Let N be a sufficiently large integer. In this paper, it is proved that, with at most O(N4/9+ε) exceptions, all even positive integers up to N can be represented in the form p21+p32+p33+p34+p35+p36, where p1,p2,p3,p4,p5,p6 are prime numbers.
Citation: Jinjiang Li, Yiyang Pan, Ran Song, Min Zhang. Exceptional set in Waring–Goldbach problem for sums of one square and five cubes[J]. AIMS Mathematics, 2022, 7(2): 2940-2955. doi: 10.3934/math.2022162
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Let N be a sufficiently large integer. In this paper, it is proved that, with at most O(N4/9+ε) exceptions, all even positive integers up to N can be represented in the form p21+p32+p33+p34+p35+p36, where p1,p2,p3,p4,p5,p6 are prime numbers.
Waring's problem of mixed powers concerns the representation of sufficiently large integer n in the form
n=xk11+xk22+⋯+xkss. |
Among the most interesting cases of mixed powers is that of establishing the representations of sufficiently large integer as the sum of one square and s positive cubes for each s>1, i.e.,
n=x2+y31+y32⋯+y3s. | (1.1) |
In 1930, Stanley [7] showed that (1.1) is solvable for s>6. Afterwards, Stanley [8] and Watson [12] solved the cases s=6 and s=5, respectively. It should be emphasized that Stanley [7] obtained the asymptotic formula for s>6, while Sinnadurai [6] obtained the asymptotic formula for s=6. But in [12], Watson only proved a quite weak lower bound for the number of representation (1.1) with s=5. In 1986, Vaughan [9] enhanced Watson's result and derived a lower bound with the expected order of magnitude. In 2002, Wooley [13] illustrated that, although the excepted asymptotic formula of (1.1) with s=5 can not be established by the technique currently available, the exceptional set is extremely sparse. To be specific, let E1(N) denote the number of integer n⩽N which can not be represented as one square and five positive cubes with expected asymptotic formula, then Wooley [13] showed that E1(N)≪Nε.
In view of the results of Vaughan [9] and Wooley [13], it is reasonable to conjecture that, for every sufficiently large even integer N, the following equation
N=p21+p32+p33+p34+p35+p36 | (1.2) |
is solvable. Here and below, the letter p, with or without subscript, denotes a prime number. But this conjecture is perhaps out of reach at present. However, it is possible to replace a variable by an almost–prime. In 2014, Cai [1] proved that, for every sufficiently large even integer N, the following equation
N=x2+p31+p32+p33+p34+p35 | (1.3) |
is solvable with x being an almost–prime P36 and the pj(j=1,2,3,4,5) primes. Later, in 2018, Li and Zhang [3] enhanced the result of Cai [1] and showed that (1.3) is solvable with x being an almost–prime P6 and the pj(j=1,2,3,4,5) primes. Recently, in 2021, Xue, Zhang and Li [15] consider the problem (1.2) with almost equal prime variables, i.e.,
{n=p21+p32+p33+p34+p35+p36,n∈[N−6U,N+6U],|p21−N6|⩽U,|p3i−N6|⩽U,i=2,3,4,5,6, | (1.4) |
where U=N1−δ+ε with δ>0 hoped to be as large as possible. Let E(N,U) denote the number of all positive even integers n satisfying
N−6U⩽n⩽N+6U, |
which can not be represented as (1.4). One wants to show that there exists δ∈(0,1) such that
E(N,U)≪U1−ε,U=N1−δ+ε. | (1.5) |
In [15], they proved that δ⩽8/225.
In this paper, we shall investigate the exceptional set of the problem (1.2) and establish the following result.
Theorem 1.1. Let E(N) denote the number of positive even integers n up to N, which can not be represented as
n=p21+p32+p33+p34+p35+p36. | (1.6) |
Then, for any ε>0, we have
E(N)≪N49+ε. |
We will establish Theorem 1.1 by using a pruning process into the Hardy–Littlewood circle method. In the treatment of the integrals over minor arcs, we will employ the methods, which is developed by Wooley in [14], combining with the new estimates for exponential sum over primes developed by Zhao [16]. For the treatment of the integrals on the major arcs, we shall prune the major arcs further and deal with them respectively. The full details will be explained in the following revelent sections.
Notation. Throughout this paper, let p, with or without subscripts, always denote a prime number; ε always denotes a sufficiently small positive constant, which may not be the same at different occurrences. As usual, we use φ(n),d(n) and ω(n) to denote the Euler's function, Dirichlet's divisor function and the number of distinct prime factors of n, respectively. Also, we use χmodq to denote a Dirichlet character modulo q, and χ0modq the principal character. e(x)=e2πix; f(x)≪g(x) means that f(x)=O(g(x)); f(x)≍g(x) means that f(x)≪g(x)≪f(x). N is a sufficiently large integer and n∈[N/2,N], and thus logN≍logn. The letter c, with or without subscripts or superscripts, always denote a positive constant, which may not be the same at different occurrences.
Let N be a sufficiently large positive integer. By a splitting argument, it is sufficient to consider the even integers n∈(N/2,N]. For the application of the Hardy–Littlewood method, we need to define the Farey dissection. Let A>0 be a sufficiently large fixed number, which will be determined at the end of the proof. We set
Q0=logAN,Q1=N16,Q2=N56,I0=[−1Q2,1−1Q2]. |
By Dirichlet's lemma on rational approximation (for instance, see Lemma 12 on page 104 of Pan and Pan [4]), each α∈[−1/Q2,1−1/Q2] can be written as the form
α=aq+λ,|λ|⩽1qQ2, | (2.1) |
for some integers a,q with 1⩽a⩽q⩽Q2 and (a,q)=1. Define
M(q,a)=[aq−Q1qN,aq+Q1qN],M=⋃1⩽q⩽Q1⋃1⩽a⩽q(a,q)=1M(q,a),M0(q,a)=[aq−Q2000qN,aq+Q2000qN],M0=⋃1⩽q⩽Q1000⋃1⩽a⩽q(a,q)=1M0(q,a),m1=I0∖M,m2=M∖M0. |
Then we obtain the Farey dissection
I0=M0∪m1∪m2. | (2.2) |
For k=2,3, we define
fk(α)=∑Xk<p⩽2Xke(pkα), |
where Xk=(N/16)1k. Let
R(n)=∑n=p21+p32+p33+p34+p35+p36X3<p2,…,p6⩽2X3X2<p1⩽2X21. |
From (2.2), one has
R(n)=∫10f2(α)f53(α)e(−nα)dα=∫1−1Q2−1Q2f2(α)f53(α)e(−nα)dα={∫M0+∫m1+∫m2}f2(α)f53(α)e(−nα)dα. |
In order to prove Theroem 1.1, we need the two following propositions:
Proposition 2.1. For n∈[N/2,N], there holds
∫M0f2(α)f53(α)e(−nα)dα=S(n)J(n)+O(n7/6log7n), | (2.3) |
where S(n) is the singular series defined in (4.1), which is absolutely convergent and satisfies
0<c∗⩽S(n)≪1 | (2.4) |
for any integer n satisfying n≡0(mod2) and some fixed constant c∗>0; while J(n) is defined by (4.5) and satisfies
J(n)≍n7/6log6n. |
The proof of (2.3) in Proposition 2.1 will be demonstrated in Section 4. For the property (2.4) of singular series, we shall give the proof in Section 5.
Proposition 2.2. Let Z(N) denote the number of integers n∈[N/2,N] satisfying n≡0(mod2) such that
2∑j=1|∫mjf2(α)f53(α)e(−nα)dα|≫n7/6log7n. |
Then we have
Z(N)≪N49+ε. |
The proof of Proposition 2.2 will be given in section 6. The remaining part of this section is devoted to establishing Theorem 1.1 by using Proposition 2.1 and Proposition 2.2.
Proof of Theorem 1.1. From Proposition 2.2, we deduce that, with at most O(N49+ε) exceptions, all even integers n∈[N/2,N] satisfy
2∑j=1|∫mjf2(α)f53(α)e(−nα)dα|≪n7/6log7n, |
from which and Proposition 2.1, we conclude that, with at most O(N49+ε) exceptions, all even integers n∈[N/2,N] can be represented in the form p21+p32+p33+p34+p35+p36, where p1,p2,p3,p4,p5,p6 are prime numbers. By a splitting argument, we get
E(N)≪∑0⩽ℓ≪logNZ(N2ℓ)≪∑0⩽ℓ≪logN(N2ℓ)49+ε≪N49+ε. |
This completes the proof of Theorem 1.1.
Lemma 3.1. Suppose that α is a real number, and that |α−a/q|⩽q−2 with (a,q)=1. Let β=α−a/q. Then we have
fk(α)≪dδk(q)(logx)c(X1/2k√q(1+N|β|)+X4/5k+Xk√q(1+N|β|)), |
where δk=12+logklog2 and c is a constant.
Proof. See Theorem 1.1 of Ren [5].
Lemma 3.2. Suppose that α is a real number, and that there exist a∈Z and q∈N with
(a,q)=1,1⩽q⩽Qand|qα−a|⩽Q−1. |
If P12⩽Q⩽P52, then one has
∑P<p⩽2Pe(p3α)≪P1−112+ε+q−16P1+ε(1+P3|α−a/q|)1/2. |
Proof. See Lemma 8.5 of Zhao [16].
Lemma 3.3. For α∈m1, we have
f3(α)≪N1136+ε. |
Proof. By Dirichlet's rational approximation (2.1), for α∈m1, one has Q1⩽q⩽Q2. From Lemma 3.2, we get
f3(α)≪X1112+ε3+X1+ε3Q−161≪N1136+ε. |
This completes the proof of Lemma 3.3.
For 1⩽a⩽q with (a,q)=1, set
I(q,a)=[aq−1qQ0,aq+1qQ0],I=⋃1⩽q⩽Q02q⋃a=−q(a,q)=1I(q,a). | (3.1) |
For α∈m2, by Lemma 3.1, we have
f3(α)≪N13logcNq12−ε(1+N|λ|)1/2+N415+ε=V3(α)+N415+ε, | (3.2) |
say. Then we obtain the following lemma.
Lemma 3.4. We have
∫I|V3(α)|2dα=∑1⩽q⩽Q02q∑a=−q(a,q)=1∫I(q,a)|V3(α)|2dα≪N−13log2AN. |
Proof. We have
∑1⩽q⩽Q02q∑a=−q(a,q)=1∫I(q,a)|V3(α)|2dα≪∑1⩽q⩽Q0q−1+ε2q∑a=−q(a,q)=1∫|λ|⩽1Q0N23logcN1+N|λ|dλ≪∑1⩽q⩽Q0q−1+ε2q∑a=−q(a,q)=1(∫|λ|⩽1NN23logcNdλ+∫1N⩽|λ|⩽1Q0N23logcNN|λ|dλ)≪N−13logcN⋅∑1⩽q⩽Q0q−1+εφ(q)≪N−13Q1+ε0logcN≪N−13log2AN. |
This completes the proof of Lemma 3.4.
In this section, we shall concentrate on establishing Proposition 2.1. We first introduce some notations. For a Dirichlet character χmodq and k=2,3, we define
Ck(χ,a)=q∑h=1¯χ(h)e(ahkq),Ck(q,a)=Ck(χ0,a), |
where χ0 is the principal character modulo q. Let χ2,χ(1)3,χ(2)3,χ(3)3,χ(4)3,χ(5)3 be Dirichlet characters modulo q. Define
B(n,q,χ2,χ(1)3,χ(2)3,χ(3)3,χ(4)3,χ(5)3)=q∑a=1(a,q)=1(C2(χ2,a)5∏i=1C3(χ(i)3,a))e(−anq), |
B(n,q)=B(n,q,χ0,χ0,χ0,χ0,χ0,χ0), |
and write
A(n,q)=B(n,q)φ6(q),S(n)=∞∑q=1A(n,q). | (4.1) |
Lemma 4.1. For (a,q)=1 and any Dirichlet character χmodq, there holds
|Ck(χ,a)|⩽2q1/2dβk(q) |
with βk=(logk)/log2.
Proof. See the Problem 14 of Chapter VI of Vinogradov [11].
Lemma 4.2. Let Ck(q,a) be defined as above. Then there holds
∑q⩽x|B(n,q)|φ6(q)≪logx. | (4.2) |
Proof. By Lemma 4.1, we have
B(n,q)≪q∑a=1(a,q)=1|C2(q,a)C53(q,a)|≪q3φ(q)d9(q). |
Therefore, the left–hand side of (4.2) is
≪∑q⩽xq3φ(q)d9(q)φ6(q)≪∑q⩽x(loglogq)5d9(q)q2≪(loglogx)5∑q⩽xd9(q)q2≪logx. |
This completes the proof of Lemma 4.2.
In order to treat the integral on the major arcs, we write fk(α) as follows:
fk(α)=∑Xk<p⩽2Xk(p,q)=1e(pk(aq+λ))+O(logq)=q∑ℓ=1(ℓ,q)=1e(aℓkq)∑Xk<p⩽2Xkp≡ℓ(modq)e(pkλ)+O(logN). |
For the innermost sum on the right–hand side of the above equation, by Siegel–Walfisz theorem, we have
∑Xk<p⩽2Xkp≡ℓ(modq)e(pkλ)=∫2XkXke(ukλ)dπ(u,q,ℓ)=∫2XkXke(ukλ)d(1φ(q)∫u2dtlogt+O(ue−c√logu))=1φ(q)∫2XkXke(ukλ)logudu+O(Xke−c√logN)=vk(λ)φ(q)+O(Xke−c√logN), |
say. Therefore, we have
fk(α)=Ck(q,a)φ(q)vk(λ)+O(Xke−c√logN), |
and thus
f2(α)f53(α)=C2(q,a)C53(q,a)φ6(q)v2(λ)v53(λ)+O(N136e−c√logN). |
Then we derive that
∫M0f2(α)f53(α)e(−nα)dα=∑1⩽q⩽Q1000q∑a=1(a,q)=1e(−anq)∫Q2000qN−Q2000qN(C2(q,a)C53(q,a)φ6(q)v2(λ)v53(λ)+O(N136e−c√logN))e(−nλ)dλ=∑1⩽q⩽Q1000B(n,q)φ6(q)∫Q2000qN−Q2000qNv2(λ)v53(λ)e(−nλ)dλ+O(N76e−c√logN). | (4.3) |
Noting that
vk(λ)=∫(2Xk)kXkkx1k−1e(λx)logxdx, |
hence the innermost integral in (4.3) can be written as
∫Q2000qN−Q2000qN(∫(2X2)2X22x−12e(λx)logxdx)(∫(2X3)3X33x−23e(λx)logxdx)5e(−nλ)dλ. | (4.4) |
By using the elementary estimate
vk(λ)=∫(2Xk)kXkkx1k−1e(λx)logxdx≪N1k−1logNmin(N,1|λ|), |
we know that if we extend the interval of the integral in (4.4) to [−1/2,1/2], then the resulting error is
≪∫12Q2000qNN−236log6N⋅dλλ6≪N−236log6N⋅q5N5Q10000≪N76(logN)500A≪n76(logn)500A. |
Hence we obtain
∫Q2000qN−Q2000qNv2(λ)v53(λ)e(−nλ)dλ=J(n)+O(n76(logn)500A), |
where
J(n)=∫12−12(∫(2X2)2X22x−12e(λx)logxdx)(∫(2X3)3X33x−23e(λx)logxdx)5e(−nλ)dλ=∫(2X2)2X22∫(2X3)3X33∫(2X3)3X33∫(2X3)3X33∫(2X3)3X33∫(2X3)3X33∫12−12x−121(x2x3x4x5x6)−23(logx1)(logx2)⋯(logx6)×e((x1+x2+x3+x4+x5+x6−n)λ)dλdx1⋯dx6≍X−12X−103(logN)6∫(2X2)2X22∫(2X3)3X33⋯∫(2X3)3X33∫12−12e((x1+x2+⋯+x6−n)λ)dλdx1⋯dx6≍X−12X−103(logN)6N5≍N76(logN)6≍n76(logn)6. | (4.5) |
Therefore, by (5.4), (4.3) becomes
∫M0f2(α)f53(α)e(−nα)dα=S(n)J(n)+O(n76log7n), |
which completes the proof of Proposition 2.1.
In this section, we shall concentrate on investigating the properties of the singular series which appear in Proposition 2.1.
Lemma 5.1. Let p be a prime and pα‖. For (a, p) = 1 , if \ell\geqslant\gamma(p) , we have C_k(p^\ell, a) = 0 , where
\begin{equation*} \gamma(p) = \begin{cases} \alpha+2, & if ~~p\not = 2~~ or ~~p = 2, \, \alpha = 0; \\ \alpha+3, & if ~~p = 2, \, \alpha > 0. \\ \end{cases} \end{equation*} |
Proof. See Lemma 8.3 of Hua [2].
For k\geqslant1 , we define
\begin{equation*} S_k(q, a) = \sum\limits_{m = 1}^qe\Bigg(\frac{am^k}{q}\Bigg). \end{equation*} |
Lemma 5.2. Suppose that (p, a) = 1 . Then
\begin{equation*} S_k(p, a) = \sum\limits_{\chi\in\mathscr{A}_k}\overline{\chi(a)}\tau(\chi), \end{equation*} |
where \mathscr{A}_k denotes the set of non–principal characters \chi modulo p for which \chi^k is principal, and \tau(\chi) denotes the Gauss sum
\begin{equation*} \sum\limits_{m = 1}^p\chi(m)e\Bigg(\frac{m}{p}\Bigg). \end{equation*} |
Also, there hold |\tau(\chi)| = p^{1/2} and |\mathscr{A}_k| = (k, p-1)-1 .
Proof. See Lemma 4.3 of Vaughan [10].
Lemma 5.3. For (p, n) = 1 , we have
\begin{equation} \left|\sum\limits_{a = 1}^{p-1}\frac{S_2(p, a)S_3^5(p, a)}{p^6}e\Bigg(-\frac{an}{p}\Bigg)\right| \leqslant32 p^{-\frac{5}{2}}. \end{equation} | (5.1) |
Proof. We denote by \mathcal{S} the left–hand side of (5.1). By Lemma 5.2, we have
\begin{equation*} \mathcal{S} = \frac{1}{p^6}\sum\limits_{a = 1}^{p-1}\Bigg(\sum\limits_{\chi_2\in\mathscr{A}_2}\overline{\chi_2(a)}\tau(\chi_2)\Bigg) \Bigg(\sum\limits_{\chi_3\in\mathscr{A}_3}\overline{\chi_3(a)}\tau(\chi_3)\Bigg)^5 e\Bigg(-\frac{an}{p}\Bigg). \end{equation*} |
If |\mathscr{A}_k| = 0 for some k\in\{2, 3\} , then \mathcal{S} = 0 . If this is not the case, then
\begin{align*} \mathcal{S} = & \, \frac{1}{p^6}\sum\limits_{\chi_2\in \mathscr{A}_2} \sum\limits_{\chi_3^{(1)}\in \mathscr{A}_3}\sum\limits_{\chi_3^{(2)}\in \mathscr{A}_3} \sum\limits_{\chi_3^{(3)}\in \mathscr{A}_3}\sum\limits_{\chi_3^{(4)}\in \mathscr{A}_3} \sum\limits_{\chi_3^{(5)}\in \mathscr{A}_3} \\ &\quad \quad\times \tau\big(\chi_2\big) \tau\big(\chi_3^{(1)}\big) \tau\big(\chi_3^{(2)}\big)\tau\big(\chi_3^{(3)}\big)\tau\big(\chi_3^{(4)}\big) \tau\big(\chi_3^{(5)}\big) \\ &\quad \quad\times \sum\limits_{a = 1}^{p-1} \overline{\chi_2(a)\chi_3^{(1)}(a)\chi_3^{(2)}(a)\chi_3^{(3)}(a) \chi_3^{(4)}(a)\chi_3^{(5)}(a)} e\Bigg(-\frac{an}{p}\Bigg). \end{align*} |
From Lemma 5.2, the sextuple outer sums have not more than \big((2, p-1)-1\big)\times\big((3, p-1)-1\big)^5 \leqslant1\times2^5 = 32 terms. In each of these terms, we have
\begin{equation*} \Big|\tau\big(\chi_2\big)\tau\big(\chi_3^{(1)}\big)\tau\big(\chi_3^{(2)}\big)\tau\big(\chi_3^{(3)}\big) \tau\big(\chi_3^{(4)}\big)\tau\big(\chi_3^{(5)}\big)\Big| = p^3. \end{equation*} |
Since in any one of these terms
\begin{equation*} \overline{\chi_2(a)\chi_3^{(1)}(a)\chi_3^{(2)}(a)\chi_3^{(3)}(a)\chi_3^{(4)}(a)\chi_3^{(5)}(a)} \end{equation*} |
is a Dirichlet character \chi\!\pmod p , the inner sum is
\begin{equation*} \sum\limits_{a = 1}^{p-1}\chi(a)e\Bigg(-\frac{an}{p}\Bigg) = \overline{\chi(-n)} \sum\limits_{a = 1}^{p-1}\chi(-an)e\Bigg(-\frac{an}{p}\Bigg) = \overline{\chi(-n)}\tau(\chi). \end{equation*} |
From the fact that \tau(\chi^0) = -1 for principal character \chi^0\bmod p , we have
\begin{equation*} \big|\overline{\chi(-n)}\tau(\chi)\big|\leqslant p^{\frac{1}{2}}. \end{equation*} |
By the above arguments, we obtain
\begin{equation*} \big|\mathcal{S}\big|\leqslant\frac{1}{p^6}\cdot 32\cdot p^3\cdot p^{\frac{1}{2}} = 32p^{-\frac{5}{2}}. \end{equation*} |
This completes the proof of Lemma 5.3.
Lemma 5.4. Let \mathcal{L}(p, n) denote the number of solutions of the congruence
\begin{equation*} x_1^2+x_2^3+x_3^3+x_4^3+x_5^3+x_6^3\equiv n \!\!\pmod p, \qquad 1\leqslant x_1, x_2, \dots, x_6\leqslant p-1. \end{equation*} |
Then, for n\equiv 0\!\pmod 2 , we have \mathcal{L}(p, n) > 0 .
Proof. We have
\begin{equation*} p\cdot\mathcal{L}(p, n) = \sum\limits_{a = 1}^pC_2(p, a)C_3^5(p, a)e\Bigg(-\frac{an}{p}\Bigg) = (p-1)^6+E_p, \end{equation*} |
where
\begin{equation*} E_p = \sum\limits_{a = 1}^{p-1}C_2(p, a)C_3^5(p, a)e\Bigg(-\frac{an}{p}\Bigg). \end{equation*} |
By Lemma 5.2, we obtain
\begin{equation*} |E_p|\leqslant (p-1)(\sqrt{p}+1)(2\sqrt{p}+1)^5. \end{equation*} |
It is easy to check that |E_p| < (p-1)^6 for p\geqslant13 . Therefore, we obtain \mathcal{L}(p, n) > 0 for p\geqslant 13 . For p = 2, 3, 5, 7, 11 , we can check \mathcal{L}(p, n) > 0 directly.
Lemma 5.5. A(n, q) is multiplicative in q .
Proof. By the definition of A(n, q) in (4.1), we only need to show that B(n, q) is multiplicative in q . Suppose q = q_1q_2 with (q_1, q_2) = 1 . Then we have
\begin{align} & B(n, q_1q_2) = \sum\limits_{\substack{a = 1\\(a, q_1q_2) = 1}}^{q_1q_2} C_2(q_1q_2, a)C_3^5(q_1q_2, a) e\Bigg(-\frac{an}{q_1q_2}\Bigg) \\ = & \sum\limits_{\substack{a_1 = 1\\(a_1, q_1) = 1}}^{q_1}\sum\limits_{\substack{a_2 = 1\\(a_2, q_2) = 1}}^{q_2} C_2\big(q_1q_2, a_1q_2+a_2q_1\big) C_3^5\big(q_1q_2, a_1q_2+a_2q_1\big) e\Bigg(-\frac{a_1n}{q_1}\Bigg)e\Bigg(-\frac{a_2n}{q_2}\Bigg). \end{align} | (5.2) |
For (q_1, q_2) = 1 and k\in\{2, 3\} , there holds
\begin{align} C_k(q_1q_2, a_1q_2+a_2q_1) = & \, \, \sum\limits_{\substack{m = 1\\(m, q_1q_2) = 1}}^{q_1q_2}e\Bigg(\frac{(a_1q_2+a_2q_1)m^k}{q_1q_2}\Bigg) \\ = & \, \, \sum\limits_{\substack{m_1 = 1\\(m_1, q_1) = 1}}^{q_1}\sum\limits_{\substack{m_2 = 1\\(m_2, q_2) = 1}}^{q_2} e\Bigg(\frac{(a_1q_2+a_2q_1)(m_1q_2+m_2q_1)^k}{q_1q_2}\Bigg) \\ = &\, \, \sum\limits_{\substack{m_1 = 1\\(m_1, q_1) = 1}}^{q_1} e\Bigg(\frac{a_1(m_1q_2)^k}{q_1}\Bigg) \sum\limits_{\substack{m_2 = 1\\(m_2, q_2) = 1}}^{q_2} e\Bigg(\frac{a_2(m_2q_1)^k}{q_2}\Bigg) \\ = & \, \, C_k(q_1, a_1)C_k(q_2, a_2). \end{align} | (5.3) |
Putting (5.3) into (5.2), we deduce that
\begin{align*} & \, B(n, q_1q_2) \nonumber \\ = & \sum\limits_{\substack{a_1 = 1\\(a_1, q_1) = 1}}^{q_1} C_2(q_1, a_1)C_3^5(q_1, a_1) e\Bigg(-\frac{a_1n}{q_1}\Bigg) \sum\limits_{\substack{a_2 = 1\\(a_2, q_2) = 1}}^{q_2} C_2(q_2, a_2)C_3^5(q_2, a_2)e\Bigg(-\frac{a_2n}{q_2}\Bigg) \nonumber \\ = & \, \, B(n, q_1)B(n, q_2). \end{align*} |
This completes the proof of Lemma 5.5.
Lemma 5.6. Let A(n, q) be as defined in (4.1). Then
\rm{(i)} We have
\begin{equation} \sum\limits_{q > Z}A(n, q)\ll Z^{-\frac{3}{2}+\varepsilon}d(n). \end{equation} | (5.4) |
\rm{(ii)} There exists an absolute positive constant c^* > 0 , such that, for n\equiv0\pmod2 , there holds
\begin{equation*} 0 < c^*\leqslant \mathfrak{S}(n)\ll1. \end{equation*} |
Proof. From Lemma 5.5, we know that B(n, q) is multiplicative in q . Therefore, there holds
\begin{equation} B(n, q) = \prod\limits_{p^t\|q}B(n, p^t) = \prod\limits_{p^t\|q}\sum\limits_{\substack{a = 1\\ (a, p) = 1}}^{p^t} C_2(p^t, a)C_3^5(p^t, a)e\Bigg(-\frac{an}{p^t}\Bigg). \end{equation} | (5.5) |
From (5.5) and Lemma 5.1, we deduce that B(n, q) = \prod\limits_{p\|q}B(n, p) or 0 according to whether q is square–free or not. Thus, one has
\begin{equation} \sum\limits_{q = 1}^{\infty}A(n, q) = \sum\limits_{\substack{q = 1\\ \rm{$q$ square–free}}}^\infty A(n, q). \end{equation} | (5.6) |
Write
\begin{equation*} \mathcal{R}(p, a): = C_2(p, a)C_3^5(p, a)-S_2(p, a)S_3^5(p, a). \end{equation*} |
Then
\begin{equation} A(n, p) = \frac{1}{(p-1)^6}\sum\limits_{a = 1}^{p-1}S_2(p, a)S_3^5(p, a)e\Bigg(-\frac{an}{p}\Bigg) +\frac{1}{(p-1)^6}\sum\limits_{a = 1}^{p-1}\mathcal{R}(p, a)e\Bigg(-\frac{an}{p}\Bigg). \end{equation} | (5.7) |
Applying Lemma 4.1 and noticing that S_k(p, a) = C_k(p, a)+1 , we get S_k(p, a)\ll p^{\frac{1}{2}} , and thus \mathcal{R}(p, a)\ll p^{\frac{5}{2}} . Therefore, the second term in (5.7) is \leqslant c_1p^{-\frac{5}{2}} . On the other hand, from Lemma 5.3, we can see that the first term in (5.7) is \leqslant2^6\cdot 32p^{-\frac{5}{2}} = 2048p^{-\frac{5}{2}} . Let c_2 = c_1+2048 . Then we have proved that, for p\nmid n , there holds
\begin{equation} |A(n, p)|\leqslant c_2p^{-\frac{5}{2}}. \end{equation} | (5.8) |
Moreover, if we use Lemma 4.1 directly, it follows that
\begin{align*} \big|B(n, p)\big| = & \, \Bigg|\sum\limits_{a = 1}^{p-1}C_2(p, a)C_3^5(p, a)e\bigg(-\frac{an}{p}\bigg)\Bigg| \leqslant \sum\limits_{a = 1}^{p-1} \Big|C_2(p, a)C_3^5(p, a)\Big| \\ \leqslant & \, \, (p-1)\cdot 2^6\cdot p^3\cdot 486 = 31104p^3(p-1), \end{align*} |
and therefore
\begin{equation} \big|A(n, p)\big| = \frac{|B(n, p)|}{\varphi^6(p)}\leqslant\frac{31104p^3}{(p-1)^5} \leqslant\frac{2^5\cdot31104p^3}{p^5} = \frac{995328}{p^2}. \end{equation} | (5.9) |
Let c_3 = \max(c_2, 995328) . Then for square–free q , we have
\begin{align*} \big|A(n, q)\big| = & \, \Bigg(\prod\limits_{\substack{p|q \\ p\nmid n}}\big|A(n, p)\big|\Bigg) \Bigg(\prod\limits_{\substack{p|q \\ p|n}}\big|A(n, p)\big|\Bigg)\leqslant \Bigg(\prod\limits_{\substack{p|q \\ p\nmid n}}\big(c_3p^{-\frac{5}{2}}\big)\Bigg) \Bigg(\prod\limits_{\substack{p|q \\ p|n}}\big(c_3p^{-2}\big)\Bigg) \\ = & \, \, c_3^{\omega(q)}\Bigg(\prod\limits_{p|q}p^{-\frac{5}{2}}\Bigg) \Bigg(\prod\limits_{p|(n, q)}p^{\frac{1}{2}}\Bigg) \ll q^{-\frac{5}{2}+\varepsilon}(n, q)^{\frac{1}{2}}. \end{align*} |
Hence, by (5.6), we obtain
\begin{align*} \sum\limits_{q > Z}|A(n, q)| \ll & \, \, \sum\limits_{q > Z}q^{-\frac{5}{2}+\varepsilon}(n, q)^{\frac{1}{2}} = \sum\limits_{d|n}\sum\limits_{q > \frac{Z}{d}}(dq)^{-\frac{5}{2}+\varepsilon}d^{\frac{1}{2}} = \sum\limits_{d|n}d^{-2+\varepsilon}\sum\limits_{q > \frac{Z}{d}}q^{-\frac{5}{2}+\varepsilon} \\ \ll & \, \, \sum\limits_{d|n}d^{-2+\varepsilon}\Bigg(\frac{Z}{d}\Bigg)^{-\frac{3}{2}+\varepsilon} = Z^{-\frac{3}{2}+\varepsilon}\sum\limits_{d|n}d^{-\frac{1}{2}} \ll Z^{-\frac{3}{2}+\varepsilon}d(n), \end{align*} |
which proves (5.4), and hence gives the absolutely convergence of \mathfrak{S}(n) . In order to prove (ii) of Lemma 5.6, by Lemma 5.5, we first note that
\begin{align} \mathfrak{S}(n) = & \, \, \prod\limits_p\Bigg(1+\sum\limits_{t = 1}^\infty A\big(n, p^t\big)\Bigg) = \prod\limits_p\big(1+A(n, p)\big) \\ = & \, \, \Bigg(\prod\limits_{p\leqslant c_3}\big(1+A(n, p)\big)\Bigg) \Bigg(\prod\limits_{\substack{p > c_3\\ p\nmid n}}\big(1+A(n, p)\big)\Bigg) \Bigg(\prod\limits_{\substack{p > c_3\\ p| n}}\big(1+A(n, p)\big)\Bigg). \end{align} | (5.10) |
From (5.8), we have
\begin{equation} \prod\limits_{\substack{p > c_3\\ p\nmid n}}\big(1+A(n, p)\big)\geqslant\prod\limits_{p > c_3}\Bigg(1-\frac{c_3}{p^{5/2}}\Bigg)\geqslant c_4 > 0. \end{equation} | (5.11) |
By (5.9), we obtain
\begin{equation} \prod\limits_{\substack{p > c_3\\ p| n}}\big(1+A(n, p)\big)\geqslant \prod\limits_{p > c_3}\Bigg(1-\frac{c_3}{p^2}\Bigg) \geqslant c_5 > 0. \end{equation} | (5.12) |
On the other hand, it is easy to see that
\begin{equation*} 1+A(n, p) = \frac{p\cdot \mathcal{L}(p, n)}{\varphi^6(p)}. \end{equation*} |
By Lemma 5.4, we know that \mathcal{L}(p, n) > 0 for all p with n\equiv0\!\pmod 2 , and thus 1+A(n, p) > 0 . Therefore, there holds
\begin{equation} \prod\limits_{p\leqslant c_3}\big(1+A(n, p)\big)\geqslant c_6 > 0. \end{equation} | (5.13) |
Combining the estimates (5.10)–(5.13), and taking c^* = c_4c_5c_6 > 0 , we derive that
\begin{equation*} \mathfrak{S}(n)\geqslant c^* > 0. \end{equation*} |
Moreover, by (5.8) and (5.9), we have
\begin{equation*} \mathfrak{S}(n)\leqslant \prod\limits_{p\nmid n}\Bigg(1+\frac{c_3}{p^{5/2}}\Bigg)\cdot \prod\limits_{p|n}\Bigg(1+\frac{c_3}{p^2}\Bigg)\ll 1. \end{equation*} |
This completes the proof Lemma 5.6.
In this section, we shall give the proof of Proposition 2.2. We denote by \mathcal{Z}_j(N) the set of integers n satisfying n\in[N/2, N] and n\equiv0 \!\pmod 2 for which the following estimate
\begin{equation} \Bigg|\int_{\mathfrak{m}_j}f_2(\alpha)f_3^5(\alpha)e(-n\alpha)\mathrm{d}\alpha\Bigg| \gg\frac{n^{\frac{7}{6}}}{\log^7n} \end{equation} | (6.1) |
holds. For convenience, we use \mathcal{Z}_j to denote the cardinality of \mathcal{Z}_j(N) for abbreviation. Also, we define the complex number \xi_j(n) by taking \xi_j(n) = 0 for n\not\in\mathcal{Z}_j(N) , and when n\in\mathcal{Z}_j(N) by means of the equation
\begin{equation} \Bigg|\int_{\mathfrak{m}_j}f_2(\alpha)f_3^5(\alpha)e(-n\alpha)\mathrm{d}\alpha\Bigg| = \xi_j(n)\int_{\mathfrak{m}_j}f_2(\alpha)f_3^5(\alpha)e(-n\alpha)\mathrm{d}\alpha. \end{equation} | (6.2) |
Plainly, one has |\xi_j(n)| = 1 whenever \xi_j(n) is nonzero. Therefore, we obtain
\begin{equation} \sum\limits_{n\in\mathcal{Z}_j(N)}\xi_j(n) \int_{\mathfrak{m}_j}f_2(\alpha)f_3^5(\alpha)e(-n\alpha)\mathrm{d}\alpha = \int_{\mathfrak{m}_j}f_2(\alpha)f_3^5(\alpha)\mathcal{K}_j(\alpha)\mathrm{d}\alpha, \end{equation} | (6.3) |
where the exponential sum \mathcal{K}_j(\alpha) is defined by
\begin{equation*} \mathcal{K}_j(\alpha) = \sum\limits_{n\in\mathcal{Z}_j(N)}\xi_j(n)e(-n\alpha). \end{equation*} |
For j = 1, 2 , set
\begin{equation*} I_j = \int_{\mathfrak{m}_j}f_2(\alpha)f_3^5(\alpha)\mathcal{K}_j(\alpha)\mathrm{d}\alpha. \end{equation*} |
From (6.1)–(6.3), we derive that
\begin{equation} I_j\gg \sum\limits_{n\in\mathcal{Z}_j(N)}\frac{n^{\frac{7}{6}}}{\log^7n} \gg \frac{\mathcal{Z}_jN^{\frac{7}{6}}}{\log^7N}, \qquad j = 1, 2. \end{equation} | (6.4) |
By Lemma 2.1 of Wooley [14] with k = 2 , we know that, for j = 1, 2 , there holds
\begin{equation} \int_0^1\big|f_2(\alpha)\mathcal{K}_j(\alpha)\big|^2\mathrm{d}\alpha \ll N^\varepsilon\Big(\mathcal{Z}_jN^{\frac{1}{2}}+\mathcal{Z}_j^2\Big). \end{equation} | (6.5) |
It follows from Cauchy's inequality, Lemma 2.5 of Vaughan [10], Lemma 3.3 and (6.5) that
\begin{align} I_1 \ll & \, \, \Bigg(\sup\limits_{\alpha\in\mathfrak{m}_1}\big|f_3(\alpha)\big|\Bigg) \times\Bigg(\int_0^1\big|f_3(\alpha)\big|^8\mathrm{d}\alpha\Bigg)^{\frac{1}{2}} \Bigg(\int_0^1\big|f_2(\alpha)\mathcal{K}_1(\alpha)\big|^2\mathrm{d}\alpha\Bigg)^{\frac{1}{2}} \\ \ll & \, \, N^{\frac{11}{36}+\varepsilon}\cdot \Big(N^{\frac{5}{3}+\varepsilon}\Big)^{\frac{1}{2}}\cdot \Big(N^\varepsilon\Big(\mathcal{Z}_1N^{\frac{1}{2}}+\mathcal{Z}_1^2\Big)\Big)^{\frac{1}{2}} \\ \ll & \, \, N^{\frac{41}{36}+\varepsilon}\Big(\mathcal{Z}_1^{\frac{1}{2}}N^{\frac{1}{4}}+\mathcal{Z}_1\Big) \ll \mathcal{Z}_1^{\frac{1}{2}}N^{\frac{25}{18}+\varepsilon} +\mathcal{Z}_1N^{\frac{41}{36}+\varepsilon}. \end{align} | (6.6) |
Combining (6.4) and (6.6), we get
\begin{equation*} \mathcal{Z}_1N^{\frac{7}{6}}\log^{-7}N\ll I_1\ll \mathcal{Z}_1^{\frac{1}{2}}N^{\frac{25}{18}+\varepsilon} +\mathcal{Z}_1N^{\frac{41}{36}+\varepsilon}, \end{equation*} |
which implies
\begin{equation} \mathcal{Z}_1\ll N^{\frac{4}{9}+\varepsilon}. \end{equation} | (6.7) |
Next, we give the upper bound for \mathcal{Z}_2 . By (3.2), we obtain
\begin{align} I_2 \ll &\, \, \int_{\mathfrak{m}_2}\big|f_2(\alpha)f_3^4(\alpha)V_3(\alpha)\mathcal{K}_2(\alpha)\big| \mathrm{d}\alpha +N^{\frac{4}{15}+\varepsilon}\times\int_{\mathfrak{m}_2} \big|f_2(\alpha)f_3^4(\alpha)\mathcal{K}_2(\alpha)\big|\mathrm{d}\alpha = I_{21}+I_{22}, \end{align} | (6.8) |
say. For \alpha\in\mathfrak{m}_2 , then either one has Q_0^{100} < q\leqslant Q_1 or Q_0^{100} < N|q\alpha-a|\leqslant NQ_2^{-1} = Q_1 . Therefore, by Lemma 3.1, we get
\begin{equation} \sup\limits_{\alpha\in\mathfrak{m}_2}\big|f_2(\alpha)\big|\ll X_2^{\frac{4}{5}+\varepsilon}+ \frac{X_2(\log N)^c}{(q(1+N|\alpha-a/q|))^{\frac{1}{2}-\varepsilon}}\ll \frac{X_2(\log N)^c}{Q_0^{50-\varepsilon}} \ll \frac{N^{\frac{1}{2}}}{\log ^{40A}N}. \end{equation} | (6.9) |
In view of the fact that \mathfrak{m}_2\subseteq\mathcal{I} , where \mathcal{I} is defined by (3.1), Cauchy's inequality, the trivial estimate \mathcal{K}_2(\alpha)\ll\mathcal{Z}_2 , Theorem 4 of Hua (See [2], P 19), Lemma 3.4 and (6.9), we obtain
\begin{align} I_{21}\ll& \, \mathcal{Z}_2\cdot\sup\limits_{\alpha\in\mathfrak{m}_2}\big|f_2(\alpha)\big|\times \Bigg(\int_0^1\big|f_3(\alpha)\big|^8\mathrm{d}\alpha\Bigg)^{\frac{1}{2}} \Bigg(\int_{\mathcal{I}}\big|V_3(\alpha)\big|^2\mathrm{d}\alpha\Bigg)^{\frac{1}{2}} \\ \ll & \, \mathcal{Z}_2\cdot\Bigg(\frac{N^{\frac{1}{2}}}{\log^{40A}N}\Bigg)\cdot \big(N^{\frac{5}{3}}\log^c N\big)^{\frac{1}{2}}\cdot\big(N^{-\frac{1}{3}}\log^{2A}N\big)^{\frac{1}{2}} \ll \frac{\mathcal{Z}_2N^{\frac{7}{6}}}{\log^{20A}N}, \end{align} | (6.10) |
where the parameter A is chosen sufficiently large for above bound to work. Moreover, it follows from Cauchy's inequality, (6.5) and Theorem 4 of Hua (See [2], P 19) that
\begin{align} I_{22} \ll & \, N^{\frac{4}{15}+\varepsilon}\times \Bigg(\int_0^1\big|f_3(\alpha)\big|^8\mathrm{d}\alpha\Bigg)^{\frac{1}{2}} \Bigg(\int_0^1\big|f_2(\alpha)\mathcal{K}_2(\alpha)\big|^2\mathrm{d}\alpha\Bigg)^{\frac{1}{2}} \\ \ll & \, N^{\frac{4}{15}+\varepsilon}\cdot \Big(N^{\frac{5}{3}+\varepsilon}\Big)^{\frac{1}{2}}\cdot \Big(N^\varepsilon\Big(\mathcal{Z}_2N^{\frac{1}{2}}+\mathcal{Z}_2^2\Big)\Big)^{\frac{1}{2}} \\ \ll & \, N^{\frac{11}{10}+\varepsilon}\cdot\Big(\mathcal{Z}_2^{\frac{1}{2}}N^{\frac{1}{4}}+\mathcal{Z}_2\Big) \ll \mathcal{Z}_2^{\frac{1}{2}}N^{\frac{27}{20}+\varepsilon} +\mathcal{Z}_2 N^{\frac{11}{10}+\varepsilon}. \end{align} | (6.11) |
Combining (6.4), (6.8), (6.10) and (6.11), we deduce that
\begin{equation*} \frac{\mathcal{Z}_2 N^{\frac{7}{6}}}{\log^{7}N}\ll I_2 = I_{21}+I_{22}\ll \frac{\mathcal{Z}_2N^{\frac{7}{6}}}{\log^{20A}N}+\mathcal{Z}_2^{\frac{1}{2}}N^{\frac{27}{20}+\varepsilon} +\mathcal{Z}_2 N^{\frac{11}{10}+\varepsilon}, \end{equation*} |
which implies
\begin{equation} \mathcal{Z}_2\ll N^{\frac{11}{30}+\varepsilon}. \end{equation} | (6.12) |
From (6.7) and (6.12), we have
\begin{equation*} \mathcal{Z}(N)\ll \mathcal{Z}_1+\mathcal{Z}_2\ll N^{\frac{4}{9}+\varepsilon}. \end{equation*} |
This completes the proof of Proposition 2.2.
The authors would like to express the most sincere gratitude to the referee for his/her patience in refereeing this paper. This work is supported by the National Natural Science Foundation of China (Grant Nos. 11901566, 12001047, 11971476, 12071238), the Fundamental Research Funds for the Central Universities (Grant No. 2021YQLX02), the National Training Program of Innovation and Entrepreneurship for Undergraduates (Grant No. 202107010), the Undergraduate Education and Teaching Reform and Research Project for China University of Mining and Technology (Beijing) (Grant No. J210703), and the Scientific Research Funds of Beijing Information Science and Technology University (Grant No. 2025035).
The authors declare that no conflict of interest exists in this manuscript.
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