#### Abstract

In this paper, we consider divisor problems related to Hecke eigenvalues in three dimensions. We establish upper bounds and asymptotic formulas for these problems on average.

#### 1. Introduction

Let be the full modular group. Let denote the set of primitive holomorphic forms with even integral weight for . Then consists of common eigenfunctions of all Hecke operators . The Hecke eigenfunction has the following Fourier series expansion:where denotes the -th normalized eigenvalue. It is known that , as a function of , is real-valued and multiplicative. Moreover, for all integers , Deligne [1] showed thatwhere is the divisor function. Also, for every prime ,

Now we introduce some specific automorphic -functions. Define the Hecke -function attached to asfor . Moreover, the Rankinâ€“Selberg -function attached to can be defined as

Then, can be rewritten as

As usual, denotes the Riemann zeta-function. The symmetric square -function attached to can be defined as, for ,which can also be expressed in the Dirichlet series

The symmetric square -function has a functional equation and could be analytic continued to an entire function over the whole complex plane. We refer to works of Hecke in [2], Gelbert and Jacquet [3], Kim [4], and Kim and Shahidi [5,6] for these properties. Therefore, the symmetric square -function could be seen as a general -function in the sense of Perelli [7].

In number theory, considering the properties and average behaviors of the Fourier coefficients is meaningful and interesting. Some classical problems concern mean values of these Fourier coefficients and related problems with the corresponding automorphic -functions (see [1, 8â€“29], etc.). Here, we just give a brief history for general divisor problems related to these Fourier coefficients.

Let be an integer greater than one, and

In particular, we have and . In 1927, Hecke [30] showed that

Subsequently, this upper bound was improved by many scholars (for example, see [12, 21, 24]). In this direction, the best result so far was obtained by Wu [24] who showed thatwhere

Rankin [20] and Selberg [22] establishedwhere is a positive constant depending on . Later, Kanemitsu, Sankaranarayanan, and Tanigawa [31] studied general divisor problems and provedwhere is an integer and ; comes from a residue and takes the form ; denotes a polynomial of with degree . Fomenko [32] also proved the same result for the sum . After that, Kanemitsu, Sankaranarayanan, and Tanigawaâ€™s result was improved by LÃ¼ [33], and some general results were obtained (see [34â€“37], etc.).

In this paper, we consider divisor problems related to Hecke eigenvalues in three dimensions motivated by the above results and IviÄ‡â€™s work on three-dimensional divisor problems (see, e.g., [38]). We first introduce some notation. For any fixed integer , we define

We are interested in studying the average behaviors of sumswhich can be seen as divisor problems related to Hecke eigenvalues in three dimensions. We establish the following results.

Theorem 1. *Let , and be fixed integers satisfying . Then, for any , one has*

Theorem 2. *Let , and be fixed integers satisfying . Then, for any , one haswhere*

#### 2. Some Lemmas

To prove our theorems, we need to quote some lemmas, which include the individual and averaged subconvexity bounds for the Riemann zeta-function and the symmetric square -function. And from the following Lemma 1, we know that the Rankinâ€“Selberg -function could be decomposed into the product of the Riemann zeta-function and the corresponding symmetric square -function.

Lemma 1. *For , one has*

*Proof. *By comparing the Euler products of two sides of (20) and applying (3), we can get this lemma. This lemma can also be found in [33].

Lemma 2. *For any , one hasuniformly for , and the subconvexity bound in the critical stripwhere .*

*Proof. *These results are proved by Good [11].

Lemma 3. *For any , one hasuniformly for , and the subconvexity bound in the critical stripwhere .*

*Proof. *The twelfth mean value estimate (23) is due to Heath-Brown [39]. The subconvexity bound (24) is due to Bourgain [40].

Lemma 4. *For any , one hasuniformly for , and the subconvexity bound in the critical stripwhere .*

*Proof. *The first result follows from Perelliâ€™s mean value theorem with (see [7]), and the second one is due to Nunes [19].

#### 3. Proof of Theorem 1

In this section, we shall complete the proof of Theorem 1. Note that

Then, by (27) and Perronâ€™s formula (see Proposition 5.54 in [41]), we can obtainwhere and is a parameter to be chosen later.

We shift the line of the integral of (28) to the line . Then, Cauchyâ€™s residue theorem shows that

The following work is to estimate , and . The estimates for the integrals over the horizontal segments are similar, so we handle and first. To get this goal, we consider three cases , , and .

We first consider the case . To estimate and , we divide the integral interval into the following eight intervals , some of which may be empty, and use Lemma 2.

Interval :

In this interval, we have

Interval :

In this interval, we have

Interval :

This interval is an empty set noting that .

Interval :

In this interval, we have

Next, we handle . We have

Then, by Lemma 2 and applying Cauchyâ€™s inequality, we can obtain

According to (29), (37), and (39), we have

By taking in (40), we can obtainwhich proves the first result of Theorem 1.

For the case , to estimate and , we still divide the integral interval into four corresponding short intervals , which are different from ones for the case . In fact, the corresponding short intervals become empty sets in the current situation. However, we still can estimate similar to the corresponding parts of the case and get

The estimate of becomes the following at the current case by noting .

Recalling (29), we have

By taking in (44), we can getwhich proves the second result of Theorem 1.

For the case , to estimate and , we also divide the integral interval into four corresponding short intervals , which are different from ones for the case . In fact, the corresponding short intervals become empty sets in the current situation. However, we still can estimate similarly to the corresponding parts of the case and get

The estimate of becomes the following by noting .

Recalling (29), we have

By taking in (48), we can getwhich proves the third result of Theorem 1.

#### 4. Proof of Theorem 2

In this section, we shall prove Theorem 2, the process of which is more complicated than Theorem 1. Note that

Then, by (50) and Perronâ€™s formula (see Proposition 5.54 in [41]), we havewhere and is a parameter which will be decided later.

In view of (20), the points and are the only three possible simple poles of the integrand in the rectangle . The corresponding possible residues at , and are equal torespectively.

We move the integral line of the integral in (28) to the parallel segment with . We first consider the case . In this situation, the points , , and are all poles of the integrand in the rectangle . Therefore, by Cauchyâ€™s residue theorem, we can obtain

Now, the remaining work is to handle these three terms , and . In addition, the estimates of these integrals on the horizontal parts are analogous, and so we deal with and firstly. To estimate and , similarly to the method estimating and , we still divide the integral interval into the following four short intervals and apply Lemmas 3 and 4.

Interval :

In this interval, we have

Interval :

In this interval, we have

Interval :

This interval is an empty set noting that .

Interval :

In this interval, we have

From (55)â€“(60), we can obtain

Now, we turn to estimate , and we have

Then, using Lemmas 3 and 4 and HÃ¶lderâ€™s inequality, we obtain

By putting (53), (61), and (63) together, we have

By taking in (64), we can obtainwhich proves the first result in Theorem 2.

For the case , we use a similar argument to the first corresponding case. In this case, the points and are the two simple poles of the integrand in the rectangle . Then, by Cauchyâ€™s residue theorem we have

To estimate , we still divide the integral interval into four short intervals , which are different from ones for the case . In fact, the corresponding short intervals become empty sets in this situation. However, we still can estimate by following a similar argument to the corresponding parts of the case and get

The estimate of becomes the following at the current case by noting .