# Counting prime points in a bounded region

Let $R$ be a semi-algebraic, compact region in $\mathbb{R}^n$ with positive Lebesgue measure. Let $N(R) = \# (R \cap \mathbb{Z}^n)$. Davenport's lemma asserts that we have

$$\displaystyle N(R) = \operatorname{Vol}(R) + O(\max\{\operatorname{Vol} \overline{R}\}),$$

where the maximum is taken over all projections $\overline{R}$ of $R$ onto coordinate subspaces obtained by setting some coordinates to zero. Is there an analogous statement for

$$\displaystyle N'(R) = \# (R \cap \mathfrak{P}^n),$$

where $\mathfrak{P}$ denotes the set of primes and their negatives (so that $N'(R)$ counts the number of lattice points in $R$ whose coordinates are all primes or the negative of a prime)?

• No, not even in one dimension. Fixing Vol$(R)$, there are intervals of that length with no primes whatsoever, and intervals of that length with lots of primes. Feb 16 '17 at 9:16
• However one has the prime number theorem, which basically says that the number of primes in an interval is roughly equal to the 'weighted length' of the interval, where the weight is $1/\log x$. Feb 16 '17 at 16:50
• That's not accurate. The weight is not 1 over the log of the interval length, but rather 1 over the log of the size of the integers in the interval. And that difference is precisely related to my initial comments (long intervals with no primes at all), and equally related to the fact that no such asymptotic formula can exist. Feb 16 '17 at 17:34

The best known version of the prime number theorem in short intervals is $\pi(x+y)-\pi(x)\sim\frac{y}{\log x}$, provided that $x^{7/12}<y=o(x)$. So you can cut a connected set $C\subseteq[0,x]^k$ into cubes with side length $x^{7/12}$, and discard all cubes intersecting the boundary. In this way you obtain that the number of lattice points with prime coordinates equals $(1+o(1))\int_C \frac{dx_1 \dots dx_k}{\log x_1\dots\log x_k} + \mathcal{O}(x^{7/12}|\partial C|)$. You now need some bound on the surface of your set, which was already done by Davenport.