Polynomial root density in a sieve

For an integer polynomial $F$ and squarefree $d$, let ${\rho }_F(d)$ count the roots of $F$ modulo $d$. The Chinese remainder theorem makes ${\rho }_F$ multiplicative, andThus $g(d)={\rho }_F(d)/d$ and $r_d=O({\rho }_F(d))$ define a sieve distribution.

For the polynomial $F(n)=n(n+2)$, take $\rho (2)=1$ and $\rho (p)=2$ at odd primes. On odd squarefree integers, the summand is at least $2^{\omega (d)}/d$. Put $y=\sqrt{z}$ and $L(y)={\sum }_{a\le y,a\text{ odd}}1/a=\frac{1}{2}\log y+O(1)$. The weight of odd pairs $a,b\le y$ is $L(y{)}^2$. The pairs for which $a$ is not squarefree, $b$ is not squarefree, or $(a,b)>1$ each have weight at most $L(y{)}^2{\sum }_{p\text{ odd}}{p}^{-2}$. Indeed, the weighted count of multiples of an odd $p^k$ is $p^{-k}L(y/p^k)\le p^{-k}L(y)$. Moreover,Hence the odd coprime squarefree pairs have weight at least $L(y{)}^2/6$. Their products $d=ab\le z$ are squarefree, and $2^{\omega (d)}$ counts all ordered coprime factorizations of $d$. This proves the displayed lower bound needed for the Selberg upper-bound sieve.

For $F(n)=n(n+2)$, every odd prime has two forbidden residue classes. Use Selberg sieve weights supported on odd squarefree integers $d\le R=x^{1/8}$. Their main quadratic form is $1/J$, where $J={\sum }_{d\le R}{\prod }_{p\mid d}2/(p-2)$. The truncated Euler-product lower bound applied to primes up to $R^{1/8}$ gives $J\gg {\log }^{2}R$: the mean logarithmic divisor size is $2{\sum }_{3\le p\le R^{1/8}}(\log p)/p<\frac{1}{2}\log R$, by the Mertens first theorem, and the full product is $\asymp {\log }^{2}R$, by the Mertens second theorem. The optimal Selberg weights have modulus at most one, so the remainder is at most $O({\sum }_{d,e\le R}[d,e])=O(R^4)$. The finitely many possible pairs with $p\le R$ contribute $O(R)$. Thus the count is $O(x/J+R^4+R)$, proving the bound.

For distinct natural numbers $h_1,\dots ,h_r$,Exclude the fixed primes dividing some $h_i-h_j$ so that the polynomial ${\prod }_i(n+h_i)$ has exactly $r$ roots modulo every remaining prime, then apply the Selberg upper-bound sieve with level $x^{1/4}$.