A Bound for a Sum of Products of Two Characters and Its Application

Abstract

Using the exponent pair method, a bound is derived for the sum ${\displaystyle \sum _{m^a{n}^b\le x}}{\chi }_1^a\left(m\right){\chi }_2^b\left(n\right)$, where $a,b$ are fixed positive integers, ${\chi }_1,{\chi }_2$ are primitive Dirichlet characters modulo $q_1$ and $q_2$, respectively, and ${\chi }_1^a,{\chi }_2^b$ are not principal characters. As an application, an estimate for the error term in an asymptotic formula for the number of square-full integers simultaneously belonging to two arithmetic progressions is obtained.

1. Introduction

The technique of character sums has proved useful in various fields of mathematics, particularly in dealing with counting problems in analytic number theory (see, for example, [1,2,3,4] (Chapter 8 in [3])). Character sums over products of more than one character have also proved useful in deriving a number of asymptotic estimates (see, for example, [5,6,7,8,9,10,11]). Because of such versatile applications, in [12], the problem of finding good bounds for character sums of the form

$$\begin{array}{c}\hfill S_{{\chi }_1,{\chi }_2}\left(x\right)=\sum _{mn\le x}{\chi }_1\left(m\right){\chi }_2\left(n\right)\end{array}$$

(1)

were investigated. In the present work, we proceed further to derive a good bound for more general sums than those found in (1). As a possible application, we look at the problem of finding the number of square-full integers simultaneously belonging to two arithmetic progressions and use our main theorem to obtain a good bound for the error term for its asymptotic estimate.

Our main result is:

Theorem 1.

Let $a,b$ be two fixed positive integers with $1\le a<b$. Let $q_1,q_2$ be two distinct positive integers. Let ${\chi }_1,{\chi }_2$ be two primitive Dirichlet characters modulo $q_1$ and $q_2$, respectively, which are subject to the condition that both ${\chi }_1^a$ and ${\chi }_2^b$ are not principal characters. For a positive real x, define

$$\begin{array}{c}\hfill V_{a,b}({\chi }_1,{\chi }_2;x)=\sum _{m^a{n}^b\le x}{\chi }_1\left(m^a\right){\chi }_2\left(n^b\right).\end{array}$$

(2)

Then, for $2\le q_1<q_2$ and $q_1^{-10/3}{q}_2^{13/3}<x^{1/(a+b)}$, we have

$$V_{a,b}({\chi }_1,{\chi }_2;x)=\left\{\begin{array}{cc}O\left(x^{2/(3a+3b)}{q}_1^{5/9}{q}_2^{7/9}\right)\hfill & \mathrm{if}2a>b,\hfill \\ O\left(x^{2/9a}{q}_1^{7/9}{q}_2^{5/9}logx\right)\hfill & \mathrm{if}2a=b,\hfill \\ O\left(x^{2/(5a+2b)}{q}_1^{(3a+2b)/(5a+2b)}{q}_2^{5a/(5a+2b)}\right)\hfill & \mathrm{if}2a<b.\hfill \end{array}\right.$$

Before embarking upon the proof of Theorem 1, let us consider some of its special cases that indicate its origin and importance.

• The work in [12] deals with the case $a=b=1$.
• When ${\chi }_1={\chi }_2=:\chi$ is a character modulo q and $a=1,b=2$, Theorem 1 yields

  $$V_{1,2}(\chi ,\chi ;x)=O(x^{2/9}{q}^{4/3}logx),$$

  which appears as a bound in Equation (64) of [2].
• When ${\chi }_1={\chi }_2=:\chi$ is a character modulo q and $a=2,b=3$, Theorem 1 yields

  $$V_{2,3}(\chi ,\chi ;x)=O\left(x^{2/15}{q}^{4/3}\right),$$

  which is an improvement of the bound in Equation (15) of [4].

To begin our proof, let

$$\psi \left(x\right)=x-\lfloor x\rfloor -\frac{1}{2}.$$

We need the following lemmas.

Lemma 1.

Let χ be a primitive character modulo $q,q\ge 2$. For a real $z>1$, we have

$$\begin{array}{c}\hfill \sum _{a\le z}\chi \left(a\right)=\sum _{j\le q}\chi \left(j\right)⌊\frac{z}{q}-\frac{j}{q}+1⌋.\end{array}$$

Proof. 

From the periodicity of the primitive character modulo q, we find that

$$\begin{array}{cc}{\displaystyle \sum _{a\le z}}\chi \left(a\right)\hfill & ={\displaystyle \sum _{j\le q}}{\displaystyle \underset{a\equiv jmodq}{{\displaystyle {\displaystyle \sum _{a\le z}}}}}\chi \left(a\right)={\displaystyle \sum _{j\le q}}{\displaystyle \underset{a\equiv jmodq}{{\displaystyle \sum _{a\le z}}}}\chi \left(j\right)\hfill \\ & ={\displaystyle \sum _{j\le q}}\chi \left(j\right){\displaystyle \underset{a\equiv jmodq}{{\displaystyle \sum _{a\le z}}}}1={\displaystyle \sum _{j\le q}}\chi \left(j\right)⌊\frac{z}{q}-\frac{j}{q}+1⌋.\hfill \end{array}$$

□

As elaborated in [13] (Section 2.3, Chapter 2), the notion of an exponent pair, which is crucial in our analysis, is defined as follows. Let $A>1/2,B\ge 1,1<h\le B$, and suppose that

$$\sum _{B<n\le B+h}{e}^{2\pi if\left(n\right)}=O\left(A^{\kappa }{B}^{\lambda }\right)$$

for some pair $(\kappa ,\lambda )$ of real numbers satisfying $0\le \kappa \le 1/2\le \lambda \le 1$, and for any differentiable real-valued function f satisfying

$$A\ll \left|f^{\prime }\left(x\right)\right|\ll A\mathrm{when}x\in [B,2B].$$

Then, we call $(\kappa ,\lambda )$ an exponent pair.

The next lemma is Lemma 17 from [2].

Lemma 2.

Let $x,\eta ,\alpha ,\omega$ be real numbers, j and q be positive integers with $x\ge 1,\alpha >0,\eta \ge 1,1\le j\le q$, $(k,\ell )$ be an exponent pair with $k>0$, and let

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

where ω is independent of n. Then,

$$R(x,\eta ,\alpha ;q,j;\omega )=O\left(1\right)+O(x^{\frac{-1}{2}}{\eta }^{1+\frac{\alpha }{2}}{q}^{-1})+\left\{\begin{array}{cc}O(x^{\frac{k}{k+1}}{\eta }^{\frac{\ell -\alpha k}{k+1}}{q}^{\frac{-\ell }{k+1}})\hfill & \mathrm{if}\ell >\alpha k,\hfill \\ O(x^{\frac{k}{k+1}}log\eta q^{\frac{-\alpha k}{k+1}})\hfill & \mathrm{if}\ell =\alpha k,\hfill \\ O({\left(x{q}^{-\alpha }\right)}^{\frac{k}{1+(1+\alpha )k-\ell }})\hfill & \mathrm{if}\ell <\alpha k,\hfill \end{array}\right.$$

where the O-constants only depend on α.

2. Proof of Theorem 1

Proof. 

For $x>1$, we have

$$\begin{array}{cc}\hfill V_{a,b}({\chi }_1,{\chi }_2;x)& =\sum _{m\le x^{1/(a+b)}}{\chi }_1\left(m^a\right)\sum _{n\le {(x/m^a)}^{1/b}}{\chi }_2\left(n^b\right)+\sum _{n\le x^{1/(a+b)}}{\chi }_2\left(n^b\right)\sum _{m\le {(x/n^b)}^{1/a}}{\chi }_1\left(m^a\right)\hfill \\ \hfill & -\sum _{m\le x^{1/(a+b)}}{\chi }_1\left(m^a\right)\sum _{n\le x^{1/(a+b)}}{\chi }_2\left(n^b\right).\hfill \end{array}$$

In view of Lemma 1, we obtain

$$\begin{array}{cc}\hfill & V_{a,b}({\chi }_1,{\chi }_2;x)\hfill \\ \hfill & =\sum _{j\le q_2}{\chi }_2^b\left(j\right)\sum _{m\le x^{1/(a+b)}}{\chi }_1^a\left(m\right)⌊\frac{x^{1/b}}{q_2{m}^{a/b}}-\frac{j}{q_2}+1⌋\hfill \\ \hfill & +\sum _{h\le q_1}{\chi }_1^a\left(h\right)\sum _{n\le x^{1/(a+b)}}{\chi }_2^b\left(n\right)⌊\frac{x^{1/a}}{q_1{n}^{b/a}}-\frac{h}{q_1}+1⌋\hfill \\ \hfill & -\sum _{h\le q_1}\sum _{j\le q_2}{\chi }_1^a\left(h\right){\chi }_2^b\left(j\right)⌊\frac{x^{1/(a+b)}}{q_2}-\frac{j}{q_2}+1⌋⌊\frac{x^{1/(a+b)}}{q_1}-\frac{h}{q_1}+1⌋\hfill \end{array}$$

Because $\lfloor x\rfloor =x-\psi \left(x\right)-\frac{1}{2},{\sum }_{j\le q}\chi \left(j\right)=0$ for non-principal characters $\chi$ and $\psi \left(x\right)=\psi (x+1)$, we have

$$\begin{array}{cc}\hfill V_{a,b}({\chi }_1,{\chi }_2;x)& =\sum _{j\le q_2}{\chi }_2^b\left(j\right)\sum _{m\le x^{1/(a+b)}}{\chi }_1^a\left(m\right)\left(\frac{x^{1/b}}{q_2{m}^{a/b}}-\frac{j}{q_2}+\frac{1}{2}-\psi \left(\frac{x^{1/b}}{q_2{m}^{a/b}}-\frac{j}{q_2}\right)\right)\hfill \\ \hfill & +\sum _{h\le q_1}{\chi }_1^a\left(h\right)\sum _{n\le x^{1/(a+b)}}{\chi }_2^b\left(n\right)\left(\frac{x^{1/a}}{q_1{n}^{b/a}}-\frac{h}{q_1}+\frac{1}{2}-\psi \left(\frac{x^{1/a}}{q_1{n}^{b/a}}-\frac{h}{q_1}\right)\right)\hfill \\ \hfill & -\sum _{j\le q_2}\sum _{h\le q_1}{\chi }_1^a\left(h\right){\chi }_2^b\left(j\right)\left(\frac{x^{1/(a+b)}}{q_2}-\frac{j}{q_2}+\frac{1}{2}-\psi \left(\frac{x^{1/(a+b)}}{q_2}-\frac{j}{q_2}\right)\right)\hfill \\ \hfill & \times \left(\frac{x^{1/(a+b)}}{q_1}-\frac{h}{q_1}+\frac{1}{2}-\psi \left(\frac{x^{1/(a+b)}}{q_1}-\frac{h}{q_1}\right)\right)\hfill \\ \hfill & =-\sum _{j\le q_2}{\chi }_2^b\left(j\right)\sum _{m\le x^{1/(a+b)}}{\chi }_1^a\left(m\right)\left(\frac{j}{q_2}+\psi \left(\frac{x^{1/b}}{q_2{m}^{a/b}}-\frac{j}{q_2}\right)\right)\hfill \\ \hfill & -\sum _{h\le q_1}{\chi }_1^a\left(h\right)\sum _{n\le x^{1/(a+b)}}{\chi }_2^b\left(n\right)\left(\frac{h}{q_1}+\psi \left(\frac{x^{1/a}}{q_1{n}^{b/a}}-\frac{h}{q_1}\right)\right)\hfill \\ \hfill & -\sum _{j\le q_2}\sum _{h\le q_1}{\chi }_1^a\left(h\right){\chi }_2^b\left(j\right)\left(\frac{j}{q_2}+\psi \left(\frac{x^{1/(a+b)}}{q_2}-\frac{j}{q_2}\right)\right)\left(\frac{h}{q_1}+\psi \left(\frac{x^{1/(a+b)}}{q_1}-\frac{h}{q_1}\right)\right)\hfill \\ \hfill & =:-T_1-T_2-T_3.\hfill \end{array}$$

(3)

We separately analyze each of the three terms $T_1,T_2,T_3$, which appear in (3). It is easy to see that

$$\begin{array}{cc}\hfill T_3& =\sum _{j\le q_2}\sum _{h\le q_1}{\chi }_1^a\left(h\right){\chi }_2^b\left(j\right)\left(\frac{j}{q_2}+\psi \left(\frac{x^{1/(a+b)}}{q_2}-\frac{j}{q_2}\right)\right)\left(\frac{h}{q_1}+\psi \left(\frac{x^{1/(a+b)}}{q_1}-\frac{h}{q_1}\right)\right)\hfill \\ \hfill & =O\left(q_1{q}_2\right).\hfill \end{array}$$

As for $T_1$, by applying Lemma 1 to the first part of the second sum, and applying the periodicity of character to the second part, we get

$$\begin{array}{cc}\hfill T_1& =\sum _{j\le q_2}{\chi }_2^b\left(j\right)\frac{j}{q_2}\sum _{h\le q_1}{\chi }_1^a\left(h\right)⌊\frac{x^{1/(a+b)}}{q_1}-\frac{h}{q_1}+1⌋\hfill \\ \hfill & +\sum _{j\le q_2}{\chi }_2^b\left(j\right)\sum _{h\le q_1}{\chi }_1^a\left(h\right)\sum _{\begin{array}{c}m\le x^{1/(a+b)}\\ m\equiv hmod{q}_1\end{array}}\psi \left(\frac{x^{1/a}}{q_2{m}^{a/b}}-\frac{j}{q_2}\right).\hfill \end{array}$$

Repeating the above steps to the first part, i.e, replacing $\lfloor x\rfloor =x-\psi \left(x\right)-1/2$, using ${\sum }_{j\le q}\chi \left(j\right)=0$ for non-principal characters, and $\psi \left(x\right)=\psi (x+m)$, $m\in \mathbb{Z}$, the first part is equal to

$$\begin{array}{c}\hfill \sum _{j\le q_2}{\chi }_2^b\left(j\right)\frac{j}{q_2}\sum _{h\le q_1}{\chi }_1^a\left(h\right)\left(-\frac{h}{q_1}+\psi \left(\frac{x^{1/(a+b)}}{q_1}-\frac{h}{q_1}\right)\right)=O\left(q_1{q}_2\right)\end{array}$$

yielding

$$-T_1=-\sum _{j\le q_2}{\chi }_2^b\left(j\right)\sum _{h\le q_1}{\chi }_1^a\left(h\right)\sum _{\begin{array}{c}m\le x^{1/(a+b)}\\ m\equiv hmod{q}_1\end{array}}\psi \left(\frac{x^{1/a}}{q_2{m}^{a/b}}-\frac{j}{q_2}\right)+O\left(q_1{q}_2\right).$$

Proceeding with $T_2$ in the same manner as that of $T_1$, we get

$$-T_2=-\sum _{h\le q_1}{\chi }_1^a\left(h\right)\sum _{j\le q_2}{\chi }_2^b\left(j\right)\sum _{\begin{array}{c}n\le x^{1/(a+b)}\\ n\equiv jmod{q}_2\end{array}}\psi \left(\frac{x^{1/a}}{q_1{n}^{b/a}}-\frac{h}{q_1}\right)+O\left(q_1{q}_2\right).$$

Thus,

$$\begin{array}{cc}\hfill V_{a,b}({\chi }_1,{\chi }_2;x)& =-\sum _{j\le q_2}\sum _{h\le q_1}{\chi }_2^b\left(j\right){\chi }_1^a\left(h\right)\sum _{\begin{array}{c}m\le x^{1/(a+b)}\\ m\equiv hmod{q}_1\end{array}}\psi \left(\frac{x^{1/b}}{q_2{m}^{a/b}}-\frac{j}{q_2}\right)\hfill \\ \hfill & -\sum _{j\le q_2}\sum _{h\le q_1}{\chi }_2^b\left(j\right){\chi }_1^a\left(h\right)\sum _{\begin{array}{c}n\le x^{1/(a+b)}\\ n\equiv jmod{q}_2\end{array}}\psi \left(\frac{x^{1/a}}{q_1{n}^{b/a}}-\frac{h}{q_1}\right)+O\left(q_1{q}_2\right)\hfill \\ \hfill & =:-S_1-S_2+O\left(q_1{q}_2\right).\hfill \end{array}$$

To estimate $S_1$ and $S_2$, we use the exponent pair $(k,\ell )=(2/7,4/7)$ in Lemma 2; we note that from $a<b$ we have $ka/b=2a/7b<4/7=\ell$ and keep in mind for the rest of the proof that $2\le q_1<q_2<q_1^{-10/3}{q}_2^{13/3}<x^{1/(a+b)}$. Using Lemma 2, we have

$$\begin{array}{cc}\hfill S_1& =\sum _{j\le q_2}\sum _{h\le q_1}{\chi }_2^b\left(j\right){\chi }_1^a\left(h\right)R\left(\frac{x^{1/b}}{q_2},x^{1/(a+b)},\frac{a}{b},q_1,h,\frac{-j}{q_2}\right)\hfill \\ \hfill & \ll \sum _{j\le q_2}\sum _{h\le q_1}\left(x^{1/(2a+2b)}{q}_1^{-1}{q}_2^{1/2}+x^{2/(3a+3b)}{q}_1^{-4/9}{q}_2^{-2/9}\right)+O\left(q_1{q}_2\right),\hfill \\ \hfill & =O\left(x^{1/(2a+2b)}{q}_2^{3/2}+x^{2/(3a+3b)}{q}_1^{5/9}{q}_2^{7/9}\right)+O\left(q_1{q}_2\right)\hfill \\ \hfill & =O\left(x^{2/(3a+3b)}{q}_1^{5/9}{q}_2^{7/9}\right),\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill S_2& =\sum _{j\le q_2}\sum _{h\le q_1}{\chi }_2^b\left(j\right){\chi }_1^a\left(h\right)R\left(\frac{x^{1/a}}{q_1},x^{1/(a+b)},\frac{b}{a},q_2,j,\frac{-h}{q_1}\right)\hfill \\ \hfill & \ll O\left(q_1{q}_2\right)+O\left(x^{1/(2a+2b)}{q}_1^{3/2}\right)+\left\{\begin{array}{cc}O\left(x^{2/(3a+3b)}{q}_1^{7/9}{q}_2^{5/9}\right)\hfill & \mathrm{if}2a>b,\hfill \\ O\left(x^{2/9a}{q}_1^{7/9}{q}_2^{5/9}logx\right)\hfill & \mathrm{if}2a=b,\hfill \\ O\left(x^{2/(5a+2b)}{q}_1^{(3a+2b)/(5a+2b)}{q}_2^{5a/(5a+2b)}\right)\hfill & \mathrm{if}2a<b\hfill \end{array}\right.\hfill \\ \hfill & =\left\{\begin{array}{cc}O\left(x^{2/(3a+3b)}{q}_1^{7/9}{q}_2^{5/9}\right)\hfill & \mathrm{if}2a>b,\hfill \\ O\left(x^{2/9a}{q}_1^{7/9}{q}_2^{5/9}logx\right)\hfill & \mathrm{if}2a=b,\hfill \\ O\left(x^{2/(5a+2b)}{q}_1^{(3a+2b)/(5a+2b)}{q}_2^{5a/(5a+2b)}\right)\hfill & \mathrm{if}2a<b.\hfill \end{array}\right.\hfill \end{array}$$

Combining these estimates, the theorem follows. □

3. Application

In this section, we illustrate a possible use of our theorem to derive a good error term in the problem of finding an asymptotic estimate of the number of square-full integers belonging simultaneously to two arithmetic progressions. An integer $n>1$ is called square-full if in its canonical prime representation each prime appears with exponent $\ge 2$; the integer 1 is square-full by convention. For $n\in \mathbb{N}$, let

$$g\left(n\right):=\left\{\begin{array}{cc}1\hfill & \mathrm{if}n\mathrm{is}\mathrm{square}-\mathrm{full},\hfill \\ 0\hfill & \mathrm{otherwise},\hfill \end{array}\right.$$

denote the characteristic function of square-full integers. Let $q_1,q_2$ be two relatively prime positive integers and let ${\ell }_i\in \{1,2,\dots ,q_i-1\}$ with $gcd({\ell }_i,q_i)=1(i=1,2)$. For $x>1$, define

$$G(x;{\ell }_1,{\ell }_2;q_1,q_2):=\sum _{\begin{array}{c}n\le x\\ \begin{array}{c}n\equiv {\ell }_{1}mod{q}_1\\ n\equiv {\ell }_{2}mod{q}_2\end{array}\end{array}}g\left(n\right),$$

(4)

which counts the number of square-full integers ($\le x$) simultaneously belonging to two arithmetic progressions. As evidenced from the proof of the Chinese remainder theorem ([14] Theorem 5.26), the set of solutions of the system of two congruences $n\equiv {\ell }_{1}mod{q}_1,n\equiv {\ell }_{2}mod{q}_2$ is contained in an arithmetic progression; as the number of square-full integers belonging to an arithmetic progression has a substantial proportion over $\mathbb{N}$ [1,4], the problem of seeking for an asymptotic estimate for $G(x;{\ell }_1,{\ell }_2;q_1,q_2)$ is non-trivial.

The orthogonality relation for Dirichlet characters mod $q_1,q_2$, Theorem 6.16 from [14] shows that

$$\begin{array}{cc}\hfill G(x;{\ell }_1,{\ell }_2;q_1,q_2)& =\frac{1}{\varphi \left(q_1\right)\varphi \left(q_2\right)}\sum _{{\chi }_{1}mod{q}_1}\sum _{{\chi }_{2}mod{q}_2}\overline{{\chi }_1}\left({\ell }_1\right)\overline{{\chi }_2}\left({\ell }_2\right)\sum _{n\le x}g\left(n\right){\chi }_1\left(n\right){\chi }_2\left(n\right).\hfill \end{array}$$

(5)

For brevity, let

$$\begin{array}{c}\hfill T(x;{\chi }_1,{\chi }_2):=\sum _{n\le x}g\left(n\right){\chi }_1\left(n\right){\chi }_2\left(n\right).\end{array}$$

Because each positive integer is square-full if and only if it can be written uniquely as $r^2{m}^3$, with m being square-free ([3] Lemma 8.3.1), we have

$$\begin{array}{c}\hfill T(x;{\chi }_1,{\chi }_2)=\sum _{r^2{m}^3\le x}{\mu }^2\left(m\right){\chi }_1\left(r^2{m}^3\right){\chi }_2\left(r^2{m}^3\right).\end{array}$$

Because ${\mu }^2\left(m\right)={\sum }_{d^2\mid m}\mu \left(d\right)$, we get

$$\begin{array}{cc}& T(x;{\chi }_1,{\chi }_2)=\sum _{r^2{m}^3\le x}\sum _{d^2\mid m}\mu \left(d\right){\chi }_1\left(r^2{m}^3\right){\chi }_2\left(r^2{m}^3\right)=\sum _{r^2{\left(d^{2}t\right)}^3\le x}\mu \left(d\right){\chi }_1\left(r^2{\left(d^{2}t\right)}^3\right){\chi }_2\left(r^2{\left(d^{2}t\right)}^3\right)\hfill \\ \hfill & =\sum _{d\le x^{1/6}}\mu \left(d\right){\chi }_1^6\left(d\right){\chi }_2^6\left(d\right)\sum _{r^2{t}^3\le x/d^6}{\chi }_1\left(r^2{t}^3\right){\chi }_2\left(r^2{t}^3\right).\hfill \end{array}$$

Using $\alpha \left(n\right)$ to denote the number of representations of each $n\in \mathbb{N}$ in the form $n=r^2{t}^3$, we get

$$\begin{array}{c}\hfill T(x;{\chi }_1,{\chi }_2)=\sum _{d\le x^{1/6}}\mu \left(d\right){\chi }_1^6\left(d\right){\chi }_2^6\left(d\right)\sum _{n\le x/d^6}\alpha \left(n\right){\chi }_1\left(n\right){\chi }_2\left(n\right).\end{array}$$

(6)

Our immediate task now is to bound the sum ${\sum }_n\alpha \left(n\right){\chi }_1\left(n\right){\chi }_2\left(n\right)$. It is easily checked using the Euler product formula, of which the Dirichlet series of the function $\alpha \left(n\right){\chi }_1\left(n\right){\chi }_2\left(n\right)$ is

$$\begin{array}{c}\hfill \sum _{n=1}^{\infty }\alpha \left(n\right){\chi }_1\left(n\right){\chi }_2\left(n\right)n^{-s}=L(2s,{\chi }_1^2{\chi }_2^2)L(3s,{\chi }_1^3{\chi }_2^3).\end{array}$$

Perron’s formula ([15] Theorem, p. 13) tells us that the main term of ${\sum }_{n\le x}\alpha \left(n\right){\chi }_1\left(n\right){\chi }_2\left(n\right)$ is

$$\underset{s=1/2,1/3}{\mathrm{Res}}L(2s,{\chi }_1^2{\chi }_2^2)L(3s,{\chi }_1^3{\chi }_2^3)\frac{x^s}{s},$$

with contribution from the cases where either ${\chi }_1^2{\chi }_2^2$ or ${\chi }_1^3{\chi }_2^3$ is a principal character. The dominating error term of ${\sum }_{n\le x}\alpha \left(n\right){\chi }_1\left(n\right){\chi }_2\left(n\right)$ is obtained by considering the cases where ${\chi }_1^2{\chi }_2^2$ and ${\chi }_1^3{\chi }_2^3$ are non-principal characters. As an example, let us complete the calculation when ${\chi }_1$ is a cubic character mod $q_1$ and ${\chi }_2$ is a quadratic character mod $q_2$. In this case, we obtain

$$\begin{array}{cc}\hfill & \sum _{n=1}^{\infty }\alpha \left(n\right){\chi }_1\left(n\right){\chi }_2\left(n\right)n^{-s}=L(2s,{\chi }_1^2{\chi }_2^2)L(3s,{\chi }_1^3{\chi }_2^3)\hfill \\ \hfill & =\prod _p{(1-{\chi }_1^2\left(p\right){\chi }_2^2\left(p\right)p^{-2s})}^{-1}{(1-{\chi }_1^3\left(p\right){\chi }_2^3\left(p\right)p^{-3s})}^{-1}\hfill \\ \hfill & =\prod _{\begin{array}{c}p\\ p\nmid q_1{q}_2\end{array}}{(1-{\chi }_1^2\left(p\right)p^{-2s})}^{-1}{(1-{\chi }_2^3\left(p\right)p^{-3s})}^{-1}\hfill \\ \hfill & =\left(\prod _p{(1-{\chi }_1^2\left(p\right)p^{-2s})}^{-1}{(1-{\chi }_2^3\left(p\right)p^{-3s})}^{-1}\right)\left(\prod _{\begin{array}{c}p\\ p\mid q_1{q}_2\end{array}}(1-{\chi }_1^2\left(p\right)p^{-2s})(1-{\chi }_2^3\left(p\right)p^{-3s})\right)\hfill \\ \hfill & =L(2s,{\chi }_1^2)L(3s,{\chi }_2^3)C(q_1,q_2),\hfill \end{array}$$

where

$$C(q_1,q_2):=\prod _{\begin{array}{c}p\\ p\mid q_1{q}_2\end{array}}(1-{\chi }_1^2\left(p\right)p^{-2s})(1-{\chi }_2^3\left(p\right)p^{-3s})$$

is a constant depending only on $q_1,q_2$. Putting

$$\begin{array}{c}\hfill L(2s,{\chi }_1^2)L(3s,{\chi }_2^3)=\sum _{n=1}^{\infty }\beta \left(n\right)n^{-s},\end{array}$$

we have

$$\beta \left(n\right)=\sum _{n=n_1^2{n}_2^3}{\chi }_1^2\left(n_1\right){\chi }_2^3\left(n_2\right).$$

Applying Theorem 1 with $a=2,b=3$, for $2\le q_1<q_2<q_1^{-10/3}{q}_2^{13/3}<x^{1/5}$, we have

$$\begin{array}{cc}\hfill \frac{1}{C(q_1,q_2)}\sum _{n\le x}\alpha \left(n\right){\chi }_1\left(n\right){\chi }_2\left(n\right)& =\sum _{n\le x}\beta \left(n\right)=\sum _{n_1^2{n}_2^3\le x}{\chi }_1^3\left(n_1\right){\chi }_2^2\left(n_2\right)=O(x^{2/15}{q}_1^{5/9}{q}_2^{7/9}).\hfill \end{array}$$

(7)

In view of (5)–(7), the error term in the estimation of $G(x;{\ell }_1,{\ell }_2;q_1,q_2)$ is $O(x^{1/6}{q}_1^{5/9}{q}_2^{7/9}/|C(q_1,q_2)|).$