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

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