Let $p_n$ be the $n$th prime and $\alpha$ an irrational number. Vinogradov proved that the sequence $\{p_n \alpha\}$ is equidistributed. Is it known whether the sequence $\{p_n^2 \alpha \}$ is equidistributed?


1 Answer 1


Yes - this follows from a general theorem of Bergelson, Kolesnik, Madritsch, Son, and Tichy (Theorem 2.1 in https://people.math.osu.edu/bergelson.1/BKMS_PrimePowers.pdf):

Let $\xi(x)=\sum_{j=1}^m\alpha_j x^{\theta_j}$ be a polynomial with real coefficients $\alpha_i\in\mathbb{R}$ such that $0<\theta_1<\cdots <\theta_m$ and either

a) at least one of the $\theta_j$ is not at integer, or

b) at least one of the $\alpha_j$ is irrational.

If $\xi$ satisfies at least one of these conditions then $(\xi(p))$ is uniformly distributed.

Your special case $\xi(x)=\alpha x^2$ was probably known earlier, but I can't find a reference for that at the moment.

EDIT: This theorem in the case $\theta_j$ all integers was proved by Rhin in 1973 using Vinogradov's method, see https://mathscinet.ams.org/mathscinet-getitem?mr=323731. According to that Mathscinet review, this result was also implicit in Vinogradov's book.

  • $\begingroup$ No, then - if all the powers are integers then the additional constraint that at least one $\alpha_j$ is irrational must be imposed. If at least one is not an integer, then it doesn't matter what the coefficients are, and the result holds even if they're all rational. $\endgroup$ Dec 23, 2018 at 13:41
  • 1
    $\begingroup$ I have rephrased the statement to hopefully make the condition a little clearer. $\endgroup$ Dec 23, 2018 at 14:03

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