Exceptional set in Waring–Goldbach problem for sums of one square and ﬁve cubes

: Let N be a su ﬃ ciently large integer. In this paper, it is proved that, with at most O (cid:0) N 4 / 9 + ε (cid:1) exceptions, all even positive integers up to N can be represented in the form p 21 + p 32 + p 33 + p 34 + p 35 + p 36 , where p 1 , p 2 , p 3 , p 4 , p 5 , p 6 are prime numbers.


Introduction and main result
Waring's problem of mixed powers concerns the representation of sufficiently large integer n in the form n = x k 1 1 + x k 2 2 + · · · + x k s s . 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 = x 2 + y 3 1 + y 3 2 · · · + y 3 s . (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 E 1 (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 E 1 (N) N ε .
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 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 is solvable with x being an almost-prime P 36 and the p j ( 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 almostprime P 6 and the p j ( 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., where U = N 1−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 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 Then, for any ε > 0, we have E(N) N 4 9 +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 χ mod q to denote a Dirichlet character modulo q, and χ 0 mod q the principal character.
. N is a sufficiently large integer and n ∈ [N/2, N], and thus log N log n. The letter c, with or without subscripts or superscripts, always denote a positive constant, which may not be the same at different occurrences.
2. Outline of the proof of Theorem 1.1 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 By Dirichlet's lemma on rational approximation (for instance, see Lemma 12 on page 104 of Pan and Pan [4]), each α ∈ [−1/Q 2 , 1 − 1/Q 2 ] can be written as the form for some integers a, q with 1 a q Q 2 and (a, q) = 1. Define M(q, a), M 0 (q, a), Then we obtain the Farey dissection In order to prove Theroem 1.1, we need the two following propositions: where S(n) is the singular series defined in (4.1), which is absolutely convergent and satisfies for any integer n satisfying n ≡ 0 (mod 2) and some fixed constant c * > 0; while J(n) is defined by (4.5) and satisfies J(n) n 7/6 log 6 n .
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.
Then we have 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.
This completes the proof of Theorem 1.1.
Then we have where δ k = 1 2 + log k log 2 and c is a constant. Proof. See Theorem 1.1 of Ren [5].
Suppose that α is a real number, and that there exist a ∈ Z and q ∈ N with (a, q) = 1, 1 q Q and |qα − a| Q −1 .
For 1 a q with (a, q) = 1, set For α ∈ m 2 , by Lemma 3.1, we have say. Then we obtain the following lemma.
Lemma 3.4. We have This completes the proof of Lemma 3.4.

Proof of Proposition 2.1
In this section, we shall concentrate on establishing Proposition 2.1. We first introduce some notations. For a Dirichlet character χ mod q and k = 2, 3, we define where χ 0 is the principal character modulo q. Let  with β k = (log k)/ log 2.
Proof. See the Problem 14 of Chapter VI of Vinogradov [11]. Proof. By Lemma 4.1, we have Therefore, the left-hand side of (4.2) is This completes the proof of Lemma 4.2.
In order to treat the integral on the major arcs, we write f k (α) as follows: For the innermost sum on the right-hand side of the above equation, by Siegel-Walfisz theorem, we have say. Therefore, we have and thus Then we derive that Noting that hence the innermost integral in (4.3) can be written as we know that if we extend the interval of the integral in (4.4) to [−1/2, 1/2], then the resulting error is

The singular series
In this section, we shall concentrate on investigating the properties of the singular series which appear in Proposition 2.1. Let p be a prime and p α k. For (a, p) Proof. See Lemma 8.3 of Hua [2].
For k 1, we define Proof. We denote by S the left-hand side of (5.1). By Lemma 5.2, we have If |A k | = 0 for some k ∈ {2, 3}, then S = 0. If this is not the case, then From Lemma 5.2, the sextuple outer sums have not more than (2, p − 1) − 1 × (3, p − 1) − 1 5 1 × 2 5 = 32 terms. In each of these terms, we have Since in any one of these terms is a Dirichlet character χ (mod p), the inner sum is From the fact that τ(χ 0 ) = −1 for principal character χ 0 mod p, we have By the above arguments, we obtain This completes the proof of Lemma 5.3.
Lemma 5.4. Let L(p, n) denote the number of solutions of the congruence Then, for n ≡ 0 (mod 2), we have L(p, n) > 0.
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 1 q 2 with (q 1 , q 2 ) = 1. Then we have For (q 1 , q 2 ) = 1 and k ∈ {2, 3}, there holds Hence, by (5.6), we obtain q>Z |A(n, q)| q>Z q − 5 2 +ε (n, q) holds. For convenience, we use Z j to denote the cardinality of Z j (N) for abbreviation. Also, we define the complex number ξ j (n) by taking ξ j (n) = 0 for n Z j (N), and when n ∈ Z j (N) by means of the equation Plainly, one has |ξ j (n)| = 1 whenever ξ j (n) is nonzero. Therefore, we obtain n∈Z j (N) where the exponential sum K j (α) is defined by ξ j (n)e(−nα).