1 Introduction

For a natural number \(k\ge 2,\) we say an integer n is k-free if \(p^{k} \not \mid n\) for any prime p. The distribution of k-free numbers is an important theme for number theory scholars. It is well known that

$$\begin{aligned} \sum _{n\le x} \mu _{k}(n)= \frac{x}{\zeta (k)}+ O(x^{1/(k+ 1)+ \varepsilon }), \end{aligned}$$

proved by Montgomery and Vanghan [13], where \(\mu _{k}(n)\) denotes the characteristic function of the k-free integers and \(\zeta (k)\) is the usual Riemann zeta function. There also exist many articles that consider the k-free values of polynomials. In the case of \(k= 2,\) we mention Estermann’s work [8], where it is showed that for an absolute constant a

$$\begin{aligned} \sum _{1\le x\le H} \mu ^{2}(x^{2}+ 1)= a H+ O(H^{2/3+ \varepsilon }). \end{aligned}$$
(1.1)

Heath-Brown [9] obtained (1.1) with the error term replaced by \(O(H^{7/12+ \varepsilon }).\) In 1932, Carlitz [2] considered the polynomial \(x (x+ 1)\) and he proved

$$\begin{aligned} \sum _{1\le x\le H} \mu ^{2}(x)\, \mu ^{2}(x+ 1)= \prod _{p} \left( 1- \frac{2}{p^{2}}\right) H+ O(H^{2/3+ \varepsilon }). \end{aligned}$$
(1.2)

Later, the exponent \(2/3+ \varepsilon \) was improved to \(7/11+ \varepsilon \) by Heath-Brown [10] using the square sieve, and to \((26+ \sqrt{433})/81+ \varepsilon \) by Reuss [15].

Evaluating square-free values of the polynomial in multiple variables is an essential generalization that has attracted many authors, including Dimitrov, Tolev, Zhou and Chen. In 2012, Tolev [16] studied the square-free values of the polynomial \(x^{2}+ y^{2}+ 1,\) and he proved

$$\begin{aligned} \sum _{1\le x, y\le H} \mu ^{2}(x^{2}+y^{2}+ 1)= \prod _{p} \left( 1- \frac{\lambda (p^{2})}{p^{4}}\right) H^{2}+ O(H^{3/4+ \varepsilon }), \end{aligned}$$

where \(\lambda (q)\) is the number of the integer solutions to the following congruence equation:

$$\begin{aligned} x^{2}+y^{2}+ 1\equiv 0\ (\bmod \,q),\ \ 1\le x, y\le q. \end{aligned}$$

Recently, by using Tolev’s method and some estimate for the Salié sum, Zhou and Ding [17] got an asymptotic formula for \(\Sigma _{1\le x, y, z\le H} \, \mu ^{2}(x^{2}+y^{2}+ z^{2}+ k).\) On the other hand, motivated by Carlitz’s result (1.2), consecutive square-free values of polynomials in multiple variables have also been considered. Dimitrov [6, 7] found asymptotic formulas for consecutive square-free numbers of the form \(x^{2}+y^{2}+ 1, x^{2}+y^{2}+ 2\) and respectively of the form \(x^{2}+ 1, x^{2}+ 2.\) Chen [3] considered the consecutive square-free numbers \(x_{1}^{2}+ \cdots + x_{k}^{2}+ 1, x_{1}^{2}+ \cdots + x_{k}^{2}+ 2\) and showed

$$\begin{aligned} \sum _{1\le x_{1}, \ldots , x_{k}\le H} \mu ^{2}(x_{1}^{2}+ \cdots + x_{k}^{2}+ 1)\, \mu ^{2}(x_{1}^{2}+ \cdots + x_{k}^{2}+ 2)= c H^{k}+ O(H^{k- \frac{1}{2}- \frac{1}{2k}+ \varepsilon }) \end{aligned}$$
(1.3)

for any given integer \(k\ge 3\) and an absolute constant c.

For the k-free values of polynomials, the classical problem is to study the following:

$$\begin{aligned} S_{k}(H)= \sum _{x\le H} \mu _{k}(x) \mu _{k}(x+ 1). \end{aligned}$$

In 1932, Carlitz [2] obtained

$$\begin{aligned} S_{k}(H)= c_{1} H+ O(H^{2(k+ 1)+ \varepsilon }), \end{aligned}$$

where the exponent \(2(k+ 1)+ \varepsilon \) may be considered as trivial. Later on, Brandes [1] derived an improvement upon the trivial exponent which is of order \(1/k^{2}\) as \(k\rightarrow \infty .\) Using the approximative determinant method of Heath-Brown, Dictmann and Marmon [5] obtained the exponent \(14/9k+ \varepsilon ,\) which sharpens previous bound for \(k\ge 6.\) The k-free values of the polynomial \((x+ a_{1})(x+ a_{2})\) were considered by Mirsky [12]. Recently, Chen and Wang [4] studied the r-free values of \(x^{2}+y^{2}+ z^{2}+ k\) and gave an asymptotic formula.

Let

$$\begin{aligned} R(H, k)= \sum _{1\le x, y, z\le H} \mu _{k}(x^{2}+y^{2}+ z^{2}+ 1)\, \mu _{k}(x^{2}+y^{2}+ z^{2}+ 2) \end{aligned}$$
(1.4)

and

$$\begin{aligned} \lambda (q_{1}, q_{2}; m, n, l)= \sum _{\begin{array}{c} 1\le x, y, z\le q_{1} q_{2}\\ x^{2}+ y^{2}+ z^{2}+ 1\equiv 0 \pmod {q_{1}} \\ x^{2}+ y^{2}+ z^{2}+ 2 \equiv 0 \pmod {q_{2}} \end{array}} e_{q_{1} q_{2}} (m x+ n y+ lz). \end{aligned}$$
(1.5)

For simplicity, we also define

$$\begin{aligned} \lambda (q_{1}, q_{2})= \lambda (q_{1}, q_{2}; 0, 0, 0),\,\lambda (q_{1}, q_{2}; m)= \lambda (q_{1}, q_{2}; m, 0, 0) \end{aligned}$$
(1.6)

and

$$\begin{aligned} \lambda (q_{1}, q_{2}; m, n)= \lambda (q_{1}, q_{2}; m, n, 0). \end{aligned}$$

Inspired by the above results, we shall study the k-free values of the polynomials \(x^{2}+ y^{2}+ z^{2}+ 1\) and \(x^{2}+ y^{2}+ z^{2}+ 2\) by following the method in Dimitrov [6] and pruning some details referring to Chen and Wang [4]. The key ingredient is still the estimation of \(\lambda (q_{1}, q_{2}; m, n, l)\), which we give with the help of the elementary properties of Gauss sums and Salié sums. We prove the following.

Theorem 1.1

For any given integer \(k\ge 2,\) the asymptotic formula

$$\begin{aligned} R(H, k)= \sigma _{k} H^{3}+ O(H^{\frac{3}{2}+ \frac{3}{2k}+ \varepsilon }+ H^{2}) \end{aligned}$$

holds. Here

$$\begin{aligned} \sigma _{k}= \prod _{p} \left( 1- \frac{\lambda (p^{k}, 1)+ \lambda (1, p^{k})}{p^{3k}}\right) . \end{aligned}$$

In the case of \(k= 2,\) we obtain from Theorem 1.1 the following:

Theorem 1.2

If \(\varepsilon > 0\) is an arbitrary positive number, then

$$\begin{aligned} R(H, 2)= \prod _{p} \left( 1- \frac{\lambda (p^{2}, 1)+ \lambda (1, p^{2})}{p^{6}}\right) H^{3}+ O(H^{9/4+ \varepsilon }). \end{aligned}$$

Theorem 1.2 improves upon (1.3) with \(k= 3.\)

2 Notations and preliminary lemmas

Throughout this paper, H is a sufficiently large positive number and m, n, l denote integers. By \(\varepsilon \) we denote an arbitrary positive number which may have different values in different places. As usual, \(\mu (n)\) denotes the Möbius function; \(\tau (n)\) and \(\omega (n)\) represent the number of positive divisors of n and the number of distinct prime factors of n, respectively. Instead of \(m\equiv n \pmod {d}\) we write for simplicity \(m\equiv n \,(d).\) (m, n, l) denotes the greatest common divisor of m, n, l and \(\Vert \xi \Vert \) denotes the distance from \(\xi \) to its nearest integer. We write \(e(t)= \exp (2 \pi it)\) and \(e_{q}(t)= e(t/q).\) For any x and q such that \((x, q)= 1\), we denote by \(\overline{x_{q}}\) the inverse of n modulo q. If we can understand the value of the modulus from the context, then we write for simplicity \(\overline{x}.\) For any odd q, \(\left( \frac{\cdot }{q}\right) \) is the Jacobi symbol.

We define the Gauss sum and the Salié sum as follows:

$$\begin{aligned} G(q; n, m)= \sum _{1\le x\le q} e_{q}(n x^{2}+ m x),\ \ G(q; n)= \sum _{1\le x\le q} e_{q}(n x^{2}) \end{aligned}$$
(2.1)

and, for odd integers q, 

$$\begin{aligned} S(q;n, m)= \sum _{{\mathop {(x, q)= 1}\limits ^{1\le x\le q}}} \left( \frac{x}{q}\right) e_{q}(n x+ m \overline{x}). \end{aligned}$$
(2.2)

In what follows, we present some lemmas used in the proof of the theorems. First we quote some important properties of the Gauss sum.

Lemma 2.1

For the Gauss sum, the following hold:

(1) If \((q_{1}, q_{2})= 1,\) then

$$\begin{aligned} G(q_{1} q_{2}; m_{1} q_{2}+ m_{2} q_{1}, n)= G(q_{1}; m_{1} q_{2}^{2}, n) \,G(q_{2}; m_{2} q_{1}^{2}, n). \end{aligned}$$

(2) If \((q, m)= d,\) then

$$\begin{aligned} G(q; m, n) ={\left\{ \begin{array}{ll} d\, G(q/d; m/d, n/d), &{} \textrm{if}\ d \mid n,\\ 0, &{} \textrm{if}\ d \not \mid n. \end{array}\right. } \end{aligned}$$

(3) If \((q, 2 m)= 1,\) then

$$\begin{aligned} G(q; m, n)= e_{q} \left( - \overline{(4 m)} n^{2}\right) \left( \frac{m}{q}\right) G(q; 1). \end{aligned}$$

Proof

See Lemma 3.1 of [6]. \(\square \)

In the next lemma, we present the upper bound result of the Salié sum.

Lemma 2.2

If q is an odd integer, then

$$\begin{aligned} \mid S(q; n, m)\mid \le 2^{\omega (q)} \sqrt{q}. \end{aligned}$$

Proof

See p. 524 in [11]. \(\square \)

Lemma 2.3

If \((q_{1}' q_{1}'', q_{2}' q_{2}'')= (q_{1}', q_{1}'')= (q_{2}', q_{2}'')= 1,\) then

$$\begin{aligned} \lambda \left( q_{1}' q_{1}'', q_{2}' q_{2}''; m, n, l\right)&= \lambda \left( q_{1}', q_{2}'; m \overline{(q_{1}'' q_{2}'')}_{q_{1}' q_{2}'}, n \overline{(q_{1}'' q_{2}'')}_{q_{1}' q_{2}'}, l\overline{(q_{1}'' q_{2}'')}_{q_{1}' q_{2}'}\right) \\&\quad \times \lambda \left( q_{1}'', q_{2}''; m \overline{(q_{1}' q_{2}')}_{q_{1}'' q_{2}''}, n \overline{(q_{1}' q_{2}')}_{q_{1}'' q_{2}''}, l\overline{(q_{1}' q_{2}')}_{q_{1}'' q_{2}''}\right) . \end{aligned}$$

Proof

Let

$$\begin{aligned} x= x_{1} q_{1}'' q_{2}''+ x_{2} q_{1}' q_{2}',\,y= y_{1} q_{1}'' q_{2}''+ y_{2} q_{1}' q_{2}',\,z= z_{1} q_{1}'' q_{2}''+ z_{2} q_{1}' q_{2}', \end{aligned}$$

where \(1\le x_{1}, y_{1}, z_{1}\le q_{1}' q_{2}'\) and \(1\le x_{2}, y_{2}, z_{2}\le q_{1}'' q_{2}''.\)

From the Chinese remainder theorem, we obtain

$$\begin{aligned} \lambda \left( q_{1}' q_{1}'', q_{2}' q_{2}''; m, n, l\right)&= \sum _{\begin{array}{c} 1\le x, y, z\le q_{1}' q_{1}'' q_{2}' q_{2}''\\ x^{2}+ y^{2}+ z^{2}+ 1\equiv 0\ (q_{1}' q_{1}'')\\ x^{2}+ y^{2}+ z^{2}+ 2 \equiv 0\ (q_{2}' q_{2}'') \end{array}} e_{q_{1}' q_{1}'' q_{2}' q_{2}''} (m x+ n y+ lz)\nonumber \\&= \sum _{\begin{array}{c} 1\le x_{1}, y_{1}, z_{1}\le q_{1}' q_{2}' \\ (q_{1}'' q_{2}'' x_{1})^{2}+ (q_{1}'' q_{2}'' y_{1})^{2}+ (q_{1}'' q_{2}'' z_{1})^{2}+ 1\equiv 0\ (q_{1}' )\\ (q_{1}'' q_{2}'' x_{1})^{2}+ (q_{1}'' q_{2}'' y_{1})^{2}+ (q_{1}'' q_{2}'' z_{1})^{2}+ 2 \equiv 0\ (q_{2}' ) \end{array}} e_{q_{1}' q_{2}'} (m x_{1}+ n y_{1}+ lz_{1})\nonumber \\&\quad \times \sum _{\begin{array}{c} 1\le x_{2}, y_{2}, z_{2}\le q_{1}'' q_{2}'' \\ (q_{1}' q_{2}' x_{2})^{2}+ (q_{1}' q_{2}' y_{2})^{2}+ (q_{1}' q_{2}' z_{2})^{2}+ 1\equiv 0\ (q_{1}'' )\\ (q_{1}' q_{2}' x_{2})^{2}+ (q_{1}' q_{2}' y_{2})^{2}+ (q_{1}' q_{2}' z_{2})^{2}+ 2 \equiv 0\ (q_{2}'' ) \end{array}} e_{q_{1}'' q_{2}''} (m x_{2}+ n y_{2}+ lz_{2}). \end{aligned}$$
(2.3)

By using the substitutions \(q_{1}'' q_{2}'' x_{1}\rightarrow x_{1}, q_{1}'' q_{2}'' y_{1}\rightarrow y_{1}, q_{1}'' q_{2}'' z_{1}\rightarrow z_{1},\) we infer

$$\begin{aligned}&\sum _{\begin{array}{c} 1\le x_{1}, y_{1}, z_{1}\le q_{1}' q_{2}' \\ (q_{1}'' q_{2}'' x_{1})^{2}+ (q_{1}'' q_{2}'' y_{1})^{2}+ (q_{1}'' q_{2}'' z_{1})^{2}+ 1\equiv 0\ (q_{1}' )\\ (q_{1}'' q_{2}'' x_{1})^{2}+ (q_{1}'' q_{2}'' y_{1})^{2}+ (q_{1}'' q_{2}'' z_{1})^{2}+ 2 \equiv 0\ (q_{2}' ) \end{array}} e_{q_{1}' q_{2}'} (m x_{1}+ n y_{1}+ lz_{1})\nonumber \\&\quad = \sum _{\begin{array}{c} \overline{(q_{1}'' q_{2}'')} _{q_{1}' q_{2}'} x_{1}, \overline{(q_{1}'' q_{2}'')} _{q_{1}' q_{2}'} y_{1}, \overline{(q_{1}'' q_{2}'')} _{q_{1}' q_{2}'} z_{1} ( \,q_{1}' q_{2}') \\ x_{1}^{2}+ y_{1}^{2}+ z_{1}^{2}+ 1\equiv 0\ (q_{1}' )\\ x_{1}^{2}+ y_{1}^{2}+ z_{1}^{2}+ 2 \equiv 0\ (q_{2}' ) \end{array}} e_{q_{1}' q_{2}'} \big (m \overline{(q_{1}'' q_{2}'')} _{q_{1}' q_{2}'}x_{1}+ n \overline{(q_{1}'' q_{2}'')} _{q_{1}' q_{2}'} y_{1}\nonumber \\&\qquad + l \overline{(q_{1}'' q_{2}'')} _{q_{1}' q_{2}'}z_{1}\big )= \lambda \left( q_{1}', q_{2}'; m \overline{(q_{1}'' q_{2}'')}_{q_{1}' q_{2}'}, n \overline{(q_{1}'' q_{2}'')}_{q_{1}' q_{2}'}, l\overline{(q_{1}'' q_{2}'')}_{q_{1}' q_{2}'}\right) . \end{aligned}$$
(2.4)

Similarly,

$$\begin{aligned}&\sum _{\begin{array}{c} 1\le x_{2}, y_{2}, z_{2}\le q_{1}'' q_{2}'' \\ (q_{1}' q_{2}' x_{2})^{2}+ (q_{1}' q_{2}' y_{2})^{2}+ (q_{1}' q_{2}' z_{2})^{2}+ 1\equiv 0\ (q_{1}'' )\\ (q_{1}' q_{2}' x_{2})^{2}+ (q_{1}' q_{2}' y_{2})^{2}+ (q_{1}' q_{2}' z_{2})^{2}+ 2 \equiv 0\ (q_{2}'' ) \end{array}} e_{q_{1}'' q_{2}''} (m x_{2}+ n y_{2}+ lz_{2})\nonumber \\&\quad = \lambda \left( q_{1}'', q_{2}''; m \overline{(q_{1}' q_{2}')}_{q_{1}'' q_{2}''}, n \overline{(q_{1}' q_{2}')}_{q_{1}'' q_{2}''}, l\overline{(q_{1}' q_{2}')}_{q_{1}'' q_{2}''}\right) . \end{aligned}$$
(2.5)

Thus, Lemma 2.3 follows immediately from (2.3)–(2.5). \(\square \)

Now we give an upper bound estimate of \(\lambda (q_{1}, q_{2}; m, n, l)\) by applying Lemmas 2.1, 2.2 and 2.3.

Lemma 2.4

If \(8\not \mid q_{1} q_{2}\) and \((q_{1}, q_{2})= 1,\) then

$$\begin{aligned} \lambda (q_{1}, q_{2}; m, n, l)\ll 64\, q_{1} q_{2}\, \tau (q_{1} q_{2})\, 2^{\omega (q_{1})} 2^{\omega (q_{2})} (q_{1} q_{2}, m, n, l). \end{aligned}$$

Proof

Case 1. \(2\not \mid q_{1} q_{2}.\)

In view of the orthogonality relations

$$\begin{aligned} \frac{1}{q} \sum _{1\le h\le q} e_{q}(h t) = {\left\{ \begin{array}{ll} 1, &{} \textrm{if}\ t\equiv 0\,(q),\\ 0, &{} \textrm{otherwise,} \end{array}\right. } \end{aligned}$$

we obtain from (1.5), (2.1) and Lemma 2.1

$$\begin{aligned}&\lambda (q_{1}, q_{2}; m, n, l)\\&= \frac{1}{q_{1} q_{2}} \sum _{1\le x, y, z\le q_{1} q_{2}} e_{q_{1} q_{2}} (m x+ n y+ l z) \\&\qquad \times \sum _{1\le h_{1}\le q_{1} } e_{q_{1}} (h_{1} (x^{2}+ y^{2}+ z^{2}+ 1)) \sum _{1\le h_{2}\le q_{2} } e_{q_{2}} (h_{2} (x^{2}+ y^{2}+ z^{2}+ 2))\\&\quad = \frac{1}{q_{1} q_{2}} \sum _{1\le h_{1}\le q_{1} } e_{q_{1}} (h_{1} ) \sum _{1\le h_{2}\le q_{2} } e_{q_{2}} (2 h_{2} )\, G(q_{1} q_{2}; h_{1}q_{2}+ h_{2} q_{1}, m)\\&\qquad \times G(q_{1} q_{2}; h_{1}q_{2}+ h_{2} q_{1}, n)\,G(q_{1} q_{2}; h_{1}q_{2}+ h_{2} q_{1}, l)\\&\quad = \frac{1}{q_{1} q_{2}} \sum _{1\le h_{1}\le q_{1} } e_{q_{1}} (h_{1} )\, G(q_{1}; h_{1} q_{2}^{2}, m)\, G(q_{1}; h_{1} q_{2}^{2}, n)\, G(q_{1}; h_{1} q_{2}^{2}, l)\\&\qquad \times \sum _{1\le h_{2}\le q_{2} } e_{q_{2}} (2 h_{2} )\, G(q_{2}; h_{2} q_{1}^{2}, m)\, G(q_{2}; h_{2} q_{1}^{2}, n)\, G(q_{2}; h_{2} q_{1}^{2}, l)\\&\quad = \frac{1}{q_{1} q_{2}} \sum _{l_{1}\mid q_{1}} \sum _{{\mathop {(h_{1}, q_{1})= \frac{q_{1}}{l_{1}}}\limits ^{1\le h_{1}\le q_{1}}} } e_{q_{1}} (h_{1} )\, G(q_{1}; h_{1} q_{2}^{2}, m)\, G(q_{1}; h_{1} q_{2}^{2}, n)\, G(q_{1}; h_{1} q_{2}^{2}, l)\\&\qquad \times \sum _{l_{2}\mid q_{2}} \sum _{{\mathop {(h_{2}, q_{2})= \frac{q_{2}}{l_{2}}}\limits ^{1\le h_{2}\le q_{2}}} } e_{q_{2}} (2 h_{2} )\, G(q_{2}; h_{2} q_{1}^{2}, m)\, G(q_{2}; h_{2} q_{1}^{2}, n)\, G(q_{2}; h_{2} q_{1}^{2}, l). \end{aligned}$$

Since \((q_{1}, q_{2})= 1, (q_{i}, h_{i})= \frac{q_{i}}{l_{i}}, l_{i}\mid q_{i}\) and \(2\not \mid q_{1} q_{2},\) by Lemma 2.1

$$\begin{aligned}&\lambda (q_{1}, q_{2}; m, n, l)\\&= q_{1}^{2} q_{2}^{2} \sum _{{\mathop {\frac{q_{1}}{l_{1}}\mid (m, n, l)}\limits ^{l_{1}\mid q_{1}}}} \frac{1}{l_{1}^{3}} \sum _{{\mathop {(r_{1}, l_{1})= 1}\limits ^{1\le r_{1}\le l_{1}}}} e_{l_{1}} (r_{1}) G(l_{1}; r_{1} q_{2}^{2}, m l_{1} q_{1}^{-1})\\&\qquad \times G(l_{1}; r_{1} q_{2}^{2}, n l_{1} q_{1}^{-1})\, G(l_{1}; r_{1} q_{2}^{2}, l\, l_{1} q_{1}^{-1}) \sum _{{\mathop {\frac{q_{2}}{l_{2}}\mid (m, n, l)}\limits ^{l_{2}\mid q_{2}}}} \frac{1}{l_{2}^{3}} \sum _{{\mathop {(r_{2}, l_{2})= 1}\limits ^{1\le r_{2}\le l_{2}}}} e_{l_{2}} (2 r_{2}) \\&\qquad \times G(l_{2}; r_{2} q_{1}^{2}, m l_{2} q_{2}^{-1})\,G(l_{2}; r_{2} q_{1}^{2}, n l_{2} q_{2}^{-1})\, G(l_{2}; r_{2} q_{1}^{2}, l\, l_{2} q_{2}^{-1})\\&\quad = q_{1}^{2} q_{2}^{2} \sum _{{\mathop {\frac{q_{1}}{l_{1}}\mid (m, n, l)}\limits ^{l_{1}\mid q_{1}}}} \frac{G^{3}(l_{1}; 1)}{l_{1}^{3}} \sum _{{\mathop {(r_{1}, l_{1})= 1}\limits ^{1\le r_{1}\le l_{1}}}} \left( \frac{r_{1}}{l_{1}}\right) ^{3} e_{l_{1}}\left( r_{1}- \overline{(4 r_{1} q_{2}^{2})} (m^{2}+ n^{2}+ l^{2}) l_{1}^{2} q_{1}^{-2}\right) \\&\qquad \times \sum _{{\mathop {\frac{q_{2}}{l_{2}}\mid (m, n, l)}\limits ^{l_{2}\mid q_{2}}}} \frac{G^{3}(l_{2}; 1)}{l_{2}^{3}} \sum _{{\mathop {(r_{2}, l_{2})= 1}\limits ^{1\le r_{2}\le l_{2}}}} \left( \frac{r_{2}}{l_{2}}\right) ^{3} e_{l_{2}}\left( 2 r_{2}- \overline{(4 r_{2} q_{1}^{2})} (m^{2}+ n^{2}+ l^{2}) l_{2}^{2} q_{2}^{-2}\right) \\&\quad = q_{1}^{2} q_{2}^{2} \sum _{{\mathop {\frac{q_{1}}{l_{1}}\mid (m, n, l)}\limits ^{l_{1}\mid q_{1}}}} \frac{G^{3}(l_{1}; 1)}{l_{1}^{3}} S(l_{1}; 1, - \overline{(4 q_{2}^{2})} (m^{2}+ n^{2}+ l^{2}) l_{1}^{2} q_{1}^{-2})\\&\qquad \times \sum _{{\mathop {\frac{q_{2}}{l_{2}}\mid (m, n, l)}\limits ^{l_{2}\mid q_{2}}}} \frac{G^{3}(l_{2}; 1)}{l_{2}^{3}} S(l_{2}; 1, - \overline{(4 q_{1}^{2})} (m^{2}+ n^{2}+ l^{2}) l_{2}^{2} q_{2}^{-2}). \end{aligned}$$

It is well known that \(|G(l_{1}, 1)|= \sqrt{l_{1}},\) thus, by Lemma 2.2,

$$\begin{aligned} \lambda (q_{1}, q_{2}; m, n, l)&\ll q_{1}^{2} q_{2}^{2} \sum _{{\mathop {\frac{q_{1}}{l_{1}}\mid (m, n, l)}\limits ^{l_{1}\mid q_{1}}}} l_{1}^{-3/2} 2^{\omega (l_{1})} l_{1}^{1/2} \sum _{{\mathop {\frac{q_{2}}{l_{2}}\mid (m, n, l)}\limits ^{l_{2}\mid q_{2}}}} l_{2}^{-3/2} 2^{\omega (l_{2})} l_{2}^{1/2}\nonumber \\&\ll q_{1}^{2} q_{2}^{2} 2^{\omega (q_{1})} 2^{\omega (q_{2})} \sum _{\frac{q_{1}}{l_{1}}\mid (q_{1}, m, n, l)} l_{1}^{-1} \sum _{\frac{q_{2}}{l_{2}}\mid (q_{2}, m, n, l)} l_{2}^{-1}\nonumber \\&\ll q_{1}^{2} q_{2}^{2} 2^{\omega (q_{1})} 2^{\omega (q_{2})} \sum _{r_{1}\mid (q_{1}, m, n, l)} q_{1}^{-1} r_{1} \sum _{r_{2}\mid (q_{2}, m, n, l)} q_{2}^{-1} r_{2}\nonumber \\&\ll q_{1} q_{2} \tau (q_{1} q_{2}) 2^{\omega (q_{1})} 2^{\omega (q_{2})} (q_{1} q_{2}, m, n, l). \end{aligned}$$
(2.6)

Case 2. \(q_{1}= 2^{h} q_{1}',\) where \(2\not \mid q_{1}', h\le 2\) and \(2\not \mid q_{2}.\)

By Lemma 2.3, we have

$$\begin{aligned} \lambda (2^{h} q_{1}', q_{2}; m, n, l)&= \lambda \left( 2^{h}, 1; m \overline{(q_{1}' q_{2})}_{2^{h}}, n \overline{(q_{1}' q_{2})}_{2^{h}}, l \overline{(q_{1}' q_{2})}_{2^{h}}\right) \\&\quad \times \lambda \left( q_{1}', q_{2}; m \overline{2^{h}}_{q_{1}' q_{2}}, n \overline{2^{h}}_{q_{1}' q_{2}}, l \overline{2^{h}}_{q_{1}' q_{2}}\right) . \end{aligned}$$

A combination of the trivial estimate \(\lambda \left( 2^{h}, 1; m \overline{(q_{1}' q_{2})}_{2^{h}}, n \overline{(q_{1}' q_{2})}_{2^{h}}, l \overline{(q_{1}' q_{2})}_{2^{h}}\right) \ll 8^{h}\) and (2.6) yields

$$\begin{aligned} \lambda (2^{h} q_{1}', q_{2}; m, n, l)&\ll 8^{h} q_{1}' q_{2} \tau ( q_{1}' q_{2}) 2^{\omega (q_{1}')} 2^{\omega (q_{2})} \left( q_{1}' q_{2}, m \overline{2^{h}}_{q_{1}' q_{2}}, n \overline{2^{h}}_{q_{1}' q_{2}}, l \overline{2^{h}}_{q_{1}' q_{2}}\right) \\&\ll 64\, q_{1} q_{2}\, \tau (q_{1} q_{2})\, 2^{\omega (q_{1})} 2^{\omega (q_{2})} (q_{1} q_{2}, m, n, l). \end{aligned}$$

Case 3. \(q_{2}= 2^{h} q_{2}',\) where \(2\not \mid q_{2}', h\le 2\) and \(2\not \mid q_{1}.\)

Similarly to Case 2, we obtain

$$\begin{aligned} \lambda ( q_{1}', 2^{h} q_{2}'; m, n, l) \ll 64\, q_{1} q_{2}\, \tau (q_{1} q_{2})\, 2^{\omega (q_{1})} 2^{\omega (q_{2})} (q_{1} q_{2}, m, n, l). \end{aligned}$$

Combining the estimates for the three cases gives the proof of Lemma 2.4. \(\square \)

Lemma 2.5

If \(8\not \mid q_{1} q_{2}\) and \((q_{1}, q_{2})= 1,\) then for the sums

$$\begin{aligned} V_{1}= \sum _{1\le m\le H} \frac{\lambda (q_{1}, q_{2}; m)}{m}, \quad V_{2}= \sum _{1\le m, n\le H} \frac{\lambda (q_{1}, q_{2}; m, n)}{m n} \end{aligned}$$

and

$$\begin{aligned} V_{3}= \sum _{1\le m, n, l\le H} \frac{\lambda (q_{1}, q_{2}; m, n, l)}{m n l}, \end{aligned}$$

we have the estimates

$$\begin{aligned} V_{1}\ll (q_{1} q_{2})^{1+ \varepsilon } H^{\varepsilon }, \,V_{2}\ll (q_{1} q_{2})^{1+ \varepsilon } H^{\varepsilon }, \,V_{3}\ll (q_{1} q_{2})^{1+ \varepsilon } H^{\varepsilon }. \end{aligned}$$

Proof

By Lemma 2.4,

$$\begin{aligned} V_{1}&\ll (q_{1} q_{2})^{1+ \varepsilon } \sum _{1\le m\le H} \frac{(q_{1} q_{2}, m)}{m}\\&\ll (q_{1} q_{2})^{1+ \varepsilon } \sum _{r\mid q_{1} q_{2}} r \sum _{{\mathop {m\equiv 0\,(r)}\limits ^{m\le H}}} \frac{1}{m}\\&\ll (q_{1} q_{2})^{1+ \varepsilon } \sum _{r\mid q_{1} q_{2}} 1 \sum _{t\le H/r} \frac{1}{t}\\&\ll (q_{1} q_{2})^{1+ \varepsilon } H^{\varepsilon }. \end{aligned}$$

In a similar way, using Lemma 2.4 we obtain

$$\begin{aligned} V_{2}&\ll (q_{1} q_{2})^{1+ \varepsilon } \sum _{1\le m, n\le H} \frac{(q_{1} q_{2}, m, n)}{m n}\\&\ll (q_{1} q_{2})^{1+ \varepsilon } \sum _{1\le m, n\le H} \frac{(q_{1} q_{2}, m) (q_{1} q_{2}, n)}{m n}\\&\ll (q_{1} q_{2})^{1+ \varepsilon } H^{\varepsilon } \end{aligned}$$

and

$$\begin{aligned} V_{3}&\ll (q_{1} q_{2})^{1+ \varepsilon } \sum _{1\le m, n, l\le H} \frac{(q_{1} q_{2}, m, n, l)}{m n l}\\&\ll (q_{1} q_{2})^{1+ \varepsilon } \sum _{1\le m, n, l\le H} \frac{(q_{1} q_{2}, m) (q_{1} q_{2}, n) (q_{1} q_{2}, l)}{m n l}\\&\ll (q_{1} q_{2})^{1+ \varepsilon } H^{\varepsilon }. \end{aligned}$$

\(\square \)

Lemma 2.6

For any real number \(\sigma \) and positive integers \(N_{1}, N_{2}\) with \(N_{1}< N_{2},\) we have

$$\begin{aligned} \sum _{N_{1}< n\le N_{2}} e(\sigma n)\ll \min \{N_{2}- N_{1}, \Vert \sigma \Vert ^{-1}\}. \end{aligned}$$

Proof

See Lemma 4.7 of [14]. \(\square \)

3 Proof of Theorem 1.1

Upon using the well-known identity

$$\begin{aligned} \mu _{k}(n)= \sum _{d^{k}\mid n} \mu (d), \end{aligned}$$

we find by (1.4) that

$$\begin{aligned} R(H, k)&= \sum _{{\mathop {(d_{1}, d_{2})= 1}\limits ^{d_{1}, d_{2}}}} \mu (d_{1}) \mu (d_{2})\, \sum _{\begin{array}{c} 1\le x, y, z\le H\\ x^{2}+ y^{2}+ z^{2}+ 1\equiv 0\ (d_{1}^{k})\\ x^{2}+ y^{2}+ z^{2}+ 2 \equiv 0\ (d_{2}^{k}) \end{array}} 1\nonumber \\&= R_{1}(H)+ R_{2}(H), \end{aligned}$$
(3.1)

where

$$\begin{aligned} R_{1}(H)= & {} \sum _{{\mathop {(d_{1}, d_{2})= 1}\limits ^{d_{1} d_{2}< \xi }}} \mu (d_{1}) \mu (d_{2})\, S(H;d_{1}^{k}, d_{2}^{k}), \end{aligned}$$
(3.2)
$$\begin{aligned} R_{2}(H)= & {} \sum _{{\mathop {(d_{1}, d_{2})= 1}\limits ^{d_{1} d_{2}> \xi }}} \mu (d_{1}) \mu (d_{2})\, S(H;d_{1}^{k}, d_{2}^{k}), \nonumber \\ S(H;d_{1}^{k}, d_{2}^{k})= & {} \sum _{\begin{array}{c} 1\le x, y, z\le H\\ x^{2}+ y^{2}+ z^{2}+ 1\equiv 0\ (d_{1}^{k})\\ x^{2}+ y^{2}+ z^{2}+ 2 \equiv 0\ (d_{2}^{k}) \end{array}} 1, \end{aligned}$$
(3.3)

\(\xi \) is a parameter to be chosen so that \(H^{1/k}\le \xi \le H^{2/k}.\)

3.1 Estimation of \(R_{1}(H)\)

To estimate the contribution of \(R_{1}(H),\) we suppose that \(q_{1}= d_{1}^{k}, q_{2}= d_{2}^{k},\) where \(d_{1}\) and \(d_{2}\) are square-free, \((q_{1}, q_{2})= 1\) and \(d_{1} d_{2}\le \xi .\)

We first analyze \(S(H;q_{1}, q_{2}).\) Define

$$\begin{aligned} \Sigma (H, q_{1}, q_{2}, x)= \sum _{{\mathop {h\equiv x\,(q_{1} q_{2})}\limits ^{h\le H}}} 1. \end{aligned}$$

By orthogonality, \(\Sigma (H, q_{1}, q_{2}, x)\) may be written as

$$\begin{aligned} \Sigma (H, q_{1}, q_{2}, x)&= (q_{1} q_{2})^{-1} \sum _{1\le h\le H} \sum _{1\le t\le q_{1} q_{2}} e_{q_{1} q_{2}} ((h- x) t)\nonumber \\&= (q_{1} q_{2})^{-1} \sum _{1\le t\le q_{1} q_{2}} e_{q_{1} q_{2}} (- x t) \sum _{1\le h\le H} e_{q_{1} q_{2}} (ht)\nonumber \\&= H (q_{1} q_{2})^{-1}+ (q_{1} q_{2})^{-1} \sum _{1\le t\le q_{1} q_{2}- 1} e_{q_{1} q_{2}} (- x t) \sum _{1\le h\le H} e_{q_{1} q_{2}} (ht). \end{aligned}$$
(3.4)

From the definition of \(S(H;q_{1}, q_{2}),\) we easily see that

$$\begin{aligned} S(H;q_{1}, q_{2})= \sum _{\begin{array}{c} 1\le x, y, z\le q_{1} q_{2}\\ x^{2}+ y^{2}+ z^{2}+ 1\equiv 0\ (q_{1})\\ x^{2}+ y^{2}+ z^{2}+ 2 \equiv 0\ (q_{2}) \end{array}} \Sigma (H, q_{1}, q_{2}, x) \Sigma (H, q_{1}, q_{2}, y) \Sigma (H, q_{1}, q_{2}, z), \end{aligned}$$
(3.5)

which combined with (3.4) yields

$$\begin{aligned} S(H;q_{1}, q_{2})= \sum _{\begin{array}{c} 1\le x, y, z\le q_{1} q_{2}\\ x^{2}+ y^{2}+ z^{2}+ 1\equiv 0\ (q_{1})\\ x^{2}+ y^{2}+ z^{2}+ 2 \equiv 0\ (q_{2}) \end{array}} \left( \frac{H^{3}}{(q_{1} q_{2})^{3}}+ 3 \frac{H^{2}}{(q_{1} q_{2})^{2}} W_{1}+ 3 \frac{H}{q_{1} q_{2}} W_{2}+ W_{3}\right) , \end{aligned}$$
(3.6)

where

$$\begin{aligned} W_{1}&:= W_{1}(x; q_{1}, q_{2}, H)= \frac{1}{q_{1} q_{2}} \sum _{1\le t\le q_{1} q_{2}- 1} e_{q_{1} q_{2}} (- x t) \sum _{1\le h\le H} e_{q_{1} q_{2}} (ht),\\ W_{2}&:= W_{2}(x, y; q_{1}, q_{2}, H)= \frac{1}{(q_{1} q_{2})^{2}} \sum _{1\le t_{1}, t_{2}\le q_{1} q_{2}- 1} e_{q_{1} q_{2}} (- x t_{1}- y t_{2})\\&\quad \times \prod _{i= 1}^{2} \left( \sum _{1\le h_{i}\le H} e_{q_{1} q_{2}} (h_{i} t_{i})\right) ,\\ W_{3}&:= W_{3}(x, y, z; q_{1}, q_{2}, H)= \frac{1}{(q_{1} q_{2})^{3}} \sum _{1\le t_{1}, t_{2}, t_{3}\le q_{1} q_{2}- 1} e_{q_{1} q_{2}} (- x t_{1}- y t_{2}- z t_{3}) \\&\quad \times \prod _{i= 1}^{3} \left( \sum _{1\le h_{i}\le H} e_{q_{1} q_{2}} (h_{i} t_{i})\right) . \end{aligned}$$

By exchanging the order of summations and noting the definitions of \(\lambda (q_{1}, q_{2}; m, n, l),\) we obtain

$$\begin{aligned} \sum _{\begin{array}{c} 1\le x, y, z\le q_{1} q_{2}\\ x^{2}+ y^{2}+ z^{2}+ 1\equiv 0\ (q_{1})\\ x^{2}+ y^{2}+ z^{2}+ 2 \equiv 0\ (q_{2}) \end{array}} \,W_{1}&= \frac{1}{q_{1} q_{2}} \sum _{1\le t\le q_{1} q_{2}- 1} \lambda (q_{1}, q_{2}; -t) \sum _{1\le h\le H} e_{q_{1} q_{2}} (ht)\\&:= L_{1}(q_{1}, q_{2}; H),\\ \sum _{\begin{array}{c} 1\le x, y, z\le q_{1} q_{2}\\ x^{2}+ y^{2}+ z^{2}+ 1\equiv 0\ (q_{1})\\ x^{2}+ y^{2}+ z^{2}+ 2 \equiv 0\ (q_{2}) \end{array}} \,W_{2}&= \frac{1}{(q_{1} q_{2})^{2}} \sum _{1\le t_{1}, t_{2}\le q_{1} q_{2}- 1} \lambda (q_{1}, q_{2}; -t_{1}, - t_{2})\\&\quad \times \prod _{i= 1}^{2} \left( \sum _{1\le h_{i}\le H} e_{q_{1} q_{2}} (h_{i} t_{i})\right) \\&:= L_{2}(q_{1}, q_{2}; H),\\ \sum _{\begin{array}{c} 1\le x, y, z\le q_{1} q_{2}\\ x^{2}+ y^{2}+ z^{2}+ 1\equiv 0\ (q_{1})\\ x^{2}+ y^{2}+ z^{2}+ 2 \equiv 0\ (q_{2}) \end{array}} W_{3}&= \frac{1}{(q_{1} q_{2})^{3}} \sum _{1\le t_{1}, t_{2}, t_{3}\le q_{1} q_{2}- 1} \lambda (q_{1}, q_{2}; -t_{1}, - t_{2}, - t_{3})\\&\quad \times \prod _{i= 1}^{3} \left( \sum _{1\le h_{i}\le H} e_{q_{1} q_{2}} (h_{i} t_{i})\right) \\&:= L_{3}(q_{1}, q_{2};H). \end{aligned}$$

Now we treat \(L_{1}(q_{1}, q_{2}; H).\) Employing Lemma 2.6 we get

$$\begin{aligned} \sum _{1\le h\le H} e_{q_{1} q_{2}} (ht)\ll \left\| \frac{t}{q_{1} q_{2}}\right\| ^{-1}. \end{aligned}$$

Hence by Lemma 2.5, it follows that

$$\begin{aligned} L_{1}(q_{1}, q_{2}; H)\ll \sum _{1\le t\le q_{1} q_{2}- 1} \frac{\lambda (q_{1}, q_{2}; -t)}{t}\ll (q_{1} q_{2})^{1+ \varepsilon }. \end{aligned}$$

The same estimates hold for \(L_{2}(q_{1}, q_{2}; H)\) and \(L_{3}(q_{1}, q_{2}; H).\) Gathering the estimates for \(L_{1}(q_{1}, q_{2}; H), L_{2}(q_{1}, q_{2}; H)\) and \(L_{3}(q_{1}, q_{2}; H)\) and noting (3.6), we arrive at

$$\begin{aligned} S(H;q_{1}, q_{2})= \frac{H^{3}}{(q_{1} q_{2})^{3}} \lambda (q_{1}, q_{2})+ O(H^{2} (q_{1} q_{2})^{-1+ \varepsilon }+ H (q_{1} q_{2})^{ \varepsilon }+ (q_{1} q_{2})^{1+ \varepsilon }). \end{aligned}$$
(3.7)

According to (3.2) and (3.7), \(R_{1}(H)\) can be estimated as follows:

$$\begin{aligned} R_{1}(H)&= \sum _{{\mathop {(d_{1}, d_{2})= 1}\limits ^{d_{1} d_{2}< \xi }}} \mu (d_{1}) \mu (d_{2})\,\frac{\lambda (d_{1}^{k}, d_{2}^{k})}{(d_{1} d_{2})^{3 k}} H^{3}\nonumber \\&\quad + O\left( \sum _{{\mathop {(d_{1}, d_{2})= 1}\limits ^{d_{1} d_{2}< \xi }}} ((d_{1} d_{2})^{-k+ \varepsilon } H^{2}+ H (d_{1} d_{2})^{\varepsilon }+ (d_{1} d_{2})^{k+ \varepsilon })\right) \nonumber \\&= \sum _{{\mathop {(d_{1}, d_{2})= 1}\limits ^{d_{1} d_{2}= 1}}}^{\infty } \mu (d_{1}) \mu (d_{2})\,\frac{\lambda (d_{1}^{k}, d_{2}^{k})}{(d_{1} d_{2})^{3 k}} H^{3}+ O(H^{3} \xi ^{1- k+ \varepsilon }+ H^{2}+ \xi ^{1+ k+ \varepsilon }). \end{aligned}$$
(3.8)

In the last step above, we can check that by using Lemma 2.4

$$\begin{aligned}&\sum _{{\mathop {(d_{1}, d_{2})= 1}\limits ^{d_{1} d_{2}> \xi }}} \mu (d_{1}) \mu (d_{2})\,\frac{\lambda (d_{1}^{k}, d_{2}^{k})}{(d_{1} d_{2})^{3 k}} \\&\quad \ll \sum _{{\mathop {(d_{1}, d_{2})= 1}\limits ^{d_{1} d_{2}> \xi }}} \frac{(d_{1} d_{2})^{2k+ \varepsilon }}{(d_{1} d_{2})^{3k}}\ll \sum _{n> \xi } \frac{\tau (n)}{n^{k- \varepsilon }} \ll \xi ^{1- k+ \varepsilon }. \end{aligned}$$

Put

$$\begin{aligned} \sigma _{k}= \sum _{{\mathop {(d_{1}, d_{2})= 1}\limits ^{d_{1} d_{2}= 1}}}^{\infty } \frac{ \mu (d_{1}) \mu (d_{2})\,\lambda (d_{1}^{k}, d_{2}^{k})}{(d_{1} d_{2})^{3 k}}. \end{aligned}$$
(3.9)

From (1.6), Lemma 2.3 and \((d_{1}, d_{2})= 1,\) we get

$$\begin{aligned} \lambda (d_{1}^{k}, d_{2}^{k})= \lambda (d_{1}^{k}, 1)\, \lambda (1, d_{2}^{k}). \end{aligned}$$
(3.10)

Combining (3.9) and (3.10) we obtain

$$\begin{aligned} \sigma _{k}= \sum _{d_{1} = 1}^{\infty } \frac{ \mu (d_{1}) \,\lambda (d_{1}^{k}, 1)}{d_{1} ^{3 k}} \sum _{d_{2} = 1}^{\infty } \frac{ \mu (d_{2}) \,\lambda (1, d_{2}^{k})}{d_{2} ^{3 k}} \delta _{d_{1}}(d_{2}), \end{aligned}$$
(3.11)

where

$$\begin{aligned} \delta _{d_{1}}(d_{2}) = {\left\{ \begin{array}{ll} 1,&{} \textrm{if} \, (d_{1}, d_{2})= 1,\\ 0, &{} \textrm{if}\, (d_{1}, d_{2})> 1. \end{array}\right. } \end{aligned}$$

Since the function

$$\begin{aligned} \frac{\mu (d_{2}) \lambda (1, d_{2}^{k})}{d_{2}^{3k}} \delta _{d_{1}} (d_{2}) \end{aligned}$$

is multiplicative with respect to \(d_{2},\) we have the Euler product representation:

$$\begin{aligned} \sum _{d_{2}= 1}^{\infty } \frac{\mu (d_{2}) \lambda (1, d_{2}^{k})}{d_{2}^{3k}} \delta _{d_{1}} (d_{2})&= \prod _{p\not \mid d_{1}} \left( 1- \frac{\lambda (1, p^{k})}{p^{3k}}\right) \nonumber \\&= \prod _{p} \left( 1- \frac{\lambda (1, p^{k})}{p^{3k}}\right) \prod _{p\mid d_{1}} \left( 1- \frac{\lambda (1, p^{k})}{p^{3k}}\right) ^{-1}. \end{aligned}$$
(3.12)

From (3.11) and (3.12) we infer

$$\begin{aligned} \sigma _{k}&= \sum _{d_{1} = 1}^{\infty } \frac{ \mu (d_{1}) \,\lambda (d_{1}^{k}, 1)}{d_{1} ^{3 k}} \prod _{p} \left( 1- \frac{\lambda (1, p^{k})}{p^{3k}}\right) \prod _{p\mid d_{1}} \left( 1- \frac{\lambda (1, p^{k})}{p^{3k}}\right) ^{-1}\nonumber \\&= \prod _{p} \left( 1- \frac{\lambda (1, p^{k})}{p^{3k}}\right) \sum _{d_{1} = 1}^{\infty } \frac{ \mu (d_{1}) \,\lambda (d_{1}^{k}, 1)}{d_{1} ^{3 k}} \prod _{p\mid d_{1}} \left( 1- \frac{\lambda (1, p^{k})}{p^{3k}}\right) ^{-1}\nonumber \\&= \prod _{p} \left( 1- \frac{\lambda (1, p^{k})}{p^{3k}}\right) \prod _{p} \left( 1- \frac{\lambda ( p^{k}, 1)}{p^{3k}} \left( 1- \frac{\lambda (1, p^{k})}{p^{3k}}\right) ^{-1}\right) \nonumber \\&= \prod _{p} \left( 1- \frac{\lambda ( p^{k}, 1) + \lambda (1, p^{k})}{p^{3k}}\right) . \end{aligned}$$
(3.13)

3.2 Estimation of \(R_{2}(H)\)

From (3.3), we derive by a splitting argument

$$\begin{aligned} R_{2}(H)\ll (\log H)^{2} \sum _{D_{1}\le d_{1}< 2 D_{1}} \sum _{D_{2}\le d_{2}< 2 D_{2}} \sum _{{\mathop {t d_{1}^{k}+ 1\equiv 0\ (d_{2}^{k})}\limits ^{t\le (3 H^{2}+ 1) d_{1}^{-k}}}} \sum _{{\mathop {x^{2}+ y^{2}+ z^{2}= t d_{1}^{k}- 1}\limits ^{1\le x, y, z\le H}}} 1, \end{aligned}$$

where

$$\begin{aligned} \frac{1}{2}\le D_{1}, D_{2}\le (3 H^{2}+ 2)^{1/k}, D_{1}, D_{2}> \frac{\xi }{4}. \end{aligned}$$
(3.14)

We therefore obtain

$$\begin{aligned} R_{2}(H)&\ll H^{\varepsilon } \sum _{D_{1}\le d_{1}< 2 D_{1}} \sum _{t\le (3 H^{2}+ 1) D_{1}^{-k}} \sum _{D_{2}\le d_{2}< 2 D_{2}} \sum _{{\mathop {t d_{1}^{k}+ 1= s d_{2}^{k}}\limits ^{s\le (3 H^{2}+ 2) D_{2}^{-k}}}} 1\nonumber \\&\ll H^{\varepsilon } \sum _{D_{1}\le d_{1}< 2 D_{1}} \sum _{t\le (3 H^{2}+ 1) D_{1}^{-k}} \tau (t d_{1}^{k}+ 1)\nonumber \\&\ll H^{\varepsilon } \sum _{D_{1}\le d_{1}< 2 D_{1}} \sum _{t\le (3 H^{2}+ 1) D_{1}^{-k}} 1\ll H^{2+ \varepsilon } D_{1}^{1-k}. \end{aligned}$$
(3.15)

Similarly, one can obtain

$$\begin{aligned} R_{2}(H)\ll H^{2+ \varepsilon } D_{2}^{1-k}. \end{aligned}$$
(3.16)

Hence from (3.14) to (3.16) we get

$$\begin{aligned} R_{2}(H)\ll H^{2+ \varepsilon } \xi ^{\frac{1}{2}- \frac{k}{2}}. \end{aligned}$$
(3.17)

Combining (3.1), (3.8), (3.13) and (3.17) gives

$$\begin{aligned} R(H)= \sigma _{k} H^{3}+ O(H^{3} \xi ^{1- k} + H^{2}+ \xi ^{1+ k+ \varepsilon }), \end{aligned}$$

where \(\sigma _{k}\) is defined in (3.13).

Now we obtain Theorem 1.1 by choosing \(\xi = H^{\frac{3}{2k}}.\)