# Equidistribution of $\{p_n^2α\}$

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

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.

• 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. – Thomas Bloom Dec 23 '18 at 13:41
• I have rephrased the statement to hopefully make the condition a little clearer. – Thomas Bloom Dec 23 '18 at 14:03