On the distribution of r-full primitive roots

Abstract

An asymptotic formula for the number of positive primitive roots, which are r-full integers and not exceeding a given range, is derived using properties of character sums and exponent pairs.

1 Introduction and result

For a fixed $r\in {\mathbb {N}}$, an r-full integer is a positive integer all of whose prime factors in its unique prime factorization appear with power $\ge r$. Usually, 2-full and 3-full integers are called square-full and cube-full integers, respectively. In [9, Section 8.5A], Shapiro considered the problem of counting primitive roots modulo an odd prime q that are square-full. His approach makes use of asymptotic estimates of certain character sums, which illustrates a rare phenomenon of a significant contribution arising from a quadratic character mod q. More precisely, Shapiro proved that the number of positive primitive roots mod q which are square-full and $\le x$ is equal to

$$\begin{aligned} \frac{\phi (q-1)}{q-1} (cx^{1\over 2}+O(x^{1\over 3}q^{1\over 6}(\log q)^{1\over 3}2^{\omega (q-1)})), \end{aligned}$$

(1)

where $\phi (n)$ is the Euler’s phi function, $\omega (n)$ denotes the number of distinct prime divisors of n, and

$$\begin{aligned} c=2\left( 1-q^{-1}\right) \sum _{p\ \text {square-free},\;(p\vert q)=-1}\;p^{-3\over 2}. \end{aligned}$$

In 2018, Munsch and Trudgian [6] improved on the error term in (1) and estimated c correctly by using the character estimate of Burgess in [2] and [5, Lemma 1.3] to show that such a number is equal to

$$\begin{aligned} \frac{\phi (q-1)}{q-1}\left( \frac{1}{\zeta (3)}\left( 1+\frac{1}{q}+\frac{1}{q^2}\right) ^{-1}C_q x^{1\over 2}+O(x^{1\over 3}(\log x)q^{1\over 9}(\log q)^{1\over 6}2^{\omega (q-1)})\right) , \end{aligned}$$

(2)

where $C_q \gg q^{-{1\over 8\sqrt{e}}}$. Recently, in [11], using the concept of exponent pair (in the problem of exponential sum estimates) and the lemmas used in the proof of Theorem 2.1 in [10], the second author further improved the estimate (2) with the following result: for a given odd prime $q\le x^{1\over 5}$, the number of positive primitive roots mod q which are square-full and $\le x$ is equal to

$$\begin{aligned}&\frac{\phi (q-1)}{q}\left\{ \frac{L({3\over 2},\chi _0)-L({3\over 2},\chi _1)}{L(3,\chi _0)}x^{1\over 2}+\frac{L({2\over 3},\chi _0)-\frac{1}{2}\sum _{\chi \in \Gamma _3}L({2\over 3},\chi ^2)}{L(2,\chi _0)} x^{1\over 3}\right\} \nonumber \\&\quad +O(x^{1\over 6}\phi (q-1)3^{\omega _{1,3}(q-1)}q^{{1\over 2}+\varepsilon }). \end{aligned}$$

(3)

Here, $\chi _0$ and $\chi _1\ne \chi _0$ denote, respectively, the principal and quadratic characters mod q, $\omega _{1,3}(n)$ denotes the number of distinct primes $p\equiv 1\!\pmod 3$ which are divisors of n and $\Gamma _3$ denote the set of all the cubic characters. Note that the terms with the cubic characters only occur if $3\mid (q-1)$.

A few remarks are now in order. First, the result in (3) clearly improves on (2) when $q<x^{3\over 25}$. Second, Munsch-Trudgian’s result (2) shows that for all sufficiently large q there is a positive square-full primitive root $<q$. Cohen and Trudgian [3] conjectured that this may in fact hold for $q>1052041$. In addition to the square-full case, in [11], cube-full primitive roots mod q were also treated.

It is then natural to try deriving similar asymptotic estimates for general r-full integers; this problem has indeed been posed as Exercise 1 in [9, p. 308]. The details of the square-full case ($r=2$) have already been carried out in [11]; in what follows, we consider only $r\ge 3$. Our main result reads as follows:

Theorem 1

Let r be an integer $\ge$3. Let q be an odd prime, $\varepsilon >0$ be fixed, $\chi$ be a Dirichlet character $\bmod \, q$ with associated Dirichlet L-function $L(s,\chi )$ and let $\chi _0$ be the principal character $\bmod \, q$. Let $A_{r}$ be the set of all non-principal characters $\bmod \, q$ of order d with $d\mid r$.

Then, for $q\le x^{1\over 2r+1}$, the number of r-full positive primitive roots $\bmod \, q$ that are $\le x$ is equal to

$$\begin{aligned}&\frac{\phi (q-1)}{q}\left( L\Big (\frac{r+1}{r}, \chi _0\Big )B_r\left( \frac{1}{r}, \chi _0\right) +\sum _{d\mid (q-1)} \frac{\mu (d)}{\phi (d)}\sum _{\chi \in A_{r}}L\Big (\frac{r+1}{r}, \chi \Big )B_r\left( \frac{1}{r}, \chi \right) \right) x^{1\over r}\\&\quad +\frac{\phi (q-1)}{q}\left( L\Big (\frac{r}{r+1}, \chi _0\Big ) B_r\left( \frac{1}{r+1}, \chi _0\right) \right. \\&\quad \left. +\sum _{d\mid (q-1)}\frac{\mu (d)}{\phi (d)}\sum _{\chi \in A_{r+1}}L\Big (\frac{r}{r+1}, \chi \Big )B_r\left( \frac{1}{r+1}, \chi \right) \right) x^{1\over r+1}\\&\quad + O(x^{1\over r+2}\phi (q-1)2^{\omega (q-1)}q^{{4\over 3}+\varepsilon }), \end{aligned}$$

where

$$\begin{aligned} B_r(s,\chi ) = \frac{L((r+2)s,\chi ^{r+2})\cdots L((2r-1)s,\chi ^{2r-1})}{\prod _{j=r+1}^{2r-1} L(2js,\chi ^{2j})} \Psi _r(s,\chi ) \end{aligned}$$

and

$$\begin{aligned}&\Psi _r(s,\chi )\nonumber \\&= \prod _{p\; \text {prime}} \left( 1+\frac{\chi ^{2r+3}(p)p^{-(2r+3)s}+\chi ^{2r+4}(p)p^{-(2r+4)s}+\cdots +\chi ^{3r^2-3r\over 2}(p)p^{-(3r^2-3r)s\over 2}}{1+\chi ^{r+1}(p)p^{-(r+1)s}+\cdots +\chi ^{2r-1}(p)p^{-(2r-1)s}} \right) ^{-1}. \end{aligned}$$

2 Auxiliary results and notation

As usual, let $\mu (n)$ denote the Möbius function. Let $\psi (x)=x-\lfloor x\rfloor -{1\over 2}$. For two integers a, b, the greatest common divisor of a and b is denoted $\gcd (a, b)$. As expounded in [11], the key analysis is to estimate appropriate character sums over the r-full integers. Before delving into this, let us collect now several preliminary results about character sums and related materials.

Lemma 2

Let f, g be arithmetic functions.

I. [7, Lemma 2]. Let $\omega$ and $\kappa$ be real numbers with $\omega >0,\;0<\kappa \ne 1$. Then

$$\begin{aligned} \sum _{n\le \omega }\frac{1}{n^\kappa }=\zeta (\kappa )-\frac{\omega ^{1-\kappa }}{\kappa -1}-\frac{\psi (\omega )}{\omega ^{\kappa }}+O(\omega ^{-\kappa -1}).\\ \end{aligned}$$

II. For real $x>1$, a non-principal character $\chi \bmod q$, we have

$$\begin{aligned} \sum _{k\le x}\chi (k)&=\sum _{j\le q}\chi (j)\left\lfloor \frac{x}{q}-\frac{j}{q}+1 \right\rfloor \end{aligned}$$

(4)

$$\begin{aligned} \sum _{k\le x} \chi (k)\cdot f(k)&=\sum _{j\le q} \chi (j) \sum _{\begin{array}{c} k\le x\\ k\equiv j \bmod q \end{array} } f(k). \end{aligned}$$

(5)

III. [8, Lemma 13]. Let $\omega \in {\mathbb {R}}\mathrm{,}\, \omega >0\mathrm{,}\ q\in {\mathbb {N}}\mathrm{,}\,q\ge 2$. We have

$$\begin{aligned} \sum _{n\le \omega ,\, \gcd (n,q)=1}f(n)=\sum _{d\vert q}\mu (d)\sum _{m\le \omega /d}f(md).\\ \end{aligned}$$

IV. For $\alpha > 0,\, \alpha \ne 1,$ and $0<\beta \le 1,$ we have

$$\begin{aligned} \sum _{ \begin{array}{c} n\le x\\ n\equiv \ell \bmod q \end{array} }\frac{1}{n^\alpha }=\frac{1}{q^{\alpha }}\zeta \left( \alpha ,\frac{\ell }{q}\right) +\frac{1}{1-\alpha }\cdot \frac{x^{1-\alpha }}{q}-\psi \left( \frac{x-\ell }{q}\right) x^{-\alpha }\ +O(qx^{-\alpha -1}), \end{aligned}$$

where $\zeta (\alpha ,\beta )=\sum _{n=0}^\infty (n+\beta )^{-\alpha }$.

V. [8, Lemma 17]. Let $x, \eta , \alpha , \omega$ be real numbers with $x\ge 1\mathrm{,} \alpha >0\mathrm{,} \eta \ge 1$, let j and q be positive integers with $1\le j\le q$, and let $(k,\ell )$ be an exponent pair with $k>0$ (see the notion of exponent pair in [4], Chapter 2]), and let

$$\begin{aligned} R(x,\eta ,\alpha ;q,j;\omega )=\sum _{\begin{array}{c} n\le \eta \\ n\equiv j\bmod {q} \end{array}}\psi \left( \frac{x}{n^\alpha }+\omega \right) , \end{aligned}$$

where $\omega$ is independent of n. Then

where the constants in the O-symbols depend only on $\alpha$.

VI. [9, Lemma 8.5.1][9, Lemma 8.5.1]. For a given odd prime p, the characteristic function for the set of primitive roots $\bmod p$ is where $\Gamma _d$ denotes the set of characters in the character group mod p that are of order d.

VII. Let $x,\rho ,\gamma \in {\mathbb {R}}$ with $x\ge 1\mathrm{,}\, \rho>0\mathrm{,}\,\gamma >0$ and let $1\le v\le x^{1\over \rho }$. We have

$$\begin{aligned} \sum _{m^\rho n^\gamma \le x} f(m) g(n)&=\sum _{m\le v} f(m) \sum _{n\le ({x\over m^\rho })^{1\over \gamma }} g(n) + \sum _{n\le ({x\over v^\rho })^{1\over \gamma } }g(n) \sum _{m\le ({x\over n^\gamma })^{1\over \rho }} f(m)\\&\quad -\sum _{m\le v}f(m) \sum _{n\le ({x\over v^\rho })^{1\over \gamma } }g(n). \end{aligned}$$

Proof

We need only prove parts II, IV and VII.

II. For (4), using the periodicity of $\chi \bmod q$, we have

$$\begin{aligned} \sum _{k\le x}\chi (k)&=\sum _{j\le q}\sum _{\begin{array}{c} k\le x\\ k \equiv j \bmod q \end{array} }\chi (k)=\sum _{j\le q}\sum _{\begin{array}{c} k\le x\\ k\equiv j \bmod q \end{array} } \chi (j)\\ {}&=\sum _{j\le q}\chi (j)\sum _{\begin{array}{c} k\le x\\ k\equiv j \bmod q \end{array} } 1 =\sum _{j\le q}\chi (j)\left\lfloor \frac{x}{q}-\frac{j}{q}+1 \right\rfloor . \end{aligned}$$

The proof of (5) is similar.

IV. Using [8, Lemma 14], we get

$$\begin{aligned} \sum _{ \begin{array}{c} n\le x\\ n\equiv \ell \bmod q \end{array} }\frac{1}{n^\alpha }&=q^{-\alpha }\sum _{0\le m\le \frac{x-\ell }{q}}\Big (m+\frac{\ell }{q}\Big )^{-\alpha }\\&=q^{-\alpha }\left( \zeta \left( \alpha ,\frac{\ell }{q}\right) +\frac{1}{1-\alpha }\left( \frac{x-\ell }{q}\right) ^{1-\alpha }-\psi \left( \frac{x-\ell }{q}\right) \left( \frac{x-\ell }{q}\right) ^{-\alpha }\right. \\&\quad \left. +\frac{\ell }{q}\left( \frac{x-\ell }{q}\right) ^{-\alpha }+O\left( \left( \frac{x-\ell }{q}\right) ^{-\alpha -1}\right) \right) , \end{aligned}$$

where we have used

$$\begin{aligned} \left( \frac{x-\ell }{q}\right) ^{1-\alpha }&=\left( \frac{x}{q}\right) ^{1-\alpha }-\frac{(1-\alpha )\ell }{q}\left( \frac{x}{q}\right) ^{-\alpha }+O\Big (\frac{\ell ^2}{q^2}\left( \frac{x}{q}\right) ^{-\alpha -1}\Big ),\,\,\\ \left( \frac{x-\ell }{q}\right) ^{-\alpha }&=\left( \frac{x}{q}\right) ^{-\alpha }+O\Big (\frac{\ell }{q}\left( \frac{x}{q}\right) ^{-\alpha -1}\Big ). \end{aligned}$$

VII. The proof follows from counting the number of lattice points under a hyperbolic-like region similar to that of [1, Theorem 3.17]. $\square$

3 Proof of Theorem 1

Let G(r) denote the set of all r-full integers. For a given Dirichlet character $\chi$ modulo an odd prime q, set

$$\begin{aligned} f_r(n,\chi )={\left\{ \begin{array}{ll} \chi (n),&{}\text {if}\quad n\in G(r), \\ 0,\qquad &{}\text {otherwise,} \end{array}\right. } \quad \text {and} \quad F_r(s,\chi )=\sum _{n=1}^\infty \frac{f_r(n,\chi )}{n^s}. \end{aligned}$$

(7)

Clearly, the function $f_r(n,\chi )$ is multiplicative in n, and so for $\mathfrak {R}(s)>1$, the series $F_r(s,\chi )$ has a Dirichlet product of the form

$$\begin{aligned} F_r(s,\chi )&= \prod _p (1+\chi ^r(p) p^{-rs}+\chi ^{r+1}(p) p^{-(r+1)s}+\cdots )\\&=\prod _p(1+\chi ^{r+1}(p)p^{-(r+1)s}+\cdots +\chi ^{2r-1}(p)p^{-(2r-1)s})L(rs,\chi ^r), \end{aligned}$$

where $L(s,\chi )=\prod _p \left( 1-\chi (p)\;p^{-s}\right) ^{-1}$. From the expression

$$\begin{aligned}&\frac{L(rs,\chi ^r)L((r+1)s,\chi ^{r+1})L((r+2)s,\chi ^{r+2})\cdots L((2r-1)s,\chi ^{2r-1})}{F_r(s,\chi )L((2r+2)s,\chi ^{2r+2})L((2r+4)s,\chi ^{2r+4})\cdots L((4r-2)s,\chi ^{4r-2}) }\\&\quad =\prod _{p} \frac{ (1+\chi ^{r+1}(p)p^{-(r+1)s})(1+\chi ^{r+2}(p)p^{-(r+2)s})\cdots (1+\chi ^{2r-1}(p)p^{-(2r-1)s})}{1+\chi ^{r+1}(p)p^{-(r+1)s}+\cdots +\chi ^{2r-1}(p)p^{-(2r-1)s} }\\&\quad =\prod _{p}\Big (1+\frac{ \chi ^{2r+3}(p)p^{-(2r+3)s}+\chi ^{2r+4}(p)p^{-(2r+4)s}+\cdots +\chi ^{3r^2-3r\over 2}(p)p^{-(3r^2-3r)s\over 2}}{1+\chi ^{r+1}(p)p^{-(r+1)s}+\cdots +\chi ^{2r-1}(p)p^{-(2r-1)s} }\Big )\\&\quad =\prod _p(1+a_r(p, s ;\chi )),\ \text {say}, \end{aligned}$$

we see that $a_r(p, s ;\chi )\sim p^{-(2r+3)\mathfrak {R}(s)} (p\rightarrow \infty )$, and so

$$\begin{aligned} F_r(s,\chi )&=L(rs,\chi ^r)\, \Psi _r(s,\chi )\prod _{j=r+1}^{2r-1}\frac{L(js,\chi ^j)}{L(2js,\chi ^{2j})}, \end{aligned}$$

where the product

$$\begin{aligned} \Psi _r(s,\chi ):=\prod _p(1+a_r(p, s ;\chi ))^{-1} \end{aligned}$$

has a Dirichlet series with abscissa of convergence ${1\over 2r+3}$ for $r\ge 3$.

To obtain our desired final estimate, we strategically write

$$\begin{aligned} F_r(s,\chi )=B_r(s,\chi ) D_r(s,\chi ), \end{aligned}$$

(8)

where, for $r\ge 3$,

$$\begin{aligned} D_r(s,\chi )&=\sum _{n=1}^\infty c_r(n,\chi )n^{-s}=L(rs,\chi ^r) L((r+1)s,\chi ^{r+1}) \end{aligned}$$

(9)

and

$$\begin{aligned} \sum _{n=1}^\infty b_r(n,\chi )n^{-s}&=: B_r(s,\chi )\\&= \frac{L((r+2)s,\chi ^{r+2})\cdots L((2r-1)s,\chi ^{2r-1})}{\prod _{j=r+1}^{2r-1} L(2js,\chi ^{2j})} \Psi _r(s,\chi ). \end{aligned}$$

By the well-known Perron’s formula and the fact that $L(s,\chi _0)$ has a first-order pole at $s=1$, we have

$$\begin{aligned} \sum _{n\le x}|b_r(n,\chi )|=O(x^{1\over r+2}). \end{aligned}$$

(10)

From (9), for real $x>0$, we set

$$\begin{aligned} C_r(x,\chi )=\sum _{n\le x} c_r(n,\chi ) =\sum _{n_1^rn_2^{r+1}\le x}\chi ^r(n_1) \chi ^{r+1}(n_2). \end{aligned}$$

(11)

In view of (8), we write

$$\begin{aligned} \sum _{n\le x}f_r(n,\chi )&=\sum _{n\le xq^{-(2r+1)}}b_r(n,\chi )C_r\left( {x\over n},\chi \right) \nonumber \\&\quad \, +\sum _{xq^{-(2r+1)}<n\le x}b_r(n,\chi )C_r\left( {x\over n},\chi \right) . \end{aligned}$$

(12)

First, we bound the second sum in (12). In view of (10), we have

$$\begin{aligned}&\sum _{xq^{-(2r+1)}<n\le x}b_r(n,\chi )C_r\left( {x\over n},\chi \right) \nonumber \\&\ll x^{1\over r} \sum _{xq^{-(2r+1)}<n\le x}\Big |{b_r(n,\chi )\over n^{1\over r}}\Big | \ll q^{4r+2\over r(r+2)}x^{1\over r+2}. \end{aligned}$$

(13)

Now we compute the first sum in (12). Using part VII of Lemma 2 with $\rho =r, \gamma =r+1$ and $v=x^{1\over 2r+1}$, the right-hand side of (11) can be written as

$$\begin{aligned} C_{r}(x,\chi )&=\sum _{m\le x^{1\over 2r+1}}\chi ^{r}(m)\sum _{n\le (\frac{x}{m^{r}})^{1\over r+1}}\chi ^{r+1}(n)\nonumber \\&+\sum _{n\le x^{1\over 2r+1}}\chi ^{r+1}(n)\sum _{m\le (\frac{x}{n^{r+1}})^{1\over r}}\chi ^r(m) \nonumber \\&\quad -\sum _{m\le x^{1\over 2r+1}}\chi ^r(m)\sum _{n\le x^{1\over 2r+1}}\chi ^{r+1}(n). \end{aligned}$$

(14)

Recalling the definition of $A_{r}$ and that of $A_{r+1}$ as given in the statement of Theorem 1, we analyze the sums in (14) in all possible four cases, namely, $\chi$ is the principal character $\chi _0$; $\chi$ $\in A_{r}$; $\chi$ $\in A_{r+1}$; and $\chi$ $\notin A_{r}\cup A_{r+1}\cup \{\chi _0\}$.

$\bullet$ For $\chi =\chi _0$, the principal character, from (14), we have

$$\begin{aligned} C_{r}(x,\chi _0)&=\sum _{\begin{array}{c} m\le x^{1\over 2r+1} \\ \gcd (m,q)=1 \end{array}} \sum _{\begin{array}{c} n \le (\frac{x}{m^{r}})^{1\over r+1} \\ \gcd (n,q)=1 \end{array} } 1 + \sum _{\begin{array}{c} n\le x^{1\over 2r+1} \\ \gcd (n,q)=1 \end{array}} \sum _{\begin{array}{c} m\le (\frac{x}{n^{r+1}})^{1\over r}\\ \gcd (m,q)=1 \end{array}} 1\\&-\sum _{\begin{array}{c} m\le x^{1\over 2r+1}\\ \gcd (m,q)=1 \end{array}} 1 \sum _{\begin{array}{c} n\le x^{1\over 2r+1}\\ \gcd (n,q)=1 \end{array}} 1. \end{aligned}$$

Using part III of Lemma 2 twice, the right expression becomes

$$\begin{aligned} C_{r}(x,\chi _0) =&\sum _{d\mid q}\sum _{t\mid q}\mu (d)\mu (t)\sum _{m\le x^{1\over 2r+1}t^{-1}}\left( \frac{x^{1\over r+1}}{d (tm)^{r\over r+1}} -\psi \left( \frac{x^{1\over r+1}}{d (tm)^{r\over r+1}}\right) -\frac{1}{2}\right) \\&+\sum _{d\mid q}\sum _{t\mid q}\mu (d)\mu (t)\sum _{m\le x^{1\over 2r+1}t^{-1}}\left( \frac{x^{1\over r}}{d (tm)^{r+1\over r}} -\psi \left( \frac{x^{1\over r}}{d (tm)^{r+1\over r}}\right) -\frac{1}{2}\right) \\ {}&-\left( \sum _{d\mid q}\mu (d)\left( \frac{x^{1\over 2r+1}}{d} -\psi \left( \frac{x^{1\over 2r+1}}{d}\right) -\frac{1}{2}\right) \right) ^2, \end{aligned}$$

where the last equality follows from using the definition of $\psi$. Using part I of Lemma 2 and the following identities: $\sum _{d\mid q}\mu (d)=0$ for $q>1$, $\sum _{d\mid q} {\mu (d)\over d} = {\phi (q)\over q}$, $\sum _{t\mid q} {\mu (t)\over t^\alpha }=\prod _{p\mid q}(1-p^{-\alpha })$ $(\alpha \in {\mathbb {R}})$ and $\zeta (s)\prod _{p\mid q}(1-p^{-s})=L(s, \chi _0)\, \, (s\ne 1),$ we have

$$\begin{aligned} C_{r}(x,\chi _0)&:= \frac{\phi (q)}{q}L\left( \frac{r+1}{r}, \chi _0\right) x^{1\over r}+\frac{\phi (q)}{q}L\left( \frac{r}{r+1}, \chi _0\right) x^{1\over r+1}\nonumber \\&\quad \, +O(q^{1+\varepsilon })-S_1-S_2, \end{aligned}$$

(15)

where

$$\begin{aligned} S_1=\sum _{d\mid q}\sum _{t\mid q}\mu (d)\mu (t)\sum _{m\le x^{1\over 2r+1}t^{-1}}\psi \left( \frac{x^{1\over r+1}}{d (tm)^{r\over r+1}}\right) \end{aligned}$$

and

$$\begin{aligned} S_2=\sum _{d\mid q}\sum _{t\mid q}\mu (d)\mu (t)\sum _{m\le x^{1\over 2r+1}t^{-1}}\psi \left( \frac{x^{1\over r}}{d (tm)^{r+1\over r}}\right) . \end{aligned}$$

We now use part V of Lemma 2 with the exponent pair $({2\over 7}, {4\over 7})$. For $q\le x^{1\over 2r+1}$, we have

$$\begin{aligned} S_1= & {} \sum _{d\mid q}\sum _{t\mid q}\mu (d)\mu (t) R\left( \frac{x^{1\over r+1}}{dt^{r\over r+1}}, \frac{x^{1\over 2r+1}}{t}, \frac{r}{r+1},1,0,0\right) \\\ll & {} \sum _{d\mid q}\sum _{t\mid q}(1+x^{1\over 4r+2}d^{1\over 2}t^{-1}+x^{2\over 6r+3}d^{-2\over 9}t^{-4\over 9})\\= & {} O(x^{2\over 6r+3}q^{{1\over 3}+\varepsilon }), \end{aligned}$$

and similarly, $S_2=O(x^{2\over 6r+3}q^{{1\over 3}+\varepsilon })$. Substituting these estimates of $S_1$ and $S_2$ into (15), for $q\le x^{1\over 2r+1}$, we get

$$\begin{aligned} C_{r}(x,\chi _0)&= \frac{\phi (q)}{q} L\left( \frac{r+1}{r}, \chi _0\right) x^{1\over r}+ \frac{\phi (q)}{q} L\left( \frac{r}{r+1}, \chi _0\right) x^{1\over r+1}\nonumber \\&\quad +O(x^{2\over 6r+3}q^{{1\over 3}+\varepsilon }). \end{aligned}$$

(16)

$\bullet$ For non-principal characters $\chi \in A_r$, from (14), we have

$$\begin{aligned}&C_{r}(x,\chi )=\sum _{\begin{array}{c} m\le x^{1\over 2r+1} \\ \gcd (m,q)=1 \end{array}}\sum _{n\le (\frac{x}{m^r})^{1\over r+1}}\chi ^{r+1}(n)\\&\qquad +\sum _{n\le x^{1\over 2r+1}}\chi ^{r+1}(n)\sum _{\begin{array}{c} m\le (\frac{x}{n^{r+1}})^{1\over r}\\ \gcd (m,q)=1 \end{array}}1-\sum _{\begin{array}{c} m\le x^{1\over 2r+1}\\ \gcd (m,q)=1 \end{array}} 1\sum _{n\le x^{1\over 2r+1}}\chi ^{r+1}(n). \end{aligned}$$

Using parts II and III of Lemma 2, we have

$$\begin{aligned} C_{r}(x,\chi )&=\sum _{j\le q} \chi (j) \sum _{d\mid q} \mu (d) \sum _{n\le \frac{x^{1\over 2r+1} }{d}} \left\lfloor \frac{ x^{1\over r+1} }{q(nd)^{r\over r+1}} -\frac{j}{q} +1 \right\rfloor \\&\quad +\sum _{j\le q}\chi (j)\sum _{d\mid q}\mu (d)\sum _{\begin{array}{c} n\le x^{1\over 2r+1}\\ n\equiv j \bmod q \end{array}}\left\lfloor \frac{x^{1\over r}}{d n^{r+1\over r}}\right\rfloor \\ {}&\quad -\sum _{j\le q}\chi (j)\sum _{d\mid q}\mu (d)\left\lfloor \frac{x^{1\over 2r+1}}{d }\right\rfloor \left\lfloor \frac{x^{1\over 2r+1}}{q }-\frac{j}{q}+1\right\rfloor \\&:=T_1+T_2+T_3. \end{aligned}$$

Using $\sum _{j\le q}\chi (j)=0$ for non-principal character $\chi ,\ \psi (x)=\psi (x+1)$, and for $q>1$, the identities $\sum _{d\mid q}\mu (d)=0,\ \sum _{d\mid q} {\mu (d)\over d}={\phi (q)\over q}$, the first term can be written as

$$\begin{aligned} T_1&=- \frac{1}{q} \sum _{j\le q}j\chi (j)\sum _{d\mid q}\mu (d)\left\lfloor \frac{x^{1\over 2r+1}}{d}\right\rfloor \\&\quad \, -\sum _{j\le q}\chi (j)\sum _{d\mid q}\mu (d)\sum _{n\le \frac{x^{1\over 2r+1}}{d}}\psi \left( \frac{x^{1\over r+1}}{q (nd)^{r\over r+1}}- \frac{j}{q}\right) \\&=- \frac{1}{q}\sum _{j\le q}j\chi (j)\sum _{d\mid q}\mu (d)\left\lfloor \frac{x^{1\over 2r+1}}{d}\right\rfloor -S_3, \end{aligned}$$

the last term as

$$\begin{aligned} T_3&=\frac{1}{q}\sum _{j\le q}j\chi (j)\sum _{d\mid q}\mu (d)\left\lfloor \frac{x^{1\over 2r+1}}{d }\right\rfloor \\&\quad \, +{\phi (q)\over q}x^{1\over 2r+1}\sum _{j\le q}\chi (j)\psi \left( \frac{x^{1\over 2r+1}}{q }- \frac{j}{q}\right) +O(q^{1+\varepsilon }) \end{aligned}$$

and the second term as

$$\begin{aligned} T_2&=x^{1\over r}\frac{\phi (q)}{q}\sum _{j\le q}\chi (j)\sum _{\begin{array}{c} n\le x^{1\over 2r+1}\\ n\equiv j \bmod q \end{array}} n^{-(r+1)\over r}\\&\quad \, -\sum _{j\le q}\chi (j)\sum _{d\mid q}\mu (d)\sum _{\begin{array}{c} n\le x^{1\over 2r+1}\\ n\equiv j \bmod q \end{array}} \psi \left( \frac{x^{1\over r}}{d n^{r+1\over r}}\right) \\&:=x^{1\over r}\frac{\phi (q)}{q}\sum _{j\le q}\chi (j)\sum _{\begin{array}{c} n\le x^{1\over 2r+1}\\ n\equiv j \bmod q \end{array}} n^{-(r+1)\over r}-S_4. \end{aligned}$$

Using part IV of Lemma 2 and $q^{-s}\sum _{j\le q}\chi (j)\zeta (s, \frac{j}{q})=L(s,\chi )$, the first term in $T_2$ becomes

$$\begin{aligned}&x^{1\over r} \frac{\phi (q)}{q}\sum _{j\le q}\chi (j)\sum _{\begin{array}{c} n\le x^{1\over 2r+1}\\ n\equiv j \bmod q \end{array}} n^{-(r+1)\over r}=\frac{\phi (q)}{q}L\left( \frac{r+1}{r}, \chi \right) x^{1\over r}\\&\qquad \qquad \qquad -\frac{\phi (q)}{q}\sum _{j\le q}\chi (j)\psi \left( \frac{x^{1\over 2r+1}}{q }-\frac{j}{q}\right) x^{1\over 2r+1}+O(q^{1+\varepsilon }). \end{aligned}$$

Grouping terms, we arrive at

$$\begin{aligned} C_{r}(x,\chi )&=\frac{\phi (q)}{q}L\left( \frac{r+1}{r}, \chi \right) x^{1\over r}+O(q^2)-S_3-S_4. \end{aligned}$$

(17)

By part V of Lemma 2 with the exponent pair $({2\over 7}, {4\over 7})$, for $q\le x^{1\over 2r+1}$, we have

$$\begin{aligned} S_3= & {} \sum _{j\le q}\chi (j)\sum _{d\mid q}\mu (d) R\Big (\frac{x^{1\over r+1}}{qd^{r\over r+1}}, \frac{x^{1\over 2r+1}}{d}, \frac{r}{r+1},1,1,\frac{-j}{q}\Big )\\\ll & {} \sum _{j\le q}\sum _{d\mid q}\Big (1+x^{1\over 4r+2}q^{1\over 2}d^{-1}+x^{2\over 6r+3}q^{-2\over 9}d^{-4\over 9}\Big )\\= & {} O(x^{2\over 6r+3}q^{{4\over 3}+\varepsilon }), \end{aligned}$$

and similarly, $S_4=O(x^{2\over 6r+3}q^{{4\over 3}+\varepsilon })$. Substituting these estimates of $S_3$ and $S_4$ into (17), for $q\le x^{1\over 2r+1}$, we get

$$\begin{aligned} C_r(x,\chi )&=\frac{\phi (q)}{q}L\left( \frac{r+1}{r}, \chi \right) x^{1\over r}+O(x^{2\over 6r+3}q^{{4\over 3}+\varepsilon }). \end{aligned}$$

(18)

$\bullet$ For non-principal characters $\chi \in A_{r+1}$, computation similar to the last case yields, for $q\le x^{1\over 2r+1}$,

$$\begin{aligned} C_r(x,\chi )&=\frac{\phi (q)}{q}L\left( \frac{r}{r+1}, \chi ^r\right) x^{1\over r+1}+O(x^{2\over 6r+3}q^{{4\over 3}+\varepsilon }). \end{aligned}$$

(19)

$\bullet$ For non-principal characters $\chi$ $\not \in A_{r}\cup A_{r+1}\cup \{\chi _0\}$, using (14) and (4) in part II of Lemma 2, $\lfloor x \rfloor =x-\psi (x)-{1\over 2}, \ \sum _{j\le q}\chi (j)=0$ for non-principal characters, and $\psi (x)=\psi (x+1)$, we have

$$\begin{aligned} C_{r}(x,\chi )&=-\sum _{j\le q}\chi ^{r+1}(j)\sum _{n\le x^{1\over 2r+1}}\chi ^r(n)\left( \frac{j}{q}+\psi \left( \frac{x^{1\over r+1}}{q n^{r\over r+1}}-\frac{j}{q} \right) \right) \\&\quad -\sum _{j\le q}\chi ^r(j)\sum _{m\le x^{1\over 2r+1}}\chi ^{r+1}(m)\left( \frac{j}{q}+\psi \left( \frac{x^{1\over r}}{q m^{r+1\over r}}-\frac{j}{q}\right) \right) \\&\quad -\sum _{j\le q}\sum _{h\le q}\chi ^r(j)\chi ^{r+1}(h)\left( \frac{j}{q}+\psi \left( \frac{x^{1\over 2r+1}}{q }-\frac{j}{q}\right) \right) \\&\quad \times \left( \frac{h}{q}+\psi \left( \frac{x^{1\over 2r+1}}{q }-\frac{h}{q}\right) \right) . \end{aligned}$$

Using (4) and (5) in part II of Lemma 2 and its proof, we have

$$\begin{aligned} C_{r}(x,\chi )&=-\sum _{j\le q}\sum _{h\le q}\chi ^{r+1}(j)\chi ^r(h)\sum _{\begin{array}{c} n\le x^{1\over 2r+1}\\ n\equiv h \bmod q \end{array}}\psi \left( \frac{x^{1\over r+1}}{qn^{r\over r+1} }-\frac{j}{q}\right) \\&\quad -\sum _{j\le q}\sum _{h\le q}\chi ^r(j)\chi ^{r+1}(h)\sum _{\begin{array}{c} m\le x^{1\over 2r+1}\\ m\equiv h \bmod q \end{array}}\psi \left( \frac{x^{1\over r}}{ qm^{r+1\over r} }-\frac{j}{q}\right) +O(q^2)\\&=:-S_5-S_6+O(q^2). \end{aligned}$$

Using part V of Lemma 2 with the exponent pair $({2\over 7}, {4\over 7})$ and $q\le x^{1\over 2r+1}$, we get

$$\begin{aligned} S_5= & {} \sum _{j\le q}\sum _{h\le q}\chi ^{r+1}(j)\chi ^r(h) R\left( \frac{x^{1\over r+1}}{q},x^{1\over 2r+1}, \frac{r}{r+1},q,h,\frac{-j}{q}\right) \\\ll & {} \sum _{j\le q}\sum _{h\le q}(1+x^{1\over 4r+2}q^{-1\over 2}+x^{2\over 6r+3}q^{-2\over 3})\\= & {} O(x^{2\over 6r+3}q^{4\over 3}), \end{aligned}$$

and similarly, $S_6 =O(x^{2\over 6r+3}q^{4\over 3})$. Thus,

$$\begin{aligned} C_{r}(x,\chi )&=O(x^{2\over 6r+3}q^{4\over 3}). \end{aligned}$$

(20)

Summing up from all possible cases, i.e., from (16), (18), (19) and (20), and combining with (12), (13) and (10), we have, for $q\le x^{1\over 2r+1}$,

$$\begin{aligned}&\sum _{n\le x}f_{r}(n,\chi )\nonumber \\&= {\left\{ \begin{array}{ll} \frac{\phi (q)}{q}L\Big (\frac{r+1}{r}, \chi _0\Big ) B_r(\frac{1}{r}, \chi _0)x^{1\over r}\\ +\frac{\phi (q)}{q} L\Big (\frac{r}{r+1}, \chi _0\Big ) B_r(\frac{1}{r+1}, \chi _0)x^{1\over r+1}\\ +O(x^{1\over r+2}q^{{1\over 3}+\varepsilon })&{}\text {if}\ \chi =\chi _0\\ \frac{\phi (q)}{q} L\Big (\frac{r+1}{r}, \chi \Big ) B_r(\frac{1}{r}, \chi )x^{1\over r}+O(x^{1\over r+2}q^{{4\over 3}+\varepsilon })&{}\text {if}\ \chi \in A_{r}\\ \frac{\phi (q)}{q}L\Big (\frac{r}{r+1}, \chi ^r\Big )B_r(\frac{1}{r+1}, \chi )x^{1\over r+1}+O(x^{1\over r+2}q^{{4\over 3}+\varepsilon })&{}\text {if} \ \chi \in A_{r+1}\\ O(x^{1\over r+2}q^{4\over 3})&{}\text {if}\ \chi \not \in A_{r}\cup A_{r+1}\cup \{\chi _0\}. \end{array}\right. } \end{aligned}$$

(21)

We are now ready to finish the proof. From (7) and part VI of Lemma 2, we see that for a given odd prime q, the number of r-full positive primitive roots mod q that are $\le x$ is equal to

$$\begin{aligned}&\sum _{m\le x}\frac{\phi (q-1)}{q-1} \sum _{d\mid q-1} \frac{\mu (d)}{\phi (d)} \sum _{\chi \in \Gamma _d} f_r(m,\chi )\\&\quad =\frac{\phi (q-1)}{q-1}\left( \sum _{m\le x}f_r(m,\chi _0)+\sum _{d\mid q-1,\, d>1} \frac{\mu (d)}{\phi (d)}\left( \sum _{\begin{array}{c} m\le x\\ \chi \in A_r \end{array}}f_r(m,\chi )+\sum _{\begin{array}{c} m\le x\\ \chi \in A_{r+1} \end{array}}f_r(m,\chi )\right) \right) \\&\qquad +\frac{\phi (q-1)}{q-1}\sum _{\begin{array}{c} d\mid q-1\\ d \not \mid r, r+1\\ d>1 \end{array}} \frac{\mu (d)}{\phi (d)}\sum _{\chi \in \Gamma _d}\sum _{m\le x}f_r(m,\chi ). \end{aligned}$$

We use the first three cases in (21) to get the main terms of sizes $x^{1\over r}$ and $x^{1\over r+1}$, in Theorem 1. Adopting the same reasoning as in [6], the error term is obtained from those in the first three cases in (21) together with the bound of the last sum as $O(x^{1\over r+2}2^{\omega (q-1)}q^{{4\over 3}+\varepsilon })$.