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Research article

On a Diophantine equation involving fractional powers with primes of special types

  • Received: 23 March 2024 Revised: 22 April 2024 Accepted: 25 April 2024 Published: 13 May 2024
  • MSC : 11L07, 11L20, 11N35, 11N36

  • Suppose that N is a sufficiently large real number. In this paper it is proved that for 2<c<990479, the Diophantine equation

    [pc1]+[pc2]+[pc3]+[pc4]+[pc5]=N

    is solvable in primes p1,p2,p3,p4,p5 such that each of the numbers pi+2,i=1,2,3,4,5 has at most [622739601916c] prime factors.

    Citation: Liuying Wu. On a Diophantine equation involving fractional powers with primes of special types[J]. AIMS Mathematics, 2024, 9(6): 16486-16505. doi: 10.3934/math.2024799

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  • Suppose that N is a sufficiently large real number. In this paper it is proved that for 2<c<990479, the Diophantine equation

    [pc1]+[pc2]+[pc3]+[pc4]+[pc5]=N

    is solvable in primes p1,p2,p3,p4,p5 such that each of the numbers pi+2,i=1,2,3,4,5 has at most [622739601916c] prime factors.



    For a fixed integer k1 and sufficiently large integer N, the well-known Waring-Goldbach problem is devoted to investigating the solvability of the following Diophantine equality

    N=pk1+pk2++pks (1.1)

    in prime variables p1,p2,,ps. Numerous mathematicians have derived many splendid results in this field. For instance, in 1937, Vinogradov [25] proved that such a representation of the type (1.1) exists for every sufficiently large odd integer N with k=1,s=3. Later in 1938, based upon Vinogradov's work, Hua [9] showed that (1.1) is solvable for every sufficiently large integer N satisfying that N5(mod24) with k=2,s=5.

    In 1933, Segal [21,22] studied the following anolog of the well-known Waring problem. Suppose that c>1 and cN; there exists a positive integer s=s(c) such that for every sufficiently large natural number N, the equation

    N=[mc1]+[mc2]++[mcs]

    has a solution with m1,m2,,ms integers, where [t] denotes the integral part of any tR.

    To obtain a result that is analogous to the ternary Goldbach problem, in 1995, Laporta and Tolev [13] considered the equation

    [pc1]+[pc2]+[pc3]=N, (1.2)

    where p1,p2,p3 are prime numbers, cR,c>1,NN, and [t] denotes the integral part of t. They proved that if 1<c<1716 and N is a sufficiently large integer, then the Eq (1.2) has a solution in prime numbers p1,p2,p3. Later, the upper bound of c was enlarged to

    1211,258235,137119,31132703,35813106

    by Kumchev and Nedeva [12], Zhai and Cao [27], Cai [6], Li and Zhang [15], and Baker [2], successively and respectively.

    On the other hand, as an analogue of Hua's theorem on five prime squares, Li and Zhang [14] first studied the solvability of the Diophantine equation

    [pc1]+[pc2]+[pc3]+[pc4]+[pc5]=N (1.3)

    in prime numbers p1,p2,p3,p4,p5. They proved that if 1<c<41090541999527,c2 and N is a sufficiently large integer, then the Eq (1.3) has a solution in prime numbers p1,p2,p3,p4,p5. Later this result was improved by Li [17] who enlarged the upper bound for c to 408197, and by Baker [2] who replaced 408197 by 609293.

    For any natural number r, let Pr denote an almost-prime with at most r prime factors, counted according to multiplicity. There are many papers that are devoted to the study of problems involving primes of a special type. In 1973, Chen [4] established that there exist infinitely many primes p such that p+2 has at most 2 prime factors. In 2000, Tolev [24] proved that for every sufficiently large integer N3(mod6), the equation

    p1+p2+p3=N (1.4)

    has a solution in prime numbers p1,p2,p3 such that p1+2P2,p2+2P5,p3+2P7. After that, this result was improved by some mathematicians, and the best result in this field was obtained by Matomäki and Shao [18], who showed that for every sufficiently large integer N3(mod6) the Eq (1.4) has a solution in prime numbers p1,p2,p3 such that p1+2,p2+2,p3+2P2.

    Bearing in mind the result of [18], it is natural for us to conjecture that if c is close to 1, then the Eq (1.2) is solvable in primes p1,p2,p3 such that pi+2P2. An attempt to establish this kind of the result was first made by Petrov [19], who showed that, for 1<c<1716 and every sufficiently large integer N, the Eq (1.2) is solvable in prime numbers p1,p2,p3 such that each of the numbers pi+2 has at most [951716c] prime factors, counted according to multiplicity. Recently, Li et al. [16] improved Petrov's result; they extended the range of c to 1<c<21731930 and reduced the number of prime factors of pi+2,i=1,2,3 to [1138743463860c].

    Referencing Hua's work, Tolev [24] also showed that for every sufficiently large integer N5(mod24), the equation

    p21+p22+p23+p24+p25=N (1.5)

    has a solution in prime numbers p1,p2,p3,p4,p5 such that p1+2P2,p2+2P2,p3+2P5,p4+2P5 and p5+2P8. And later in 2009, Cai and Lu [5] improved Tolev's result by showing that the Eq (1.5) has a solution in prime numbers p1,p2,p3,p4,p5 such that p1+2P2,p2+2P2,p3+2P4,p4+2P4 and p5+2P5. Motivated by Petrov [19] and Tolev [24], it is reasonable to conjecture that if N is a sufficiently large natural number and c is close to 2, then the Eq (1.3) has a solution in prime numbers p1,p2,p3,p4,p5 such that pi+2 are almost-primes of a certain fixed order.

    In this paper, we shall prove the following result.

    Theorem 1.1. Suppose that 2<c<990479 and let N be a sufficiently large natural number. Then the equation (1.3) has a solution in prime numbers p1,p2,p3,p4,p5 such that each of the numbers p1+2,p2+2,p3+2,p4+2 and p5+2 has at most [622739601916c] prime factors, counted with the multiplicity.

    Throughout this paper, the letter p, with or without subscript, always stand for prime numbers. We use ε to denote a sufficiently small positive number, and the value of ε may change from statement to statement. As usual, we use μ(n),Λ(n),φ(n) and τ(n) to denote Möbius' function, von Mangolds' function, Euler's function and the Dirichlet divisor function, respectively. We write f=O(g) or, equivalently, fg if |f|Cg for some positive number C. If we have simultaneously, that AB and BA, then we shall write AB. Moreover, we shall use (m,n) and [m,n] for the greatest common divisor and the least common multiple of the integers m and n, respectively. And we use e(α) to denote e2πiα. In addition, we define

    2<c<990479,X=(N3)1c,δ=990479c,ξ=3c252,η=4δ13,D=Xδ,z=Xη,τ=Xξc,P(z)=2<p<zp,logp=5j=1(logpj),λ±(d)Rosser's weights of orderD. (2.1)

    Lemma 2.1. Suppose that D>4 is a real number and let λ±(d) represent the Rosser functions of level D. Then we have the following properties:

    (1) For any positive integer d we have

    |λ±(d)|1,λ±(d)=0ifd>Dorμ(d)=0.

    (2) If nN then

    dnλ(d)dnμ(d)dnλ+(d). (2.2)

    (3) If zR and if

    P(z)=2<p<zp,B=2<p<z(11p),N±=dP(z)λ±(d)φ(d),s0=logDlogz, (2.3)

    then we have

    BN+B(F(s0)+O((logD)1/3)),
    BNB(f(s0)+O((logD)1/3)),

    where F(s) and f(s) denote the classical functions in the linear sieve theory that are respectively defined by

    F(s)=2eγs(1+s12log(t1)tdt),3<s5

    and

    f(s)=2eγlog(s1)s,2<s4.

    Here γ denotes the Euler constant.

    Proof. This is a special case of the work by Greaves [8].

    Lemma 2.2. Let

    Λi=d|(pi+2,P(z))μ(d),Λ±i=d|(pi+2,P(z))λ±(d),i=1,2,3,4,5.

    Then we have

    Λ1Λ2Λ3Λ4Λ5Λ1Λ+2Λ+3Λ+4Λ+5+Λ+1Λ2Λ+3Λ+4Λ+5+Λ+1Λ+2Λ3Λ+4Λ+5+Λ+1Λ+2Λ+3Λ4Λ+5+Λ+1Λ+2Λ+3Λ+4Λ54Λ+1Λ+2Λ+3Λ+4Λ+5.

    Proof. The proof is the same as in Lemma 13 of [3].

    Lemma 2.3. Suppose that f(x):[a,b]R has continuous derivatives of arbitrary order on [a,b], where 1a<b2a. Suppose further that

    |f(j)(x)|λ1a1j,j1,x[a,b].

    Then for any exponential pair (κ,λ), we have

    a<nbe(f(n))λκ1aλ+λ11.

    Proof. See (3.3.4) of [7].

    Lemma 2.4. For any complex number zn, we have

    |a<nbzn|2(1+baQ)|q|<Q(1|q|Q)a<n,n+qbzn+q¯zn,

    where Q is any positive integer.

    Proof. See Lemma 8.17 of [11].

    Lemma 2.5. Let t be a non-integer, α(0,1) and H3. Then we have

    e(α{t})=|h|Hch(α)e(ht)+O(min(1,1Ht)),

    where

    ch(α)=1e(α)2πi(h+α).

    Proof. See Lemma 12 of [1].

    Lemma 2.6. For any real number θ, we have

    min(1,1Hθ)=+h=ahe(hθ),

    where

    ahmin(log2HH,1|h|,Hh2).

    Proof. See (3) of [10].

    Lemma 2.7. Let f(x) be a real differentiable function such that f(x) is monotonic and f(x)m>0, or f(x)m<0, throughout the interval [a,b]. Then

    bae(f(x))dx1m.

    Proof. See Lemma 4.2 of [23].

    Lemma 2.8. Suppose that M>1,c>1,cZ and γ>0. Let A(M;c,γ) denote the number of solutions of the following inequalities

    |nc1+nc2nc3nc4|<γ,M<n1,n2,n3,n42M.

    Then we have

    A(M;c,γ)(γM4c+M2)Mε.

    Proof. See Theorem 2 of [20].

    The central focus of this paper is the study of the sum

    Γ=X2<p1,p2,p3,p4,p5X[pc1]+[pc2]+[pc3]+[pc4]+[pc5]=N(pi+2,P(z))=1i=1,2,3,4,5logp.

    In order to prove Theorem 1.1, we need only to show that Γ>0. By the trivial orthogonality relation given by

    10e(αh)dα={1,if h=0,0,otherwise

    we can write Γ as

    Γ=X2<p1,p2,p3,p4,p5X(pi+2,P(z))=1i=1,2,3,4,5(logp)1ττe(([pc1]+[pc2]+[pc3]+[pc4]+[pc5]N)α)dα. (3.1)

    By the definition of Λi in Lemma 2.2, we can see that

    Λi=d|(pi+2,P(z))μ(d)={1,if (pi+2,P(z))=1,0,otherwise.

    Then by Lemma 2.2 we find that

    Γ=X2<p1,p2,p3,p4,p5X(logp)Λ1Λ2Λ3Λ4Λ51ττe(([pc1]+[pc2]+[pc3]+[pc4]+[pc5]N)α)dαX2<p1,p2,p3,p4,p5X(logp)1ττe(([pc1]+[pc2]+[pc3]+[pc4]+[pc5]N)α)dα×(Λ1Λ+2Λ+3Λ+4Λ+5+Λ+1Λ2Λ+3Λ+4Λ+5+Λ+1Λ+2Λ3Λ+4Λ+5+Λ+1Λ+2Λ+3Λ4Λ+5=+Λ+1Λ+2Λ+3Λ+4Λ54Λ+1Λ+2Λ+3Λ+4Λ+5)=Γ1+Γ2+Γ3+Γ4+Γ54Γ6, (3.2)

    By the symmetric property, we have

    Γ1=Γ2=Γ3=Γ4=Γ5=X2<p1,p2,p3,p4,p5X(logp)Λ1Λ+2Λ+3Λ+4Λ+5×1ττe(([pc1]+[pc2]+[pc3]+[pc4]+[pc5]N)α)dα,Γ6=X2<p1,p2,p3,p4,p5X(logp)Λ+1Λ+2Λ+3Λ+4Λ+5×1ττe(([pc1]+[pc2]+[pc3]+[pc4]+[pc5]N)α)dα.

    Hence, by (3.1) and (3.2) we obtain

    Γ5Γ14Γ6. (3.3)

    Now define

    L±(α)=X2<pX(logp)e([pc]α)d|(p+2,P(z))λ±(d)=d|P(z)λ±(d)μX<pXd|p+2(logp)e([pc]α). (3.4)

    Consider Γ1 first. By (3.4) we can derive that

    Γ1=1ττL(α)L+(α)4e(Nα)dα=Γ11+Γ12, (3.5)

    where

    Γ11=ττL(α)L+(α)4e(Nα)dα, (3.6)
    Γ12=1ττL(α)L+(α)4e(Nα)dα. (3.7)

    Similarly, we have

    Γ6=ττL+(α)5e(Nα)dα+1ττL+(α)5e(Nα)dα=:Γ61+Γ62. (3.8)

    Now combining (3.3), (3.5) and (3.8) we get

    Γ5Γ114Γ61+(5Γ124Γ62). (3.9)

    In the following sections, we shall prove that

    5Γ114Γ61X5clog5X,Γ12,Γ62X5cε.

    In this section, we will give an asymptotic formula for the integrals Γ11 and Γ61 defined by (3.6) and (3.8), respectively. We consider the sum

    L(α)=dDλ(d)X2<pXd|p+2(logp)e([pc]α), (4.1)

    where λ(d) are real numbers satisfying

    |λ(d)|1,λ(d)=0if2|dorμ(d)=0. (4.2)

    Furthermore, we define

    I(α)=XX2e(tcα)dt. (4.3)

    Lemma 4.1. Let L(α) and I(α) be defined by (4.1) and (4.3), respectively. Suppose that ξ and δ satisfy the following conditions

    ξ+7δ<2and3ξ+6δ<2.

    Then for |α|τ, we have

    L(α)=dDλ(d)φ(d)I(α)+O(XlogAX),

    where A>0 is a sufficiently large constant.

    Proof. See Lemma 2.8 in [16].

    Lemma 4.2. Let L(α) and I(α) be defined by (4.1) and (4.3), respectively. Then we have

    (i)|α|τ|I(α)|4dαX4clog4X,(ii)|α|τ|L(α)|4dαX4clog10X,(iii)10|L(α)|4dαX2+3ε.

    Proof. By using the trivial estimate |I(α)|X and L(x)Xlog2X, (ⅰ) and (ⅱ) follow from Lemma 2.9 in [16]. For (ⅲ), we have

    10|L(α)|4dα=diDi=1,2,3,4λ(d1)λ(d2)λ(d3)λ(d4)X2<p1,p2,p3,p4Xdi|pi+2,i=1,2,3,44i=1(logpi)×10e(([pc1]+[pc2][pc3][pc4])α)dα=X2<p1,p2,p3,p4X[pc1]+[pc2]=[pc3]+[pc4]4i=1(logpi)diD,di|pi+2i=1,2,3,4λ(d1)λ(d2)λ(d3)λ(d4)(log4X)X2<n1,n2,n3,n4X[nc1]+[nc2]=[nc3]+[nc4]τ(n1+2)τ(n2+2)τ(n3+2)τ(n4+2)Xε(log4X)X2<n1,n2,n3,n4X[nc1]+[nc2]=[nc3]+[nc4]1X2εX2<n1,n2,n3,n4X|nc1+nc2nc3nc4|<41X3ε(4X4c+X2)X2+3ε,

    where Lemma 2.8 is applied in the last step.

    Let

    M±(α)=dDλ±(d)φ(d)I(α)=N±I(α).

    Then we can easily get the elementary estimate

    M±(α)|I(α)|logX. (4.4)

    By using Lemma 4.2 and (4.4) we find that

    L(α)L+(α)4M(α)M+(α)4=(L(α)M(α))L+(α)4+(L+(α)M+(α))M(α)L+(α)3+(L+(α)M+(α))M(α)M+(α)L+(α)2+(L+(α)M+(α))M(α)M+(α)2L+(α)+(L+(α)M+(α))M(α)M+(α)3XlogAX(|L+(α)|4+|L+(α)|3|I(α)|logX+|L+(α)|2|I(α)|2log2X=+|L+(α)||I(α)|3log3X+|I(α)|4log4X). (4.5)

    Let

    Jτ=ττM(α)M+(α)4e(Nα)dα. (4.6)

    Then we can derive from Lemma 4.2, (3.6), (4.5) and (4.6) that

    Γ11Jτ|α|τ|L(α)L+(α)4M(α)M+(α)4|dαXlogA4X(|α|τ|L+(α)|4dα+|α|τ|I(α)|4dα)X5clogA14X. (4.7)

    Define

    J=+I(α)5e(Nα)dα. (4.8)

    Then by the argument used in Lemma 2.13 of [16], we have

    JX5c. (4.9)

    With the help of Lemma 2.7 we can get that I(α)|α|1X1c. Hence, from (4.6) and (4.8) we find that

    |N(N+)4JJτ|(log5X)|α|>τ|I(α)|5dα(log5X)|α|>τ|α|5X55cdαX5c4ξlog5XX5cε. (4.10)

    Now combining (4.7) and (4.10) we obtain

    Γ11=N(N+)4J+O(X5clogA14X). (4.11)

    Similarly, we can prove that

    Γ61=(N+)5J+O(X5clogA14X). (4.12)

    In this section we shall consider the upper bound for the integrals Γ12 and Γ62 defined by (3.7) and (3.8), respectively. Define

    T(α,X)=dDX2<nXd|n+2e([nc]α).

    Lemma 5.1. For α(0,1), we have

    T(α,X)X2c+1320+εD720+logXαXc1.

    Proof. The proof is exactly the same as that of Lemma 2.7 in [16], where the exponential pair (κ,λ)=(19,1318) is used. One can see [16] for details.

    Lemma 5.2. Let f(n) be a complex valued function defined on n(X2,X]. Then we have

    X2<nXΛ(n)f(n)=S1S2S3,

    where

    S1=kX13μ(k)X2k<Xk(log)f(k),S2=kX23c(k)X2k<Xkf(k),S3=X13<kX23a(k)X2k<XkΛ()f(k),

    and where a(k),c(k) are real numbers satisfying

    |a(k)|τ(k),|c(k)|logk.

    Proof. The proof can be found on page 112 of [26].

    Lemma 5.3. Suppose that 2<c<990479. Let λ(d) be real numbers that satisfy (4.2) and L(α) be defined by (4.1). Then we have

    supα(τ,1τ)|L(α)|X32c4ε.

    Proof. From (4.1), it is easy to see that

    L(α)=L1(α)+O(X12+ε), (5.1)

    where

    L1(α)=dDλ(d)X2<nXd|n+2Λ(n)e([nc]α).

    Hence by (5.1) we need only to show that the estimation

    supα(τ,1τ)|L1(α)|X32c4ε (5.2)

    holds for 2<c<990479. Let H=X8479. Then, by using Lemma 2.5, we get

    L1(α)=|h|Hch(α)X2<nXΛ(n)dDd|n+2λ(d)e((h+α)nc)+O((logX)X2<nXmin(1,1Hnc)). (5.3)

    By Lemmas 2.3 and 2.6 with the exponential pair (κ,λ)=(16,23), we obtain

    (logX)X2<nXmin(1,1Hnc)=(logX)X2<nX+k=ake(knc)(logX)+k=|ak||X2<nXe(knc)|(logX)(Xlog2HH+1kH1k((kXc)16X12+XkXc)=+k>HHk2((kXc)16X12+XkXc))(logX)(XH1+H16Xc6+12+X1c)X32c4ε. (5.4)

    Next we consider the first term on the right-hand side of (5.3). We write it in the following form

    S(α)=|h|Hch(α)X2<nXΛ(n)f(n), (5.5)

    where

    f(n)=dDd|n+2λ(d)e((h+α)nc).

    By applying Lemma 5.2 we find that

    S(α)=S1S2S3, (5.6)

    where

    S1=|h|Hch(α)kX13μ(k)X2k<Xk(log)f(k), (5.7)
    S2=|h|Hch(α)kX23c(k)X2k<Xkf(k), (5.8)
    S3=|h|Hch(α)X13<kX23a(k)X2k<XkΛ()f(k), (5.9)

    and |a(k)|τ(k),|c(k)|logk. Clearly, by (5.8) we can write S2 as

    S2=S21+S22, (5.10)

    where

    S21=|h|Hch(α)kX13c(k)X2k<Xkf(k),S22=|h|Hch(α)X13<kX23c(k)X2k<Xkf(k). (5.11)

    Therefore, by (5.6) and (5.10) we have

    S(α)|S1|+|S21|+|S22|+|S3|. (5.12)

    We consider the sum S21 defined by (5.11) first. We change the order of the summation to write it in the following form

    S21=dDλ(d)|h|Hch(α)kX13c(k)X2k<Xkd|k+2e((h+α)(k)c).

    Since λ(d)=0 for 2|d, from the condition d|k+2 we have that (k,d)=1. Hence there exists an integer 0 such that k+20(modd) is equivalent to 0(modd), which means that =0+md for some integer m. Therefore, we get

    S21=dDλ(d)|h|Hch(α)kX13(k,d)=1c(k)X2kd0d<mXkd0de((h+α)kc(0+md)c). (5.13)

    By using Lemma 2.3 with the exponential pair (κ,λ)=A2BA2B(0,1)=(120,3340) we find that the sum over m in (5.13) is given by

    (|h+α|kdXc1)120(Xkd)3340+(|h+α|kdXc1)1|h+α|120Xc20+3140k3140d3140+|h+α|1k1d1X1c. (5.14)

    Then from (4.2), (5.13) and (5.14) we can obtain

    S21Xε(Xc20+1720D940H120+X1c). (5.15)

    For the sum S1 given by (5.7), we can apply partial summation to get rid of the log factor and then proceed as in the same process for S21 to get

    S1Xε(Xc20+1720D940H120+X1c). (5.16)

    Now we consider the sum S3. By a splitting argument, we can decompose S3 into O(logX) sums of the following form

    W(K)=|h|Hch(α)K<k<K1a(k)X2k<XkΛ()dDd|k+2λ(d)e((h+α)(k)c), (5.17)

    where

    K12K,X13K<K1X23. (5.18)

    We assume that KX12 first. It follows from (5.17), (5.18) and Cauchy's inequality that

    |W(K)|2XεKmaxγ(τ,H+1)K<k<K1|X2k<XkΛ()dDd|k+2λ(d)e(γ(k)c)|2. (5.19)

    Suppose that Q is an integer which satisfies

    1QXK. (5.20)

    For the inner sum over in (5.19), by applying Lemma 2.4 we can derive that

    |W(K)|2X1+εQmaxγ(τ,H+1)K<k<K1|q|Q(1|q|Q)X2k<,+qXkΛ()×d1Dd1|k+2λ(d1)e(γ(k)c)Λ(+q)d2Dd2|k(+q)+2λ(d2)e(γ(k(+q))c)X1+εQmaxγ(τ,H+1)d1Dd2Dλ(d1)λ(d2)|q|Q(1|q|Q)×X2K1<,+qXKΛ()Λ(+q)S, (5.21)

    where

    S=˜K<k~K1d1|k+2d2|k(+q)+2e(γkc((+q)cc))

    and

    ˜K=max(K,X2,X2(+q)),~K1=min(K1,X,X+q).

    Since λ(d)=0 for 2|d, we can assume that (d1d2,2)=1. Then it follows from d1|k+2 and d2|k(+q)+2 that (d1,)=(d2,+q)=1. Hence there exists an integer k0 that is dependent on ,h,d1,d2 such that the pair of conditions k+20(modd1) and k(+q)+20(modd2) is equivalent to the congruence kk0(mod[d1,d2]). Thus, we have

    S=˜Kk0[d1,d2]<m~K1k0[d1,d2]e(F(m)), (5.22)

    where

    F(m)=γ(k0+m[d1,d2])c((+q)cc). (5.23)

    For q=0, by the trivial estimate we have

    SK[d1,d2]. (5.24)

    For the case q0, by (5.23) we get

    |F(j)(m)||γ||q|c1[d1,d2]Kc1(K[d1,d2])1j,j1.

    We apply Lemma 2.3 with the following exponential pair

    (κ,λ)=A(1384+ε,5584+ε)=(13194+ε,7697+ε)

    to derive that

    S(|γ||q|c1[d1,d2]Kc1)13194+ε(K[d1,d2])7697+ε+(|γ||q|c1[d1,d2]Kc1)1Xε(|γ|13194|q|1319413194(c1)[d1,d2]139194K13194c+139194+|γ|1|q|11c[d1,d2]1K1c). (5.25)

    Note that

    d1Dd2D1[d1,d2]139194=rDd1Dd2Dr=(d1,d2)(rd1d2)139194rDk1Drk2Dr(1rk1k2)139194rDr139194(Dr)5597D5597,

    and

    d1Dd2D1[d1,d2](logD)3.

    Then we use the above two estimates, (5.21), (5.24) and (5.25) to get

    |W(K)|2X1+εQmaxγ(τ,H+1)X2K1<XKΛ()2d1Dd2DK[d1,d2]+X1+2εQmaxγ(τ,H+1)d1Dd2D0<|q|<QX2K1<,+qXKΛ()Λ(+q)×(|γ|13194|q|1319413194(c1)[d1,d2]139194K13194c+139194+|γ|1|q|11c[d1,d2]1K1c)X2+εQ1+X1+2εQ|γ0|13194K13194c+139194(d1Dd2D1[d1,d2]139194)(0<|q|<Q|q|13194)×(X2K1<XK13194(c1))+X1+2εQ|γ0|1K1c(d1Dd2D1[d1,d2])×(0<|q|<Q|q|1)(X2K1<XK1c)Xε(X2Q1+X13c+375194Q13194D5597|γ0|13194K2197+X|γ0|1Q1K1c) (5.26)

    for some γ0[τ,H+1]. We choose

    Q0=X13207(1c)D110207|γ0|13207K1469,Q=[min(Q0,XK1)].

    Then, it is easy to check that

    Q1Q10+KX1. (5.27)

    Substituting (5.27) into (5.26), we obtain

    |W(K)|2Xε(X2(Q10+KX1)+X13c+375194Q13194D5597|γ0|13194K2197=+X|γ0|1(Q10+KX1)K1c)Xε(X13c+380207D110207|γ0|13207+X53+X553181c414D110207|γ0|194207+X1c2|γ0|1),

    which implies that

    |W(K)|Xε(X13c+380414D55207|γ0|13414+X56+X553181c828D55207|γ0|97207+X12c4|γ0|12). (5.28)

    When K<X12, we can represent W(K) as follows:

    W(K)=|h|Hch(α)X2K1<XKΛ()max(K,X2)<kmin(K1,X)a(k)×dDd|k+2λ(d)e((h+α)(k)c).

    Now we have that XKX12, then, we may proceed as in (5.19)–(5.28) but with roles of k and reversed. Thus we can again derive the estimate (5.28). Consequently, we obtain

    S3Xε(X13c+380414D55207|γ0|13414+X56+X553181c828D55207|γ0|97207+X12c4|γ0|12). (5.29)

    To bound S22, we use the same methodology as for S3 to derive that

    S22Xε(X13c+380414D55207|γ0|13414+X56+X553181c828D55207|γ0|97207+X12c4|γ0|12). (5.30)

    Now combining (5.12), (5.15), (5.16), (5.29) and (5.30) and from the fact that γ0[τ,H+1], we find that

    S(α)Xε(Xc20+1720D940H120+X1c+X13c+380414D55207|γ0|13414+X56=+X553181c828D55207|γ0|97207+X12c4|γ0|12)Xε(Xc20+81519580+9δ40+X1c+X13c414+1011811017+55δ207+X56=+X553828+c4+55δ20797ξ207+X12+c4ξ2).

    Therefore, from condition (2.1) we conclude that if 2<c<990479 then

    supα(τ,1τ)|S(α)|X32c4ε. (5.31)

    With the help of (5.3)–(5.5) and (5.31), we finally obtain that

    supα(τ,1τ)|L1(α)|X32c4ε

    holds for 2<c<990479, and the proof of Lemma 5.3 is completed.

    Lemma 5.4. Suppose that 2<c<990479. Then we have

    1ττ|L(α)|5dαX5cε.

    Proof. Let G(α)=¯L(α)|L(α)|3. We have

    |1ττ|L(α)|5dα|=|dDλ(d)X2<pXd|p+2(logp)1ττe([pc]α)G(α)dα|(logX)dDX2<pXd|p+2|1ττe([pc]α)G(α)dα|(logX)dDX2<nXd|n+2|1ττe([nc]α)G(α)dα|. (5.32)

    From (5.32) and Cauchy's inequality, we get

    |1ττ|L(α)|5dα|2X(logX)3dDX2<nXd|n+2|1ττe([nc]α)G(α)dα|2=X(logX)31ττ¯G(β)dβ1ττT(αβ,X)G(α)dαX(logX)31ττ|G(β)|dβ1ττ|T(αβ,X)G(α)|dα. (5.33)

    Now

    1ττ|T(αβ,X)G(α)|dατ<α<1τ|αβ|Xc|T(αβ,X)G(α)|dα+τ<α<1τ|αβ|>Xc|T(αβ,X)G(α)|dα. (5.34)

    By the trivial bound T(α,X)XlogX and Lemma 5.3, we have

    τ<α<1τ|αβ|Xc|T(αβ,X)G(α)|dαX(logX)supα(τ,1τ)|G(α)||αβ|XcdαX1c(logX)supα(τ,1τ)|L(α)|4X72cε. (5.35)

    From Lemmas 5.1 and 5.3, we obtain

    τ<α<1τ|αβ|>Xc|T(αβ,X)G(α)|dατ<α<1τ|αβ|>Xc|L(α)|4(X2c+1320+εD720+X1clogX|αβ|)dαX2c+1320+εD72010|L(α)|4dα+X1c(logX)supα(τ,1τ)|L(α)|4|αβ|>Xc1|αβ|dαX2c+5320+7δ20+ε+X72cεX72cε, (5.36)

    where (ⅲ) of Lemma 4.2 is used. It follows from (5.34)–(5.36) that

    1ττ|T(αβ,X)G(α)|dαX72cε. (5.37)

    Combining (5.33), (5.37) and (ⅲ) of Lemma 4.3, we get

    |1ττ|L(α)|5dα|2X(logX)3X72cε10|L(α)|4dαX102cε2. (5.38)

    Now Lemma 5.4 follows from (5.38).

    We are now in a position to estimate Γ12 and Γ62. By Hölder's inequality and Lemma 5.4 we find that

    |Γ12|1ττ|L(α)||L+(α)|4dα(1ττ|L(α)|5dα)15(1ττ|L+(α)|5dα)45X5cε. (5.39)

    Similarly, for Γ62 we have

    |Γ62|1ττ|L+(α)|5dαX5cε. (5.40)

    Proposition 6.1. We have

    5Γ114Γ61X5clog5X.

    Proof. It follows from (4.11), (4.12) and Lemma 2.1(3) that

    5Γ114Γ61=(5N4N+)(N+)4J+O(X5clogA14X)(5f(logDlogz)4F(logDlogz))(1+O(log1/3D))B5J+O(X5clogA14X)=(5f(134)4F(134))B5J+O(X5cε)=40eγ13(log944545942log(t1)tdt)B5J+O(X5cε)0.001B5J+O(X5cε)X5clog5X,

    where the following trivial estimate is used:

    B1logX.

    Now according to (3.9), (5.39), (5.40) and Proposition 6.1, we obtain

    Γ(5Γ114Γ61)+O(|Γ12|+|Γ62|)X5clog5X,

    which implies that Γ>0 for a sufficiently large natural number N. Then, (1.3) would have a solution in primes p1,p2,p3,p4,p5 satisfying

    (p1+2,P(z))=(p2+2,P(z))=(p3+2,P(z))=(p4+2,P(z))=(p5+2,P(z))=1. (6.1)

    Suppose that pi+2 has l prime factors, counted with multiplicity. From (6.1) and the condition X2<piX we see that

    X+2pi+2zl=Xηl.

    Then, lη1. This means that pj+2 has at most [622739601916c] prime factors counted with multiplicity. Now Theorem 1.1 is proved.

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

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

    The author declares no conflict of interest.



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