to show that the primes in the latter set only have a fractional density. Doesn't look any easier, really? – Jyrki Lahtonen Mar 20 '12 at 11:21
@JyrkiLahtonen My idea was to apply somehow Dirichlet's theorem on arithmetic progressions...but it seems that I cannot use that theorem for this problem... – pedja Mar 20 '12 at 13:08
@pedja: That is not the source. What I am asking is did you make this problem yourself? Did you get it from someone? Do you know that there is a resolution to this and looking for a way to resolve this? – Aryabhata Mar 20 '12 at 18:07
@pedja: I know about Beatty sequences. Please mention in the question that you made it yourself... – Aryabhata Mar 20 '12 at 18:47
@pedja: The reason for this is it is easy to make up problems, but resolving them could be surprisingly hard (eg: Goldbach conjecture). You should make it clear upfront so people know what to expect. – Aryabhata Mar 20 '12 at 19:16

This is too long to be a comment, so I will write it as an answer.

Given $n\in\mathbb{N}$, define the interval $$I_n=\Bigl[\frac{n}{\sqrt3},\frac{n+1}{\sqrt3}\Bigr].$$ If $I_n$ contains an integer, then $$n=\Bigl\lfloor\sqrt3\,\Bigl\lceil\frac{n}{\sqrt3}\Bigr\rceil\Bigr\rfloor.$$ If we could prove that there are infinitely many primes $p$ such that $I_p\cap\mathbb{N}\ne\emptyset$, then we could answer the question in the affirmative. I wil give a probabilistic argument, assuming that the fractional parts of $p/\sqrt3$, where $p$ runs over all primes, are uniformly distributed in $[0,1]$. In that case, the probability that $I_p\cap\mathbb{N}\ne\emptyset$ would be the width of the interval $I_p$, which is $1/\sqrt3=0.577\dots$ Thus, about $57\%$ of all primes should satisfy $I_p\cap\mathbb{N}\ne\emptyset$. Computation shows that $576874$ of the first $10^6$ primes verify the condition.

## Edit

According to Aryabhata's comment, $\bigl\{\,\{p\,\sqrt3\,\bigr\}: p \text{ is prime}\}$ is uniformly distributed in $[0,1]$. Then there are an infinite number of primes $p$ such that $I_p\cap\mathbb{N}\ne\emptyset$ and $$p=\Bigl\lfloor\sqrt3\,\Bigl\lceil\frac{p}{\sqrt3}\Bigr\rceil\Bigr\rfloor.$$

The same argument shows that given an irrational $\alpha>0$ there are infinite primes of the form $\lfloor\alpha\,n\rfloor$.

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$\{p_n \alpha\}$ is equidistributed in $(0,1)$ for irrational $\alpha$ ($p_n$ = $n^{th}$ prime). This was shown by Vinogradov in his attempt to prove Goldbach conjecture. – Aryabhata Mar 20 '12 at 22:28
@Aryabhata Thanks for the reference. I have edited my answer. – Julián Aguirre Mar 20 '12 at 22:49