On the number of integers which form perfect powers in the way of $x(y_1^2+y_2^2+y_3^2+y_4^2) = z^k$

1.   Introduction

Counting the number of integral solutions of a certain Diophantine equation is an interesting project in number theory. Let $N_k(H)$ be the number of pairs of positive integers $x_1$, $x_2 \leqslant H$ whose product $x_1x_2$ is a perfect $k$-th power. Tolev ^[1] first established an asymptotic formula for $N_k (H)$. The proof combines Perron's formula with elementary ideas from the work of Heath-Brown and Moroz ^[2]. It is proved that for any integer $k \geqslant2$,

where $c_k > 0$ is an explicit constant depending on $k$. De la Bretèche et al. ^[3] improved this result based on the multiple Dirichlet series theory and complex analysis. They showed that there exists a constant $\theta_k > 0$ such that

where $Q$ is a polynomial of degree $k-1$ with leading coefficient $c_k$. In fact, they considered a more general case and counted the number of tuples of $n \geqslant 2$ integers whose product is a perfect $k$-th power. Precisely, they proved that there exists a constant $\theta_{n, k} > 0$, such that

where $Q_{n, k}$ is a polynomial of degree $\binom{n+k-1}{k}-n$.

It is natural to consider the analogue of the above results for polynomials. Liu and Niu ^[4] counted pairs of polynomials whose product is a cube over finite field and obtained an asymptotic formula (see [4, Theorem 1.2]) by contour integration. It is indicated that there are some differences and challenges between polynomials and integers.

Returning to the case of integers, this problem is closely related to the integral points on algebraic surface

This is a split toric surface and can also be regarded as the product of three integers forming a perfect cube. It has been studied by many authors including de la Bretèche ^[5], Fouvry ^[6], Heath-Brown and Moroz ^[2], and Salberger ^[7]. Denote by $N_S(H)$ the number of primitive integral points (i.e., $\gcd(x_0, x_1, x_2, x_3) = 1$) on $S$ satisfying $x_0\neq0$ and $\max_{0 \leqslant i \leqslant3} |x_i| \leqslant H$. The sharpest unconditional result is proved by de la Bretèche ^[5], which states

where $P$ is a polynomial of degree $6$ and $c$ is a positive constant. Now, we look at the corresponding non-split toric surface

It can be observed that $S'$ and $S$ are isomorphic over $\mathbb{Q}(\mathrm{i})$. Denote by $N_{S'}(H)$ the number of primitive integral points on $S'$ satisfying $z\neq0$ and

De la Bretèche et al. ^[8] studied this surface and proved that

where $P'$ is a cubic polynomial. Liu et al. ^[9] further considered the following case

For ease of presentation, we let

and denote by $N_4(H)$ the number of integral tuples $(x, y_1, \cdots, y_4)$, satisfying $\max\{|x|, \|\boldsymbol{y}\|\} \leqslant H$, which form nonzero perfect cubes in the way of (1.1) since the variable $z$ is completely determined by $x$ and $\boldsymbol{y}$. They showed that

In fact, they dealt with a more general case that the number of squares is a multiple of $4$ (see [9, Theorem 7.1]). Zhai ^[10] improved this result by obtaining a power-saving error term that

where $P_4$ is a quadratic polynomial. All above results are obtained by studying the corresponding multiple Dirichlet series. Liu et al. stated in ^[9] that the same idea can be applied to investigate the number of integral solutions of some higher-degree Diophantine equation like

They obtained an asymptotic formula in ^[11] for $d = 4$, and Wen ^[12] established the asymptotic formulae for any integer $d \geqslant 4$ with power-saving error terms.

We remark here that the Diophantine equations $S$, $S'$, and $S_4$ mentioned above are homogeneous, so there is an equivalent relation between rational points in projective space and integral points in affine space up to a scalar multiplication. They are actually related to another project in number theory, which is Manin's conjecture (see ^{[2,5,6,7,8,9]} for more details).

Motivated by the above work, in this paper, we mainly focus on the Diophantine equation

for $k \geqslant 2$. Similar to (1.1), we denote by $N_4^k(H)$ the number of integral tuples $(x, y_1, \cdots, y_4)$ satisfying $\max\{|x|, \|\boldsymbol{y}\|\} \leqslant H$, which form nonzero perfect $k$-th powers in the way of (1.2). Recall that the Lindelöf hypothesis (LH in brief, [13, §Ⅱ.3.4]) states that

for any $\varepsilon > 0$. Our main result is as follows.

Theorem 1.1. Let $k \geqslant 4$ be any integer. Assuming LH, then for any $\varepsilon > 0$, there exists a constant $\vartheta_k$, such that

where

and $[\alpha]$ is the integral part of $\alpha$, the implied constant only depends on $k$ and $\varepsilon$, $\mathcal{P}$ is a polynomial of degree $k-1$ given by (3.14) with leading coefficient $16\mathscr C_k$, and $\mathscr C_k$ is a positive constant given by (3.13).

Remark 1.2. The assumption of LH is just for simplifying the calculation. We claim that the above asymptotic formulae still holds for $k = 4$ unconditionally, since we can apply the fourth moment estimate of the Riemann zeta function instead of LH as in (4.8).

The case of $k = 3$ has been solved in ^[9,10] unconditionally as mentioned above. In fact, following the proof of Theorem 1.1, one can easily obtain an asymptotic formula for $k = 3$ with a power-saving error term $O(H^{3-1/5+\varepsilon})$. We shall leave that as an example to justify our result. We next give an unconditional result for $k = 2$.

Theorem 1.3. Unconditionally, we have

where $P$ is a linear polynomial given by (4.7) and the implied constant only depends on $\varepsilon$.

2.   Application of Perron's formula

We first introduce the bivariate Perron's formula [10,Lemma 2.2], which plays an important role in our proof.

Lemma 2.1. Suppose that $f(n_1, n_2)$ is a bivariate arithmetic function and its Dirichlet series

is absolutely convergent for $\Re(s_j) > \sigma_j\; (j = 1, 2)$ with some $\sigma_1$, $\sigma_2 > 0$. Let $x_1$, $x_2$, $T_1$, $T_2 \geqslant10$ be parameters such that $x_j\not\in \mathbb{N}$, and define

We have

where

Recalling the definition of $N_4^k(H)$ before, we see that

Let $r_4(n)$ be the number of representations of a positive integer $n$ as the sum of four squares

with

It is well known that

and $r^*_4(n)$ is a multiplicative arithmetic function. Let $\mathbb{1}_k$ denote an indicator function of perfect $k$-th power defined by

In view of the above problem, we can write

In order to deal with this double sum, we define the corresponding Dirichlet series

The following proposition gives the expression and convergence of $\mathcal F_k(s, w)$, which allows us to extend the double Dirichlet series to a suitable large region.

Proposition 2.2. Let $k \geqslant2$. If $\Re (s) > 1/k$ and $\Re (w) > 1+1/k$, then

where $\mathcal{H}_k(s, w)$ is an Euler product given by (2.16), which is absolutely convergent if $s$ and $w$ satisfy the conditions

Furthermore, we have

in the above region, and the implied constant is absolute.

Proof. Note that $r_4^*(n)$ is multiplicative, and (2.1) yields $r_4^*(1) = 1$ and

for any integer $\nu \geqslant1$, then we can rewrite $\mathcal F_k(s, w)$ in (2.4) as the Euler product

Here, we let $\mu+\nu = kd$ for $d \geqslant0$, according to the definition of $\mathbb{1}_k$ in (2.2). On the other hand, a simple formal calculation shows

with

and

for $k \geqslant3$. Similarly, we have

When $p \geqslant3$, in view of (2.8), we apply (2.10) with

to deduce that

where

Meanwhile, for $p = 2$, the formula (2.11) with

gives us

where

It can be observed that $\mathcal F_{k, p}(s, w)/\mathcal H_{k, p}(s, w)$ will give the Euler product of Riemann zeta functions in (2.12) and (2.14). Combining (2.9) and (2.12)–(2.15), we get (2.5) with

Next, we shall discuss the convergence. In view of the expression of $\mathcal H_{k, p}(s, w)$ in (2.13) and (2.15), when expanding $\mathcal H_{k, p}(s, w)$ into $1$ plus some monomials about $p$, we see that the power of $p$ in the denominator of each monomial is great than $1$ if

so we have

and $\mathcal H_k(s, w)$ is absolutely convergent in this region, which implies (2.7). This completes the proof. □

We next make use of bivariate Perron's formula. Suppose $H\notin \mathbb{N}$ and let $T\in[10, H^{{1}/{2}}]$ be a parameter to be chosen later. The analytic property of $\mathcal F_k(s, w)$ allows us to set

in Lemma 2.1, then we get

with

Noticing that

and following the arguments in ^[10], one can easily get

and the same result holds for $\mathfrak{J}_2(H, T)$. It follows that

It suffices to evaluate the double integral on the righthand side of (2.17).

3.   Proof of the Theorem 1.1

In this section, we shall prove the main theorem. The proof is divided into three steps. We shall first turn the double integral into some usual single integrals by Cauchy's residue theorem and then deal with the single integrals. The final step is choosing the suitable parameters.

3.1.   Application of Cauchy's residue theorem

In this subsection, we shall apply Cauchy's residue theorem to evaluate the inner integral over $w$ in (2.17) and derive the following result.

Lemma 3.1. Let $k \geqslant2$ and $N_4^k(H)$ be defined as in (2.3). Assuming LH, we have

where

with

for $1 \leqslant j \leqslant k$.

Proof. In terms of the inner integral in (2.17), we consider the domain formed by four points

In this domain, from Proposition 2.2, we easily see that the integrand function $\mathcal F_k(s, w)/(ws)$ has $k$ simple poles:

for $1 \leqslant j \leqslant k.$

Using the residue theorem for the variable $w$, we get

where $I_j(H, T)$ is given by (3.1) and

Set $s = b_1+ \mathrm{i} t$ with $-T \leqslant t \leqslant T$ and $w = u+10 \mathrm{i} T$ with $1/2k+1 \leqslant u \leqslant b_2$. By Proposition 2.2, one has $\mathcal{H}_k(s, w) \ll 1$, since the conditions in (2.6) are satisfied clearly. On the other hand, recall that LH states

then we deduce from the Phragmén-Lindelöf principle that the subconvexity bound of the Riemann zeta function under LH satisfies

This will be used several times below, and it follows that

holds uniformly for $1/2k+1 \leqslant u \leqslant b_2$. This implies

and the same bound holds for $R_3(H, T)$. Similarly, for $u = 1/2k+1$, we have

and $R_2(H, T)$ can be estimated as

Inserting the upper bounds of $R_i(H, T)$ $(i = 1, 2, 3)$ into (3.2), we obtain the required formula. □

3.2.   Evaluation of $I_j (H, T)$

In this subsection, we shall evaluate $I_j (H, T)$ for $1 \leqslant j \leqslant k$, and our main idea is Cauchy's residue theorem.

Lemma 3.2. Let $k \geqslant4$ and $I_j(H, T)$ for $1 \leqslant j \leqslant k$ be defined as in (3.1), then we have the following estimates under LH for each $I_j(H, T)$:

where $\mathcal C_{3k/2},$ $P_j(\log H)$ for $2k/3 < j < k$ and $P_k(\log H)$ are defined in (3.6), (3.9), and (3.12), respectively.

Remark 3.3. In fact, the above result still holds for $k = 3$ apart from the third formula, since the third case vanishes if $k = 3$.

Proof. Recall the definition of $I_j(H, T)$ in (3.1). We deduce from Proposition 2.2 that

It follows that

The proof of this lemma is divided into four parts based on whether $3-2k/j$ is positive, negative, or zero. Keep in mind that the letter $j$ is always an integer.

If $1 \leqslant j < 2k/3$, the condition (2.6) becomes

which implies

So, we shift the line of integration from $b_1$ to

By Proposition 2.2, the Euler product $\mathcal{H}_k(s, w_j)$ is absolutely convergent in the region $b_1 \leqslant \Re(s) \leqslant \sigma_j$, and we have $\mathcal{H}_k(s, w_j)\ll 1$. It follows from this and (3.3) that

for $b_1 \leqslant \sigma \leqslant \sigma_j$, since the real part of each zeta function is

for $0 \leqslant\ell \leqslant k-1$ and $\ell\neq j$. The above bound allows us to extend the integral of $I_j(H, T)$ on $[-T, T]$ to $(-\infty, \infty)$, so we have

where $\int_{(b_1)}$ means $\int_{b_1-i\infty}^{b_1+i\infty}$. Note that there is no pole in the region $b_1 \leqslant \Re(s) \leqslant \sigma_j$. The residue theorem tells us that

By (3.4), the integral on the righthand side can be estimated as

Combining this with (3.5), we obtain the first assertion in Lemma 3.2.

If $j = 2k/3$ is an integer, we find that

Otherwise, this case does not exist. Note that

for $s = b_1+ \mathrm{i} t$, which shows that $G_{2k/3}(b_1+ \mathrm{i} t, \chi)$ has a good convergence as a function in $t$. It follows that

where

is an absolute constant. This gives the second assertion.

If $2k/3 < j < k$, we shift the line of integration from $b_1$ to

It is easy to check that (2.6) is exactly satisfied, so we have $\mathcal{H}_k(s, w_j)\ll 1$. It follows from (3.3) again that

for $\sigma_j' < \sigma \leqslant b_1$, which yields

Note that

for $0 \leqslant\ell < j$, so there is only a pole $s = 1/k$ of order $j$ given by $G_j(s)$ in the strip $\sigma_j' \leqslant \sigma \leqslant b_1$. The residue theorem gives us

We can compute the residue in the main term as

This is a polynomial of degree $j-1$ with leading coefficient

The integral in (3.8) can be bounded by

Inserting (3.9) and the above estimate into (3.8), we get the third formula in Lemma 3.2.

Finally, it remains to deal with the case $j = k$. Now, we shift the line of integration from $b_1$ to

Noticing that $w_k = 1/k+1$ is no longer dependent on $s$, it follows from (3.3) that

since $\mathcal{H}_k(s, w_k)$ is absolutely convergent for $\sigma_k \leqslant\Re(s) \leqslant b_1$ according to (2.6). Note that there is only one pole $s = 1/k$ of order $k$ given by $G_k(s)$ in the rectangle formed by $\sigma_k\pm \mathrm{i} T$ and $b_1\pm \mathrm{i} T$. It follows from the residue theorem that

where $K_{i}(H, T)$ for $i = 1, 2, 3$ corresponds to the above three integrals, respectively, and $P_k$ is a polynomial of degree $k-1$ given by

with leading coefficient

This is a positive constant depending on $k$. By (3.10), using the same method as before, we can estimate $K_{1}(H, T)$ by

so does $K_{3}(H, T)$. As for $K_{2}(H, T)$, noting that

we get

The last formula in Lemma 3.2 is obtained followed from the above two estimates and (3.11). This completes the proof. □

3.3.   Completion the proof of Theorem 1.1

Combining Lemmas 3.1 and 3.2, we get

for $k \geqslant4$. It can be simplified to

where $[\alpha]$ is the integral part of $\alpha$ and

This is a polynomial of degree $k-1$ with leading coefficient $16\mathscr C_k$. It suffices to choose a suitable $T$ to balance the error terms. It can be observed that the error term is actually dominated by $H^{\frac{3}{2k}+\frac{2k-3}{2k[2k/3]}+\varepsilon}$ for $k \geqslant4$. Just letting $H^{9/4k^2}\ll T\ll H^{1/k}$, one can get the second formula in Theorem 1.1.

4.   Proof of Theorem 1.3

Following the proof of Theorem 1.1, we give a sketch of the proof for $k = 2$. According to Proposition 2.2, the corresponding Dirichlet series can be written as

with

which is absolutely convergent and satisfies $\mathcal H_2(s, w)\ll 1$ in the region

Applying Perron's formula, we get

for $b_1 = 1/2+\varepsilon$ and $b_2 = 3/2+\varepsilon$.

Following the arguments in Section 3, we still shift the path of integration over $w$ to $\Re(w) = 5/4$. Clearly, (4.1) is satisfied. Applying the residue theorem to evaluate the inner integral in (4.2), we can get (3.2) for $k = 2$. Note that the assumption of LH is used to bound $\mathcal F_k(s, w)$ and $G_j(s)$ in some region before. Unconditionally, we have the following well-known estimate for the Riemann zeta function (see [13, Theorem Ⅱ.3.8] for example):

Just replacing (3.3) by (4.3) and using the same method as in §3.1, we can get

with

The treatments of $I_1(H, T)$ and $I_2(H, T)$ are a little different from before. There are only two Riemann zeta functions involved here, which can be treated more carefully. As for $I_1(H, T)$, we shift the line of integration from $b_1$ to $\sigma_1 = 1-\varepsilon$ due to (4.1). It follows from (4.3) that

for $s = \sigma+ \mathrm{i} t$ and $b_1 \leqslant\sigma \leqslant \sigma_1$. Noticing that there is no pole for the integrand function in this strip, similar to (3.5), we deduce from the residue theorem that

For $s = \sigma_1+ \mathrm{i} t$, the integral above can be bounded by

It follows that

It suffices to deal with $I_2(H, T)$. In view of (4.1), we shift the path of integration from $b_1\pm \mathrm{i} T$ to $\varepsilon\pm \mathrm{i} T$. Note that there is a pole $s = 1/2$ of order $2$ in this region. It follows from the residue theorem that

where $P$ is a linear polynomial given by

and $\gamma$ is Euler's constant. We derive from (4.3) that the first and third integrals in (4.6) can be bounded by

Recall the fourth moment estimate of the Riemann zeta function (see [14, Theorem 5.1])

for $1/2 \leqslant \sigma < 1$ and the functional equation, which states

for $\sigma \leqslant1/2$. By Hölder's inequality and partial integration, we can estimate the second integral in (4.6) as follows

Combining all the above, we get

Finally, inserting (4.5) and (4.9) into (4.2), we get

Choosing $T = H^{1/3}$, the error terms are balanced to $H^{2+3/2-1/3+\varepsilon}$. This gives the required formula in Theorem 1.3.

5.   Conclusions

In this paper, we study the number of integers which form perfect powers in the way of

and the proof relies on techniques coming from complex analysis. We point out that this problem was never done before except for the case $k = 3$. It is not easy to establish a unified asymptotic formula with power-saving error terms for large $k$, so we assume the Lindelöf hypothesis for $k \geqslant4$. Theorem 1.1 gives an asymptotic formula with a power-saving error term for the number of such integers of bounded size under LH. This is the novelty of this paper. Moreover, Theorem 1.3 gives an unconditional result for $k = 2$.

Use of AI tools declaration

The author declares she has not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

The author would like to thank the anonymous referees for many useful comments on the manuscript.

Conflict of interest

The author declares no conflict of interest.