One may hope to extend the generalised Hardy-Littlewood conjecture stated (and proven in the finite complexity case in the paper linked) by Green and Tao to polynomial systems. For example given a polynomial $\displaystyle{\phi \in \mathbb{Z}[x_1,\ldots,x_d]}$ and a bounded domain $K\subset \mathbb{R}^d$ (probably a convex body would be more reasonable), one may be interested in an asymptotic for

$\displaystyle{\lvert\{x=(x_1,\ldots,x_d)\in K\cap \mathbb{Z}^d\mid \phi(x)\text{ is prime}\}\rvert.}$

We will rather try to get an asymptotic of the form

$\displaystyle{\sum_{(x_1,\ldots,x_d)\in\mathbb{Z}^d\cap K}\Lambda(\phi(x_1,\ldots,x_d))=\beta_{\infty}\prod_p\beta_p+o(\beta_{\infty})}$

where $\beta_{\infty}$ is basically the number of points with positive integer coordinates in $K$, so hopefully $\beta_{\infty}=\text{Vol}(K\cap \phi^{-1}(\mathbb{R}_+))$, and the local factors take into account the obstructions or the absence of obstructions to primality modulo $p$. Recall that $\Lambda$ classically denotes the von Mangoldt function. There are only very few non-linear polynomials for which an asymptotic for the number of prime values is available. The easiest one is obviously $\phi(x,y)=x^2+y^2$. Indeed, the primes which are sum of two squares are the primes congruent to 1 modulo 4 (and also the prime 2, but it’s a single prime, so we don’t have to care), and they are represented 8 times each as a value of $\phi$. So

$\displaystyle{\sum_{x^2+y^2\leq N}\Lambda(x^2+y^2)\sim\sum_{n\leq N,n\equiv 1\text{ mod } 4}\Lambda(n)\sim 4N.}$

Now let’s check what the conjecture would say. Here $\beta_{\infty}=\text{Vol}(\{(x,y)\mid x^2+y^2\leq N\})\sim \pi N$. What about the $\beta_p$ ? They are supposed to be $\beta_p=\frac{p}{p-1}\mathbb{P}(\phi(x,y)\neq 0\text{ mod } p)$. Now it easy to check that $\mathbb{P}(\phi(x,y)\equiv 0\text{ mod } p)=\frac{2p-1}{p^2}$ if $p\equiv 1\text{ mod } 4$, and $\mathbb{P}(\phi(x,y)\equiv 0\text{ mod } p)=\frac{1}{p^2}$ otherwise. So $\beta_p=\frac{p-1}{p}$ in the former case, and $\beta_p=\frac{p+1}{p}$ in the latter case. Thus, $\prod_p\beta_p$ doesn’t converge absolutely, in contrast with the traditional Green-Tao situation, where $\beta_p=1+O(1/p^2)$… However, if we imagine that to each prime $p\equiv 1\text{ mod } 4$ corresponds a prime $p'\equiv 3 \text{ mod } 4$ with $p\approx p'$, we could compute the product of the $\beta_p$ by grouping the factors into pairs $\beta_p\times\beta_{p'}\approx 1-1/p^2$. A bit more precisely,

$\displaystyle{\prod_{p\equiv 1\text{ mod } 4,p\leq N}(1+1/p)\sim C\sqrt{N}}$

and

$\displaystyle{\prod_{p\equiv 1\text{ mod } 4,p\leq N}(1-1/p)\sim C'/\sqrt{N}}$

which implies that

$\displaystyle{\prod_{p\leq N}\beta_p}\sim CC'$

so this product is convergent, although not absolutely. I guess the constant $C'$ is $e^{-\gamma/2}$, see Mertens theorem, and that $C=e^{\gamma/2}\prod_{p\equiv 1\text{ mod } 4}(1-p^{-2})$. I don’t really know how to compute this convergent product. We quickly notice that $\beta_2=1$. If the product could be equal to $4/\pi$, which it can’t unfortunately, being smaller than 1, we would apparently get the correct result, but this remains a quite dodgy case, which should make one cautious about stating ambitious generalisations of the Hardy-Littlewood conjecture.

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