8
$\begingroup$

Is there a real number $A$ such that $$\left \lfloor n^{A} \right \rfloor$$ is a prime number (for all natural numbers $n$)? It is obvious that $A>1+\epsilon$ from the prime number theorem.

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ Piatetski-Shapiro's theorem (mentioned briefly here) gives $A > 12/11$ (which has been improved by Rivat and Wu to $A > 243/205$). There may be an elementary argument that no such $A$ exists. $\endgroup$
    – Jeremy Rouse
    Commented Nov 18, 2017 at 15:23
  • $\begingroup$ @JeremyRouse Thanks, it was useful for me. $\endgroup$
    – user102007
    Commented Nov 18, 2017 at 20:25

1 Answer 1

Reset to default
21
$\begingroup$

No, such an $A$ does not exist. First, $A$ cannot be an integer because then $\lfloor n^A\rfloor $ is never a prime for $n\geq 2$. So, assume that $A$ is not an integer. Then by Weyl's equidistribution theorem the fractional parts of $n^{A}/2$ are equidistributed modulo $1$. In particular $\{n^A/2\}\in [0,1/2)$ infinitely often. So $\lfloor n^{A}\rfloor$ is infinitely often an even number.

Improve this answer
$\endgroup$
4
  • $\begingroup$ A little confused why you took $\frac{n^{A}}{2}$ instead of $n^{A}$. For example: $n^{1.\left ( 7 \right )}$ is not working for $\forall n$. $\forall n$ is not the same thing as infinitely $n$. You have proved that $A$ can not be an irrational number (this is cool), but what about rational? Look at the theory of sieves, there is such a sum $\sum_{n=1}^{\infty }a\left ( n \right )< \infty $, this method only works thanks to $\forall n$. $\endgroup$
    – user102007
    Commented Nov 18, 2017 at 17:09
  • $\begingroup$ @Retro The point is that proving that the fractional part of $\{ n^A / 2\}$ is infinitely often smaller than $1/2$. This gives that $\lfloor n^A \rfloor = n^A - \{ n^A \} = 2 (n^A/2 - \{ n^A/2 \} ) = 2\lfloor n^A /2 \rfloor$, so $\lfloor n^A \rfloor $ is even. Also, we only need that this holds for one $n$ such that $\lfloor n^A \rfloor > 2$, because then this expression is not a prime number, and hence it is not prime for all $n$. $\endgroup$
    – wythagoras
    Commented Nov 18, 2017 at 18:02
  • $\begingroup$ @wythagoras Whether correctly I understand that $n^{A}$ it is possible to replace on $\frac{n^{A}}{2}$ if $n$ can be extended to infinity? $\endgroup$
    – user102007
    Commented Nov 18, 2017 at 18:07
  • $\begingroup$ I will change the question and accept the answer. $\endgroup$
    – user102007
    Commented Nov 18, 2017 at 18:29

You must log in to answer this question.