Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 150 / 2

Let $\mathcal{A}=(a_n)$ be a finite nonnegative sequence, let $\mathcal{P}$ be a set of prime numbers, and defineThe sifting function isFor a finite set $A$ of integers, take $a_n$ to be the number of occurrences of $n$ in $A$; then $S(A,\mathcal{P};z)$ counts members divisible by no $p\in \mathcal{P}$ below $z$. For $d\mid P(z)$, write

Suppose the sieve distribution has the formwhere $g$ is a multiplicative arithmetic function on squarefree integers and $0\le g(p)<1$. DefineThen the Selberg upper-bound sieve statesfor $1<D\le z$, with immaterial endpoint changes under other level conventions.

To construct the weights, putand setThen ${\lambda }_1=1$. If $\gcd (n,P(z))=1$, the divisor sum ${\sum }_{d\mid n}{\lambda }_d$ equals one, and henceSumming against $a_n$ and expanding givesThe Selberg diagonalization of the positive quadratic form givesFinally, grouping the error by $m=[d,e]$ gives at most $3^{\omega (m)}$ pairs $(d,e)$ for each squarefree $m$; using $|{\lambda }_d|\le 1$ yields the stated remainder.

Letand let $W$ be the product of the finitely many primes at most $r$ or dividing some nonzero difference $h_i-h_j$. Sieve with the primes $p\nmid W$. For such a prime, the congruence $F(n)\equiv 0(\mathrm{mod}p)$ has exactly $r$ distinct roots. By the Chinese remainder theorem, the root-counting function $\rho (d)$ is multiplicative on squarefree $d$ coprime to $W$, andThus the polynomial root density in a sieve applies with

Take $z=x^{1/4}$ and $D=z$. For squarefree $d$ coprime to $W$,The supplied mean-value estimate, partial summation, and removal of square factors using $r^{\omega (n)}{\ll }_{r,ϵ}{n}^{ϵ}$ giveThe error in the Selberg upper-bound sieve isConsequently

If every $n+h_i$ is prime and all of them exceed $z$, then $F(n)$ has no prime divisor $p<z$ with $p\nmid W$, apart from a fixed finite set of divisibility cases absorbed into the implied constant. The cases with some $n+h_i\le z$ contribute $O_r(z)$. Therefore