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The Siegel-Walfisz theorem is stated in https://en.m.wikipedia.org/wiki/Siegel–Walfisz_theorem.

I want to know if it can be extended unconditionally to a modulus $q$ such that any factorization $q=\prod_{i}q_{i}$ fulfills $(q_{i},q_{j})=(q_{i}q_{j})^{\delta_{ij}/2}$ (hence for $q$ being the product of distinct primes) and $p_{i}\mid q$ implies $p_{i}\ll \log^{A} x$ or if some strong conjecture like GRH is required.

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  • $\begingroup$ Should it be $(q_iq_j)^{\delta_{ij}/2}$ by a chance? $\endgroup$
    – Wojowu
    Commented Mar 6, 2020 at 8:34
  • $\begingroup$ Right, thank you for pinpointing the typo. I'll edit. $\endgroup$ Commented Mar 6, 2020 at 9:23

1 Answer 1

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For large $q$ such a result does not hold. This follows from the results in the paper "Limitations to the Equi-Distribution of Primes I" by Friedlander and Granville. Specifically, their Proposition 1 implies that for any constant $B>1$ and arbitrarily large $Q$ (assuming GRH, all sufficiently large $Q$ work), if $q$ is a number in the range $\left(\frac{Q}{(\log Q)^{1/8}},Q\right)$ with no prime factors below $\log q$, then for some $x\in(Q(\log Q)^B,3Q(\log Q)^B)$ and $a$ relatively prime to $q$ we have $\psi(x;q,a)-\frac{x}{\varphi(q)}\neq o(x)$.

It follows that for any $A\geq 1$, the uniform estimate $\psi(x;q,a)=\frac{x}{\varphi(q)}+o(x)$ cannot hold for all $q$ satisfying your condition. In case you are curious about $A<1$, it is impossible for a number $q$ to be a product of distinct primes smaller than $\log q$. Indeed, if $q$ is a product of $N$ primes below $\log q$, we have $N\leq\log q\leq\log 2^N<N$.

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  • $\begingroup$ Your answer is instructive but there must be a typo cause the interval to which $x$ belongs is empty. $\endgroup$ Commented Mar 6, 2020 at 10:45
  • $\begingroup$ Also, the hypothesis is "a number $q$ with no prime factors below $\log q$" but does the theorem apply if some prime factors of $q$ are below $\log q$ while others are bigger? $\endgroup$ Commented Mar 6, 2020 at 10:59
  • $\begingroup$ @SylvainJULIEN Thanks, the typo was a missing factor. I don't know to what extent the result holds if $q$ has small factors. I suspect that if it has few factors below $\log q$ (something like, their product is $(\log q)^{O(1)}$), then the result remains valid. $\endgroup$
    – Wojowu
    Commented Mar 6, 2020 at 11:02

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