When the number of divisors is a quadratic residue

TL;DR

This paper investigates the behavior of a multiplicative function related to the divisor function modulo a prime, revealing dependence on quadratic residues and providing bounds for specific cases.

Contribution

It introduces a new comparison between the mean value of a divisor-related function and the classical divisor function, highlighting the influence of quadratic residue properties.

Findings

The mean value of the function depends heavily on whether 2 is a quadratic residue modulo q.

For q=5, a bound for short sums is established using advanced techniques from the geometry of numbers.

The results connect divisor functions, quadratic residues, and bounds on exponential sums.

Abstract

Let $q > 2$ be a prime number and define $\lambda_q := \left( \frac{\tau}{q} \right)$ where $\tau(n)$ is the number of divisors of $n$ and $\left( \frac{\cdot}{q} \right)$ is the Legendre symbol. When $\tau(n)$ is a quadratic residue modulo $q$, then $\left( \lambda_q \star \mathbf{1} \right) (n)$ could be close to the number of divisors of $n$. This is the aim of this work to compare the mean value of the function $\lambda_q \star \mathbf{1}$ to the well known average order of $\tau$. The proof reveals that the results depend heavily on the value of $\left( \frac{2}{q} \right)$. A bound for short sums in the case $q=5$ is also given, using profound results from the theory of integer points close to certain smooth curves.

Equations

$$L(s,\lambda )=\frac{\zeta (2s)}{\zeta (s)}\left(\sigma >1\right).$$

$$\sum _{n⩽x}\left(\lambda \star 1\right)(n)=\lfloor x^{1/2}\rfloor .$$

$$L(s,{\lambda }_3)=\frac{\zeta (3s)}{\zeta (s)}\left(\sigma >1\right)$$

$$\sum _{n⩽x}\left({\lambda }_3\star 1\right)(n)=\lfloor x^{1/3}\rfloor .$$

$$\sum _{n⩽x}\left({\lambda }_q\star 1\right)(n).$$

$$\sum _{n⩽x}\tau (n)=x\left(\log x+2\gamma -1\right)+O\left(x^{\theta +\epsilon }\right)$$

$$\sum _{n⩽x}\left({\lambda }_q\star 1\right)(n)=x\zeta (q)P_q(1)\left\{\log x+2\gamma -1+q\frac{{\zeta }^{\prime }}{\zeta }(q)+\frac{P_q^{\prime }}{P_q}(1)\right\}+O_{q,\epsilon }\left(x^{\max \left(1/c_q,\theta \right)+\epsilon }\right)$$

$$P_q(1)=\prod _p\left(1+\sum _{m=c_q}^{q-1}\left\{\left(\frac{m+1}{q}\right)-\left(\frac{m}{q}\right)\right\}\frac{1}{p^m}\right)$$

$$\frac{P_q^{\prime }}{P_q}(1)=-\sum _p\log p\left(\frac{{\sum }_{m=c_q}^{q-1}\left\{\left(\frac{m+1}{q}\right)-\left(\frac{m}{q}\right)\right\}\frac{m}{p^m}}{1+{\sum }_{m=c_q}^{q-1}\left\{\left(\frac{m+1}{q}\right)-\left(\frac{m}{q}\right)\right\}\frac{1}{p^m}}\right).$$

$$\sum _{n⩽x}\left({\lambda }_q\star 1\right)(n)=x^{1/2}\zeta \left(\frac{q}{2}\right)R_q\left(\frac{1}{2}\right)+O_{q,\epsilon }\left(x^{1/3+\epsilon }\right)$$

$$R_q\left(\frac{1}{2}\right):=\prod _p\left(1+\sum _{m=3}^{q-1}\left\{\left(\frac{m+1}{q}\right)+\left(\frac{m}{q}\right)\right\}\frac{1}{p^{m/2}}\right).$$

$$\sum _{n⩽x}\left({\lambda }_q\star 1\right)(n){\ll }_q{x}^{1/2}{e}^{-c{\left(\log x^{1/4}\right)}^{3/5}{\left(\log \log x^{1/4}\right)}^{-1/5}}.$$

$$\sum _{n⩽x}\left({\lambda }_q\star 1\right)(n){\ll }_{q,\epsilon }{x}^{1/4}{e}^{{\left(\log \sqrt{x}\right)}^{1/2}(\log \log \sqrt{x}{)}^{5/2+\epsilon }}.$$

$$\sum _{n⩽x}\left({\lambda }_7\star 1\right)(n)\doteq \text{numprint}0.454x\left(\log x+2\gamma +\text{numprint}0.784\right)+O_{\epsilon }\left(x^{1/2+\epsilon }\right).$$

$$\sum _{n⩽x}\left({\lambda }_{23}\star 1\right)(n)\doteq \text{numprint}0.899x\left(\log x+2\gamma -\text{numprint}0.678\right)+O_{\epsilon }\left(x^{131/416+\epsilon }\right).$$

$$\sum _{n⩽x}\left({\lambda }_{13}\star 1\right)(n)\doteq \text{numprint}1.969x^{1/2}+O_{\epsilon }\left(x^{1/3+\epsilon }\right).$$

$$\sum _{n⩽x}\left({\lambda }_5\star 1\right)(n)\ll x^{1/2}{e}^{-c{\left(\log x^{1/4}\right)}^{3/5}{\left(\log \log x^{1/4}\right)}^{-1/5}}.$$

$${\lambda }_q:=\left(\frac{\tau }{q}\right)$$

$$(F\star G)(n):=\sum _{d\mid n}F(d)G(n/d)$$

$$a_r(n):=\{\begin{array}{ll}1,& \text{if }n=m^r;\\ 0,& \text{otherwise}.\end{array}$$

$${\delta }_c(x):=e^{-c(\log x{)}^{3/5}(\log \log x{)}^{-1/5}}\text{and}\omega (x):=e^{{\left(\log x\right)}^{1/2}(\log \log x{)}^{5/2+\epsilon }}.$$

$$M(x):=\sum _{n⩽x}\mu (n)\text{and}L(x):=\sum _{n⩽x}\lambda (n).$$

$$L\left(s,{\lambda }_q\right)=\zeta (qs)\zeta (s)\prod _p\left(1+\sum _{m=c_q}^{q-1}\left\{\left(\frac{m+1}{q}\right)-\left(\frac{m}{q}\right)\right\}\frac{1}{p^{ms}}\right)$$

$$c_q:=\{\begin{array}{ll}2,& \mathrm{if}q\equiv \pm 7(\mathrm{mod}24);\\ ⩾4,& \mathrm{if}q\equiv \pm 1(\mathrm{mod}24).\end{array}$$

$$L\left(s,{\lambda }_q\right)=\frac{\zeta (qs)\zeta (2s)}{\zeta (s)}\prod _p\left(1+\sum _{m=d_q}^{q-1}\left\{\left(\frac{m+1}{q}\right)+\left(\frac{m}{q}\right)\right\}\frac{1}{p^{ms}}\right)$$

$$d_q:=\{\begin{array}{ll}2,& \mathrm{if}q\equiv \pm 5(\mathrm{mod}24)\mathrm{or}q=3;\\ 3,& \mathrm{if}q\equiv \pm 11(\mathrm{mod}24).\end{array}$$

$$L\left(s,{\lambda }_q\right)$$

$$L\left(s,{\lambda }_q\right)=\zeta (qs)\zeta (s)\prod _p\left(1-\frac{1}{p^s}+\left(1-\frac{1}{p^s}\right)\sum _{m=2}^{q-1}\left(\frac{m}{q}\right)\frac{1}{p^{(m-1)s}}\right)$$

$$\left(1-\frac{1}{p^s}\right)\sum _{m=2}^{q-1}\left(\frac{m}{q}\right)\frac{1}{p^{(m-1)s}}$$

$$L\left(s,{\lambda }_q\right)=\frac{\zeta (qs)\zeta (2s)}{\zeta (s)}\prod _p\left(1+\frac{1}{p^s}+\left(1+\frac{1}{p^s}\right)\sum _{m=2}^{q-1}\left(\frac{m}{q}\right)\frac{1}{p^{(m-1)s}}\right)$$

$$\left(1+\frac{1}{p^s}\right)\sum _{m=2}^{q-1}\left(\frac{m}{q}\right)\frac{1}{p^{(m-1)s}}$$

$$G_q(s)=\zeta (qs)\prod _p\left(1+\sum _{m=c_q}^{q-1}\left\{\left(\frac{m+1}{q}\right)-\left(\frac{m}{q}\right)\right\}\frac{1}{p^{ms}}\right):=\zeta (qs)P_q(s):=\sum _{n=1}^{\infty }\frac{g_q(n)}{n^s}.$$

Full text

1. Introduction and main result

If $\lambda=(-1)^{\Omega}$ is the Liouville function, then

[TABLE]

This implies the convolution identity

[TABLE]

Define $\lambda_{3}:=\left(\frac{\tau}{3}\right)$ where where $\tau(n)$ is the number of divisors of $n$ and $\left(\frac{\cdot}{3}\right)$ is the Legendre symbol modulo $3$. Then from Proposition 3 below

[TABLE]

implying the convolution identity

[TABLE]

Now let $q>2$ be a prime number and define $\lambda_{q}:=\left(\frac{\tau}{q}\right)$ where $\left(\frac{\cdot}{q}\right)$ is the Legendre symbol modulo $q$. Our main aim is to investigate the sum

[TABLE]

When $\tau(n)$ is a quadratic residue modulo $q$, one may wonder if $\left(\lambda_{q}\star\mathbf{1}\right)(n)$ has a high probability to be equal to the number of divisors of $n$. It then could be interesting to study its average order and to compare it to that of $\tau$, i.e.

[TABLE]

where $\frac{1}{4}\leqslant\theta\leqslant\frac{131}{416}$, the left-hand side being established by Hardy [5], the right-hand side being the best estimate to date due to Huxley [6]. The main result of this paper can be stated as follows.

Theorem 1**.**

Let $q>3$ be a prime number.

1. $\triangleright$

If $q\equiv\pm 1\;(\operatorname{mod}8)$

[TABLE]

where $\theta$ is defined in (1), $c_{q}$ is given in (2) and

[TABLE] 2. $\triangleright$

If $q\equiv\pm 11\;(\operatorname{mod}24)$

[TABLE]

where

[TABLE] 3. $\triangleright$

If $q\equiv\pm 5\;(\operatorname{mod}24)$, there exists $c>0$ such that

[TABLE]

Furthermore, if the Riemann hypothesis is true, then for $x$ sufficiently large

[TABLE]

Example 2**.**

[TABLE]

2. Notation

In what follows, $x\geqslant e^{4}$ is a large real number, $\varepsilon\in\left(0,\frac{1}{4}\right)$ is a small real number which does not need to be the same at each occurrence, $s:=\sigma+it\in\mathbb{C}$, $q$ always denotes an odd prime number, $\left(\frac{\cdot}{q}\right)$ is the Legendre symbol modulo $q$ and define

[TABLE]

where $\tau(n):=\sum_{d\mid n}1$. Also, $\mathbf{1}$ is the constant arithmetic function equal to $1$.

For any arithmetic functions $F$ and $G$, $L(s,F)$ is the Dirichlet series of $F$, the Dirichlet convolution product $F\star G$ is defined by

[TABLE]

and $F^{-1}$ is the Dirichlet convolution inverse of $F$. If $r\in\mathbb{Z}_{\geqslant 2}$, then

[TABLE]

For some $c>0$, set

[TABLE]

Finally, let $M(x)$ and $L(x)$ be respectively the Mertens function and the summatory function of the Liouville function, i.e.

[TABLE]

3. The Dirichlet series of $\lambda_{q}$

Proposition 3**.**

Let $q\geqslant 3$ be a prime number. For any $s\in\mathbb{C}$ such that $\sigma>1$

1. $\triangleright$

If $q\equiv\pm 1\;(\operatorname{mod}8)$

[TABLE]

where

[TABLE] 2. $\triangleright$

If $q\equiv\pm 3\;(\operatorname{mod}8)$

[TABLE]

where

[TABLE]

Proof.

Set $\chi_{q}:=\left(\frac{\cdot}{q}\right)$ for convenience. From [8, Lemma 2.1], we have

[TABLE]

If $q\equiv\pm 1\;(\operatorname{mod}8)$, then $\left(\frac{2}{q}\right)=1$ and

[TABLE]

where

[TABLE]

Similarly, if $q\equiv\pm 3\;(\operatorname{mod}8)$, then $\left(\frac{2}{q}\right)=-1$ and

[TABLE]

where

[TABLE]

We achieve the proof noting that, if $q\equiv\pm 1\;(\operatorname{mod}24)$, then $\left(\frac{3}{q}\right)-\left(\frac{2}{q}\right)=\left(\frac{4}{q}\right)-\left(\frac{3}{q}\right)=0$ and, similarly, if $q\equiv\pm 11\;(\operatorname{mod}24)$, then $\left(\frac{3}{q}\right)+\left(\frac{2}{q}\right)=0$ whereas $\left(\frac{4}{q}\right)+\left(\frac{3}{q}\right)=2$. ∎

4. Proof of Theorem 1

4.1. The case $q\equiv\pm 1\;(\operatorname{mod}8)$

For $\sigma>1$, we set

[TABLE]

First observe that $c_{q}<q$ in the case $q\equiv\pm 1\;(\operatorname{mod}24)$. Indeed, among the $q-4$ integers $m\in\{4,\dotsc,q-1\}$, it is known from [3, p.76] that there are $\frac{1}{2}(q-3)-3$ of them such that $\left(\frac{m}{q}\right)=\left(\frac{m+1}{q}\right)$. Consequently there are $\frac{1}{2}(q+1)$ integers $m\in\{4,\dotsc,q-1\}$ verifying $\left(\frac{m}{q}\right)\neq\left(\frac{m+1}{q}\right)$, and the inequality follows.

Thus this Dirichlet series is absolutely convergent in the half-plane $\sigma>\frac{1}{c_{q}}$ where $c_{q}$ is given in (2), so that

[TABLE]

By partial summation, we infer

[TABLE]

From Proposition 3, $\lambda_{q}\star\mathbf{1}=g_{q}\star\tau$. Consequently

[TABLE]

where $\theta$ is defined in (1) and where we used

[TABLE]

4.2. The case $q\equiv\pm 11\;(\operatorname{mod}24)$

For $\sigma>1$, we set

[TABLE]

Since $q>5$, this Dirichlet series is absolutely convergent in the half-plane $\sigma>\frac{1}{3}$, so that

[TABLE]

From Proposition 3, $\lambda_{q}\star\mathbf{1}=h_{q}\star a_{2}$, hence

[TABLE]

4.3. The case $q\equiv\pm 5\;(\operatorname{mod}24)$

In this case, it is necessary to rewrite $L(s,\lambda_{q})$ in the following shape.

Lemma 4**.**

Assume $q\equiv\pm 5\;(\operatorname{mod}24)$. For any $\sigma>1$, $L(s,\lambda_{q})=\dfrac{K_{q}(s)}{\zeta(s)\zeta(2s)}$ with

[TABLE]

where

[TABLE]

and

[TABLE]

The Dirichlet series $L_{q}$ is absolutely convergent in the half-plane $\sigma>\frac{1}{5}$, and the Dirichlet series $\mathcal{L}_{q}$ is absolutely convergent in the half-plane $\sigma>\frac{1}{6}$.

Proof.

From Proposition 3, we immediately get

[TABLE]

Now suppose $q>5$ and $q\equiv\pm 5\;(\operatorname{mod}24)$. In this case, $\left(\frac{3}{q}\right)+\left(\frac{2}{q}\right)=-2$ and $\left(\frac{4}{q}\right)+\left(\frac{3}{q}\right)=0$ so that we may write by Proposition 3

[TABLE]

where

[TABLE]

Assume $q\equiv\pm 19,\pm 29\;(\operatorname{mod}120)$. Then

[TABLE]

$K_{q}(s)$ can therefore be written as

[TABLE]

Similarly, if $q\equiv\pm 43,\pm 53\;(\operatorname{mod}120)$, then

[TABLE]

Hence

[TABLE]

The proof is complete. ∎

We now are in a position to prove Theorem 1 in the case $q\equiv\pm 5\;(\operatorname{mod}24)$.

Assume first that $q\equiv\pm 19,\pm 29\;(\operatorname{mod}120)$ and let $\ell_{q}(n)$ be the $n$-th coefficient of the Dirichlet series $L_{q}(s)$. From Lemma 4, $\lambda_{q}\star\mathbf{1}=\ell_{q}\star a_{4}\star a_{2}^{-1}$ and therefore

[TABLE]

Since $L(z)\ll z\delta_{c}(z)$ for some $c>0$

[TABLE]

The Dirichlet series $L_{q}(s):=\sum_{n=1}^{\infty}\ell_{q}(n)n^{-s}$ is absolutely convergent in the half-plane $\sigma>\frac{1}{5}$, consequently

[TABLE]

and by partial summation

[TABLE]

We infer that

[TABLE]

Now suppose that the Riemann hypothesis is true. By [1], which is a refinement of [9], we know that $M(z)\ll_{\varepsilon}z^{1/2}\,\omega(z)$. The method of [9, 1] may be adapted to the function $L$ yielding

[TABLE]

Observe that, for any $a\geqslant 2$, $\varepsilon>0$ and $z\geqslant e^{e^{e}}$

[TABLE]

so that $L(z)\ll_{\varepsilon}z^{1/2}\,\omega(z)$ and hence

[TABLE]

achieving the proof in that case. The case $q=5$ is similar but simpler since $\lambda_{5}\star\mathbf{1}=a_{5}\star a_{2}^{-1}$ by (3).

Finally, when $q\equiv\pm 43,\pm 53\;(\operatorname{mod}120)$, we proceed as above. Let $\nu_{q}(n)$ be the $n$-th coefficient of the Dirichlet series $\mathcal{L}_{q}(s)$. Then $\lambda_{q}\star\mathbf{1}=\nu_{q}\star a_{4}^{-1}\star a_{2}^{-1}$ from Lemma 4, so that

[TABLE]

and estimating trivially yields

[TABLE]

and we complete the proof as in the previous case. ∎

Remark 5*.*

Let us stress that a bound of the shape

[TABLE]

for all $x$ sufficiently large and small $\varepsilon>0$, is a necessary and sufficient condition for the Riemann hypothesis. Indeed, if this estimate holds, then by partial summation the series $\sum_{n=1}^{\infty}\left(\lambda_{q}\star\mathbf{1}\right)(n)n^{-s}$ is absolutely convergent in the half-plane $\sigma>\frac{1}{4}$. Consequently, the function $K_{q}(s)\zeta(2s)^{-1}$ is analytic in this half-plane. In particular, $\zeta(2s)$ does not vanish in this half-plane, implying the Riemann hypothesis, proving the necessary condition, the sufficiency being established above.

5. A short interval result for the case $q=5$

5.1. Introduction

This section deals with sums of the shape

[TABLE]

where $x^{\varepsilon}\leqslant y\leqslant x$. From Theorem 1

[TABLE]

and if the Riemann hypothesis is true, then

[TABLE]

The purpose is to improve significantly upon these estimates when $y=o(x)$, by using fine results belonging to the theory of integer points near a suitably chosen smooth curve. To this end, we need the following additional specific notation. Let $\delta\in\left(0,\frac{1}{4}\right)$, $N\in\mathbb{Z}_{\geqslant 1}$ large, $f:\left[N,2N\right]\longrightarrow\mathbb{R}$ be any map, and define $\mathcal{R}(f,N,\delta)$ to be the number of elements of the set of integers $n\in\left[N,2N\right]$ such that $\|f(n)\|<\delta$, where $\|x\|$ is the distance from $x$ to its nearest integer. Note that the trivial bound is given by

[TABLE]

5.2. Tools from the theory

In what follows, $N\in\mathbb{Z}_{\geqslant 1}$ is large and $\delta\in\left(0,\frac{1}{4}\right)$. The first result is [7, Theorem 5] with $k=5$. See also [2, Theorem 5.23 (iv)].

Lemma 6** ($5$th derivative test).**

Let $f\in C^{5}\left[N,2N\right]$ such that there exist $\lambda_{4}>0$ and $\lambda_{5}>0$ satisfying $\lambda_{4}=N\lambda_{5}$ and, for any $x\in\left[N,2N\right]$

[TABLE]

Then

[TABLE]

Remark 7*.*

The basic result of the theory is the following first derivative test (see [2, Theorem 5.6]): Let $f\in C^{1}\left[N,2N\right]$ such that there exist $\lambda_{1}>0$ such that $\left|f^{\,\prime}(x)\right|\asymp\lambda_{1}$. Then

[TABLE]

This result is essentially a consequence of the mean value theorem.

The second tool is [4, Theorem 7] with $k=3$.

Lemma 8**.**

Let $s\in\mathbb{Q}^{*}\setminus\{\pm 2,\pm 1\}$ and $X>0$ such that $N\leqslant X^{1/s}$. Then there exists a constant $c_{3}:=c_{3}(s)\in\left(0,\frac{1}{4}\right)$ depending only on $s$ such that, if

[TABLE]

then

[TABLE]

Our last result relies the short sum of $\lambda_{5}\star\mathbf{1}$ to a problem of counting integer points near a smooth curve.

Lemma 9**.**

Let $1\leqslant y\leqslant x$. Then

[TABLE]

Proof.

Using (3), we get

[TABLE]

so that

[TABLE]

and for any integers $N\in\left]\left(16y^{2}x^{-1}\right)^{1/5},\left(2x\right)^{1/5}\right]$ and $n\in\left[N,2N\right]$

[TABLE]

so that the sum does not exceed

[TABLE]

as asserted. ∎

5.3. The main result

Theorem 10**.**

Assume $y\leqslant c_{3}\,x^{11/20}$ where $c_{3}:=c_{3}\left(\frac{5}{2}\right)$ is given in (5). Then

[TABLE]

Furthermore, if $y\leqslant c_{3}\,x^{19/36}$

[TABLE]

Proof.

We split the first term in Lemma 9 into three parts, according to the ranges

[TABLE]

In the first case, we use Lemma 6 with $\lambda_{4}=\left(xN^{-13}\right)^{1/2}$ and $\lambda_{5}=\left(xN^{-15}\right)^{1/2}$ which yields

[TABLE]

For the second range, we use Lemma 8 with $X=x^{1/2}$, $s=\frac{5}{2}$ and $\delta=y\left(N^{5}x\right)^{-1/2}$. Notice that the conditions $N>2x^{1/10}$ and $y\leqslant c_{3}\,x^{11/20}$ ensure that $\delta<\frac{1}{4}$ and $N^{2}\delta\leqslant c_{3}$. We get

[TABLE]

The last range is easily treated with (4), giving

[TABLE]

Using Lemma 9, we finally get

[TABLE]

and note that $x^{-1/40}y^{1/6}+x^{-3/20}y^{1/4}+x^{-1/5}y^{2/5}\ll x^{1/12}$ as soon as $y\leqslant x^{13/20}$. This completes the proof of the first estimate, the second one being obvious. ∎

6. Acknowledgments

The author deeply thanks Prof. Kannan Soundararajan for the help he gave him to adapt his result to the function $L(x)$, and Benoit Cloitre for bringing this problem to his attention.