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Many properties of the set of the primes are inherited by any dense subset thereof, such as the existence of solutions to various linear equations. But there’s one property, recently proved, namely the existence of infinitely many gaps $p_{n+1}-p_n\leq C$ bounded by a constant, which is not inherited by dense subsets of the primes. For instance, it is possible to remove basically one half of the primes and be left with a subset of primes $\{p'_n\mid n\in\mathbb{N}\}$ such that $p'_{n+1}-p'_n\ll \log p_n$.

So instead one could ask if when coloring the primes in finitely many colours, one is guaranteed to find a colour class containing infinitely many pairs of primes whose difference is bounded by a common constant. This does not follow immediately from pigeonhole principle, which only says that one colour class contains infinitely many primes.

However, this follows easily if we recall that the work of Maynard does not only give infinitely many bounded gaps, that is $\liminf p_{n+1}-p_n<+\infty$, but also for any $m$ that $\liminf p_{n+m}-p_n < +\infty$. Let $H(m)$ be this liminf.

In other words there are infinitely many (disjoint) intervals of size $I_n=[x_n,x_n+H(m)]$ which contains at least $m+1$ primes. So if you colour the primes in $m$ colours, and you consider these intervals, you find that each one must contains at least two primes of the same colour. By pigeonhole principle again, there must be a colour for which this occurs infinitely often. Whence the existence in this colour class of infinitely gaps of size at most $H(m)$.

The squarefree numbers form a set which is well-known to have density $6/\pi^2$ in the integers, in other words their characteristic function $\mu^2$ satisfy $\sum_{n\leq x}\mu^2(n)=6/\pi^2x+o(x)$. In this set all your dream theorems about the count of linear patterns are true and easy.

Hardy-Littlewood squarefree tuples

The Hardy-Littlewood conjecture consists, in its strongest form, in an asymptotic, for every $\mathcal{H}=\{h_1,\ldots,h_k\}$, of the form $\displaystyle{\lvert\{n\leq x\mid n+h_1,\ldots,n+h_k\text{ are all prime}\}\rvert=(c(\mathcal{H})+o(1))\frac{x}{\log^kx}}$

where $c(\mathcal{H})$ is a constant possibly equal to 0. It is quite far from being proven, although interesting steps towards it have been made in the last ten years, by Goldston-Pintz-Yildirim and Maynard.

In the squarefree numbers, the analogous problem is rather easy and was settled by Mirsky in 1947. He proved that $\displaystyle{\lvert\{n\leq x\mid n+h_1,\ldots,n+h_k\text{ are all squarefree}\}\rvert=c(\mathcal{H})x+O(x^{2/3+\epsilon})}$

where $\displaystyle{c(\mathcal{H})=\prod_p(1-\frac{\nu_{\mathcal{H}}(p^2)}{p^2})}$

and $\nu_{\mathcal{H}}(p^2)$ is the cardinality of the size of the projection of the set $\mathcal{H}$ in $\mathbb{Z}/p^2\mathbb{Z}$, that is the number of forbidden classes modulo $p^2$ for $n$.

I show a quick way of deriving it with a much worse error term $O(x/\log x)$. Fix $w$, a large number; ultimately, $w=C\log x$ should be ok. Fix $W=\prod_{p\leq w}p^2$. Then the number $X_w$ of $n\leq x$ such that none of $n+h_1,\ldots,n+h_k$ have a square factor $p^2$ with $p\leq w$ is easy to compute: in $[W]$ or any interval of length $W$ there are by chinese remainder theorem this many of them $\displaystyle{ \prod_{p\leq w}(p^2-\nu_{\mathcal{H}}). }$
Thus in $[x]$, there are this many of them $\displaystyle{x/W^2\prod_{p\leq w}(p^2-\nu_{\mathcal{H}})+O(W^2)}.$

There are also people who are not squarefree, but escape our sieving by primes less than $w$. These are the guys such that at least one of the $n+h_i$ have a divisor $p^2$ with $p>w$ prime. Now there are are at most $x/p^2$ multiples of $p^2$ in $[x]$. So at most $\displaystyle{ \sum_{w
should be removed. Let’s try to balance the error terms $O(x/w)$ and $O(W^2)=\exp(2w(1+o(1))$; take $w=\frac{1}{2}(\log x-\log\log x)$.
This way both seem to be $O(x/\log x)$.

Green-Tao type theorem

We can also easily get asymptotics of the form $\displaystyle{\sum_{n\in\mathbb{Z}^d\cap K}\prod_{i\in[t]}\mu^2(\psi_i(t))=\text{Vol}(K)(\prod_p\beta_p+o(1))}$

where $K\subset [0,N]^d$ is a convex body and $\Psi$ a system of affine-linear forms of which no two are affinely related. Moreover $\beta_p$ is again the proportion of non-forbidden vectors $a=(a_1,\ldots,a_d)$ of residues modulo $p^2$. One way to do it would be to simply use the Hardy-Littlewood type asymptotics proved above coordinate by coordinate. We can also prove it quite easily, by first observing that the set of n where at least one of the $\psi_i$ vanishes has size $O(N^{d-1})$ because the 0-set of an affine-linear form is an affine hyperplane. Then as $\psi_i(n)=O(N)$, the divisors $a_i$ will have to satisfy $a_i\ll \sqrt{N}$. So now we restrict the set of $n$ to the ones where the forms don’t vanish. Then we partition the sum into one where $a_1$ is restricted to be at most $N^{\delta}$ (we fix some small $\delta >0$), and the remaining one. Using the well-known fact that $\mu^2(n)=\sum_{d^2\mid n}\mu(d)$, we remark that the remaining one equals $\displaystyle{ \sum_{N^{\delta}\leq a_1\ll \sqrt{N}}\mu(a_1)\sum_{n:a_1^2\mid \psi_1(n)}\mu^2(\psi_2(n))\cdots\mu^2(\psi_t(n))}$

Now the number of $n$ in the sum is $O(N^d/a_1^2+N^{d-1})$. Given that $\sum_{N^{\delta}\leq a_1} a_1^{-2}\ll N^{-\delta}$, we can bound this sum by $O(N^{d-\delta})$. We proceed similarly for $i=2,\ldots,t$, so that the sum to estimate is $\displaystyle{ \sum_{a_1,\ldots,a_t\leq x^\delta}\prod_i\mu(a_i)\lvert\{n\in\mathbb{Z}^d\cap K\mid \forall i \quad \psi_i(n)\equiv 0\mod a_i^2\}\rvert}$

Quite gross volume packing arguments indicate that the number of integral points (lattice points) to estimate above is $\text{Vol}(K)\alpha_{\Psi}(a_1^2,\ldots,a_t^2)+O(N^{d-1+2t\delta})$, where $\displaystyle{ \alpha_{\Psi}(k_1,\ldots,k_t)=\mathbb{E}_{n\in [\text{lcm}(k_i)]^d}\prod_i1_{k_i\mid\psi_i(n)}}$

is the density of zeroes modulo the $a_i^2$. Moreover we can extend the sum beyond $N^\delta$ at the cost of a mere $N^{O(\delta)}$. Hence finally for any $\epsilon >0$ we can write that $\displaystyle{\sum_{n\in\mathbb{Z}^d\cap K}\prod_{i\in[t]}\mu^2(\psi_i(t))=\text{Vol}(K)\prod_p\beta_p+o_\epsilon(N^{d-1+\epsilon}).}$

Indeed by multiplicativity it is easy to transform $\sum_{a_1,\ldots,a_t}\alpha_{\Psi}(a_1^2,\ldots,a_t^2)\prod_i\mu(a_i)$ into a product over primes.

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