Density of numbers where a large prime factor satisfies a congruence

Question

I am looking for an upper bound on the number of integers $n<x$ such that $n$ has a prime factor $p>\log(x)^{(1+\delta)}$ such that $p \equiv a \mod b$. Where $a,b$ are fixed and coprime and $0<\delta<1$

This is related to a proof of the fact that "most" finite groups have "large" cyclic subgroups. I am trying to extend this work to prove a related result for a particular class a groups. Which follows from a 1976 paper by Bertram, "On large cyclic subgroups of finite groups".

I believe the upper bound should be such that the density is $1$ as we send $x\rightarrow \infty$, though this is not really my field so I'm struggling to come up with a proof.

Answer 1

This is a standard sieve problem: estimate the number of integers up to x that do not possess a prime factor from a given set. The number of integers $n<x$ which do not possess such a prime factor is $O(x \log\log x/\log x)$, which implies that the set on $n<x$ that do have the property has density 1. This is a simple application of a sieve upper bound; see Halberstam and Richert's book Sieve Methods, Chapter 2, for the basics of sieve theory.

Lower bounds for the number of integers $n<x$ which do not possess such a prime factor are more delicate. When $b=4$ and $a=1$, such lower bounds are known and will be of the form $c x \log\log x/\log x$ for a positive constant $c$; this is a consequence of a theorem of Iwaniec from 1972, using a much more advanced sieve method (see also Friedlander and Iwaniec, Opera de Cribro, Chapter 14). Thus, if $a\mod b$ is contained in the progression $1 \mod 4$ then the same lower bound applies. The same method applies to b=6 and a=1 (and hence any progression $a\mod b$ contained in $1\mod 6$). However, when b=4 and a=3, it is not currently known that there are infinitely many such integers, and the same applies to b=6,a=5.