# A Note on Cube-Full Numbers in Arithmetic Progression

Authors: Mingxuan Zhong; Yuankui Ma
Journal: Journal of Mathematics
 (2021)

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## Abstract

We obtain an asymptotic formula for the cube-full numbers in an arithmetic progressionn≡lmodq, where q,l=1. By extending the construction derived from Dirichlet’s hyperbola method and relying on Kloosterman-type exponential sum method, we improve the very recent error term with x118/4029<q.


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## Body

## 1. Introduction and Main Results

Letk>1 be a fixed integer and n be a positive integer. We call n a powerful number (or k-full number) if n=1 or for a prime p dividing n, pk also divides n. Let Pk denote the set of powerful numbers. Suppose k=2,3, and this defines square-full numbers and cube-full numbers, respectively. Erdo..s and Szerkeres [1] first introduced powerful numbers and gave(1)∑n≤x,n∈Pk1=∑m=k2k−1ck,mx1/m+Δkx,where ck,m are effective constants and Δkx≪x1/k+1. From then on, many authors have studied the powerful numbers and got a lot of relevant conclusions (see [2–18] and references therein).In 2013, Liu and Zhang [19] investigated the distribution of square-full numbers in arithmetic progressions and got an asymptotic formula(2)∑n≤x,n∈P2n≡lmodq1=αl,qx1/2+Oq49/141+εx19/47+ε,under the condition of q,l=1. By utilizing the method of exponent pairs, Srichan [20] then obtained(3)∑n≤x,n∈P2n≡lmodq1=∑k=23βkℓ,qx1/k+Oq3/2+εx1/6,∑n≤x,n∈P3n≡lmodq1=∑k=35γkℓ,qx1/k+Oq127/92+εx7/46,where the error terms had been corrected by Watt [MR3265055].Recently, Chan [21] got a new asymptotic formula(4)∑n≤x,n∈P2n≡lmodq1=∑k=23δkℓ,qx1/k+Oεx1/6q1/12+x1/5q1/5qε,which improved his own result with Tsang [22]. As a critical step, he [21] mainly dealt with a sum in the form of by following closely Montgomery and Vaughan’s construction [23]. It is somewhat similar to Dirichlet’s hyperbola method shown in Figure 1.(5)∑a2b3≤xa2b3≡lmodq1.Figure 1
Dirichlet’s hyperbola method.Actually, they divided the above sum into four parts as shown in Figure2 and then discussed them separately.Figure 2
Generalized Dirichlet's hyperbola method (proof of Theorem 3 in [7]).Motivated by this idea, we turn to discuss the following sum with three parameters:(6)∑a3b4c5≤xa3b4c5≡lmodq1.By extending the construction from Montgomery and Vaughan [23], the summation (5) is divided into eight parts as shown in Figure 3.Figure 3
Generalized Dirichlet's hyperbola method in three dimensional space (proof of Theorem 2 below).Then, relying again on Kloosterman-type exponential sum method, an asymptotic formula of (5) is obtained. Finally, an asymptotic formula of cube-full numbers in an arithmetic progression is derived.Before we formulate our result we need to give some definitions that will be used below. For modulusq, we define(7)N3n;q≔#xmodq:x3≡nmodq,x,q=1,N4n;q≔#xmodq:x4≡nmodq,x,q=1,N5n;q≔#xmodq:x5≡nmodq,x,q=1.It is clear thatN3na3;q=N3n;q, N4na4;q=N4n;q, and N5na5;q=N5n;q for any a,q=1, and N3n;q=N4n;q=N5n;q=0 if n,q>1.Theorem 1.
Forl,q=1,(8)∑n≤x,n∈P3n≡lmodq1=Aqlx1/3q+Bqlx1/4q+Cqlx1/5q+Ox49/360q1/2+x71/357q3/14qε,where(9)Aql=Aq′l∑d8e9f10≥1d,e,f,q=1μdμfμed8/3e3f10/3∑g|edg,q=1μgg5/3,Bql=Bq′l∑d8e9f10≥1d,e,f,q=1μdμfμed2e9/4f5/2∑g|edg,q=1μgg5/4,Bql=Cq′l∑d8e9f10≥1d,e,f,q=1μdμfμed8/5e9/5f2∑g|edg,q=1μgg,in which Aq′l, Bq′l, and Cq′l are defined below.
Note that compared with the result in [20], we improve the error term when x118/4029<q.
The key in our proof of Theorem1 is the following.Theorem 2.
Letl,q=1, and η,λ be two parameters such that(10)110<η<4651,551<λ<13132,or110<η<67660,13132<λ<110.
Then, we have(11)∑a3b4c5≤xa3b4c5≡lmodq1=Aq′lx1/3q+Bq′lx1/4q+Cq′lx1/5q+Ox1/3−4/3ηq1/2−ε+maxxη,q1/2qεxλ,where(12)Aq′l=∑b=1∞∑c=1∞N3lb2c;qb4/3c5/3,Bq′l=ϕqq5−54∫1∞t−tt9/4dt−3−34∫1∞t−tt7/4dt−∑c=1∞N4lc3;q−ϕq/qc5/4+34∫1∞∑c=1∞∑a≤μN4lac3;q−ϕq/qμc5/4μ−7/4dμ,Cq′l=−4−45∫1∞t−tt9/5dt−32−35∫1∞t−tt8/5dtϕqq−45∫1∞∑b≤μN5lb;q−ϕq/qμμ8/5dμ+N5l;q−ϕqq−35∫1∞∑a≤μN5la2;q−ϕq/qμ8/5dμ+1225∫1∞∫1∞∑b≤λ∑a≤μN5la2b;q−ϕq/qλμμ8/5λ9/5dλdμ.


## 2. Some Lemmas

Before we start the proof, let us give a few lemmas which are needed later.Lemma 1.
Forl,q=1,(13)∑a≤uNila;q=ϕqqu+Oεq1/2+ε,i=3,4,5,∑a≤uNila2;q=ϕqqu+Oεq1/2+ε,i=3,5,∑a≤uN4la3;q=ϕqqu+Oεq1/2+ε.Proof.
The first result can be found in Lemma3 of [21]. Note that(14)∑a≤qNila;q=ϕq,i=3,4,5.
The second and third one can be proved in the same way. The proofs of the last three are slightly different. For example, by orthogonal property of additive characters, we have(15)∑a≤uN3la2;q=1q∑a≤u∑n=1q∑x=1q′enx3−lna2q,=1q∑a≤u∑n=1qe−lna2q∑y=1qenyq∑χ∈G3χy=1q∑χ∈G3∑a≤u∑y=1qχy∑n=1qeny−lna2q=∑χ∈G3∑a≤u∑y≤qy≡la2modqχy=∑a≤u∑y≤qy≡la2modqy,q=11+∑χ∈G3χ≠χ0∑a≤u∑y≤qy≡la2modqχy=∑a≤ua,q=11+O∑χ∈G3χ≠χ0∑a≤uχ2a=∑d|qμd∑a≤ud|a1+OG3q1/2+ε=ϕqqu+Oq1/2+ε,where G3 is the set of all characters χmodq such that χ3=χ0, the principal character.Lemma 2.
Forq≥1 and l,q=1,(16)∑n=1q−1n,q∑b=B+1B+L∑c=C+1C+Me−nlb2cq≪q⋅dqlogq.Proof.
Letx be the distance from x to the nearest integer; then, we have(17)∑n=1q−1n,q∑b=B+1B+L∑c=C+1C+Me−nlb2cq=∑d|qd∑n′=1q/d−1∑b=B+1B+L∑c=C+1C+Me−n′lb2cq/d,≪∑d|qd∑n′=1q/d−1∑b=B+1B+LminL,1n′b2/q/d≪∑d|qd∑b=B+1B+L∑n′≤q/Ldb2L+∑q/Ldb2<n′≤q/dq/dn′b2≪∑d|qdqd+logq⋅qd,≪q⋅dqlogq.Lemma 3.
Forl,q=1,(18)∑b≤xμb,q=1∑c≤xνc,q=1∑a≤xλa3b4c5≡lmodq1=ϕqq2xλ+μ+ν+Oεxλq1/2−ε+Oεxμ+νqε.Proof.
Using the trivial estimation of the innermost sum, we have(19)∑b≤xμb,q=1∑c≤xνc,q=1∑a≤xλa3b4c5≡lmodq1=∑b≤xμ∑c≤xνN3lb2c;qxλq+O1,=xλq∑b≤xμ∑c≤xνN3lb2c;q+Oε∑b≤xμ∑c≤xνN3lb2c;q.
Then, by orthogonal property of additive characters, we obtain(20)∑b≤xμ∑c≤xνN3lb2c;q=∑b≤xμ∑c≤xν1q∑n=1q∑x=1q′enx3−lb2cq=ϕqqxμ+ν+O1+1q∑n=1q−1∑b≤xμ∑c≤xνe−nlb2cq∑x=1q′enx3q=ϕqqxμ+ν+O1+1q∑n=1q−1∑b≤xμ∑c≤xνe−nlb2cq∑ymodqenyq∑χ∈G3χy.
Interchanging the order of summations and combining Lemma2 and Eq. (12.48) on page 324 of [24], we have(21)∑b≤xμ∑c≤xνN3lb2c;q=ϕqqxμ+ν+O1+O1q∑χ∈G3∑n=1q−1∑b≤xμ∑c≤xνe−nlb2cqn,qdn,qq=ϕqqxμ+ν+O1+OdqqG3∑n=1q−1∑b≤xμ∑c≤xνe−nlb2cqn,q=ϕqqxμ+ν+Oεq1/2+ε.
Finally, we get(22)∑b≤xμb,q=1∑c≤xνc,q=1∑a≤xλa3b4c5≡lmodq1=ϕqq2xλ+μ+ν+Oεxλq1/2−ε+Oεxμ+νqε.
By definition ofN3n;q, N3n;q≪εqε (see Lemma 2.2 in [25]).Lemma 4.
If we define(23)Sa,b;q=∑n=1q′ean3+bn¯4q,then we have Sa,b;q≪a,b,q1/2q1/2+ε.Proof.
Following much the same way as Lemma4 of [21], we first suppose q=rs with r,s=1. By the “reciprocity” formula ss¯+rr¯≡1modq, where ss¯≡1modr and rr¯≡1mods, and the additive multiplicity of exponential function(24)ean3+bn¯4q=eas¯n3+bs¯n¯4rear¯n3+br¯n¯4s,with n=sx+ry, 1≤x≤r, 1≤y≤s, x,r=1, and y,s=1, it follows that(25)Sa,b;q=∑x=1r′∑y=1s′eas¯x3+bs¯x¯4rear¯y3+br¯y¯4s=Sas¯,bs¯;rSar¯,br¯;s.
Now we just need to discuss the argument in the following cases:(I)
Prime moduliq=p case. Now Theorem 2 obtained by Moreno and Moreno [26], which is a special form of the Bombieri-Weil bound [27], implies(26)Sa,b;p≤10a,b,p1/2p1/2,provided thatax7+b/x4 is not the form of hpx−hx with hx∈F¯px, where F¯p is the algebraic closure of Fp. Let(27)ax7+bx4=fpxgpx−fxgx,withfx,gx∈F¯px and fx,gx=1. Then, we have gpx|x4, derived from(28)gpxax7+b=x4fpx−gp−1xfx.This is impossible ifp>4, by comparing the degrees of both sides of the above. If p≤4, the validity of (26) can be easily checked.(II)
Prime power moduliq=pβ case with β>1. Obviously, we only need to consider it with the assumption a,b,p=1. Following the proofs of Lemma 12.2 and 12.3 in [24] with the equation of Sa,b;q, we obtain(29)Sa,b;p2α=pα∑y=1g′y≡0modpαpα′egyp2α,(30)Sa,b;p2α+1=pα∑y=1g′y≡0modpαpα′egyp2α+1Gpy,where(31)gy=ay7+by4,Gpy=∑z=1pehyz2+g′yp−αzp,withhy=g″y/2.
Note thatg′y=3ay10−4by3/y8 and hy=3ay11+10by4/y10. Now we concentrate on the number of solutions of congruence equation g′y≡0modpα with y,p=1.
β=2α with α≥1. Then, the congruence equation is(32)3ay7−4b≡0modpα.
Ifb,p=p, then a,p=1. Relying on the properties of indices, we deduce that (32) has no solution when p>3 and one solution when p=3. Next, we assume b,p=1. If p=2 and 4‖a, then (32) has at most seven solutions. And if p>3 and a,p=1, (32) also has at most seven solutions. Then, we have(33)Sa,b;p2α≤7pα,ifa,b,p=1,where β=2α+1 with α⩾1. Firstly, in the same way, if b,p=p, by the analysis in the case β=2α, the sum in (30) is empty unless p=3 in which case one has Sa,b;32α+1≤3α+1. Now suppose b,p=1; according to Chapter 3 in [24], we know if p∤2hy, then we have Gpy≤p1/2. Therefore, if p≠2,7 (for otherwise p|3ay7+10b and p|3ay7−4b imply p|14b, a contradiction), we have Gpy≤p1/2; hence, Sa,b;p2α+1≤7pα+1/2 as there are at most seven solutions to (32). If p=2 or 7, Gpy≤7, and hence Sa,b;p2α+1≤49pα.
In any case, we have(34)Sa,b;p2α+1≤49pα+1/2,ifa,b,p=1.
Combining (25), (26), (33), and (34), we finally obtain(35)Sa,b;q≤49ωqa,b,q1/2q1/2≪a,b,q1/2q1/2+ε,which completes the proof of Lemma 4.
By applying Lemma4, we have the following.Lemma 5.
Forl,q=1,(36)NA,BK,L≔#a,b:a3b4≡lmodq,A<a≤A+K,B<b≤B+L,=ϕqKLq2+OK+Lq1/2−ε+q1/2+ε.Proof.
Note that ifa3b4≡lmodq, then a4b4≡lamodq and b≡lab¯3modq. So,(37)NA,BK,L=#a,b,w:a≡l¯w4modq,b≡lw¯3modq,A<a≤A+K,B<b≤B+L,1≤w≤q,w,q=1.
The remaining part of the proof is similar to Theorem 4 in [21].


## 3. Proof of Theorem2

Consider three positive parametersλ, μ, and ν. By extending the construction from Montgomery and Vaughan [23] as shown in Figure 3, we have(38)∑a3b4c5≤xa3b4c5≡lmodq1=∑a≤xλa,q=1∑b4c5≤x/a3b4c5≡la3¯modq1+∑b≤xμb,q=1∑a3c5≤x/b4a3c5≡lb4¯modq1+∑c≤xνc,q=1∑a3b4≤x/c5a3b4≡lc5¯modq1−∑a≤xλa,q=1∑b≤xμb,q=1∑c5≤x/a3b4c5≡la3b4¯modq1−∑b≤xμb,q=1∑c≤xνc,q=1∑a3≤x/b4c5a3≡lb4c5¯modq1−∑a≤xλa,q=1∑c≤xνc,q=1∑b4≤x/a3c5b4≡la3c5¯modq1+∑a≤xλa,q=1∑b≤xμb,q=1∑c≤xνc,q=1a3b4c5≡lmodq1+∑a>xλa,q=1∑b>xμb,q=1∑c>xνa3b4c5≡lmodqa3b4c5≡lmodq1≔T1+T2+T3−T4−T5−T6+T7+T8.First, we estimateT3.(39)T3=∑c≤xνc,q=1∑a3b4≤x/c5a3b4≡lc5¯modq1,=∑c≤xνc,q=1∑b≤xμ1b,q=1∑a3≤x/b4c5a3≡lb4c5¯modq1+∑c≤xνc,q=1∑a≤xλ1a,q=1∑b4≤x/a3c5b4≡la3c5¯modq1−∑c≤xνc,q=1∑a≤xλ1a,q=1∑b≤xμ1b4≡la3c5¯modq1+∑c≤xνc,q=1∑a>xλ1a3b4≤x/c5∑b>xμ1a3b4≡lc5¯modq1≔T31+T32−T33+T34.ForT33, we have(40)T33=ϕqq2xλ1+μ1+ν+Oxλ1q1/2−ε+Oxμ1+νqε,by Lemma 3.Then, we estimateT31 and T32. For T31, we know(41)T31=∑c≤xν∑b≤xμ1N3lb2c;qx1/3b4/3c5/3q+O1,=∑c≤xν1c5/3∑b=1∞N3lb2c;qb4/3x1/3q−∑c≤xν1c5/33ϕqq2x1/3−1/3μ1+Ox1/3−1/3μ1q1/2−ε∑c≤xν1c5/3+Oxμ1+νqε.The first term in the above formula is(42)x1/3q∑b=1∞1b4/3∑c=1∞N3lb2c;qc5/3−32ϕqq2x1/3−2/3ν∑b=1∞1b4/3+Ox1/3−5/3νq1/2−ε=x1/3q∑b=1∞∑c=1∞N3lb2c;qb4/3c5/3−32ϕqq2x1/3−2/3ν∑b=1∞1b4/3+Ox1/3−5/3νq1/2−ε.In order to simplify our final result, by using Euler’s summation formula which can be found in Theorem 3.2 in [28], the constant ∑b=1∞1/b4/3 can be rewritten as(43)∑b=1∞1b4/3=limx⟶∞∑b⩽x1b4/3=limx⟶∞−3x−1/3+4−43∫1∞t−tt7/3dt+Ox−4/3=4−43∫1∞t−tt7/3dt=6−2∫1∞t−tt7/3dt.Thus, we obtain the asymptotic formula ofT31 as(44)T31=x1/3q∑b=1∞∑c=1∞N3lb2c;qb4/3c5/3−6−2∫1∞t−tt7/3dtϕqqx1/3−2/3ν−∑c≤xν1c5/33ϕqq2x1/3−1/3μ1+Ox1/3−5/3νq1/2−ε+Ox1/3−4/3μ1q1/2−ε+xμ1+νqε.Next, we deal withT32.(45)T32=∑c≤xνc,q=1∑a≤xλ1a,q=1∑b4≤x/a3c5b4≡la3c5¯modq1=x1/4q∑c≤xν1c5/4∑a≤xλ1N4lac3;qa3/4+Oxν+λ1qε=x1/4q∑c≤xν1c5/4∑a≤xλ1N4lac3;q−ϕq/qa3/4+ϕqq∑a≤xλ11a3/4+Oxν+λ1qε=x1/4q∑c≤xν1c5/4∫1xλ11μ3/4d∑a≤μN4lac3;q−ϕqqμ+ϕqq4x1/4λ1−3−34∫1∞t−tt7/4dt+Ox−34λ1+Oxν+λ1qε=x1/4q∑c≤xν1c5/4Oq1/2−εx3/4λ1−N4lc3;q+ϕqq+34∫1∞∑a≤μN4lac3;q−ϕq/qμμ7/4dμ+ϕqq2x1/4∑c≤xν1c5/44x1/4λ1−3−34∫1∞t−tt7/4dt+Ox−3/4λ1+Oxν+λ1qε.If we letFcμ=∑a≤μN4lac3;q−ϕq/qμ, then the first term in the above formula is(46)−x1/4q∑c≤xνN4lc3;q−ϕq/qc5/4+34x1/4q∫1∞∑c≤xνFcμ/c5/4μ7/4dμ+Ox1/4−3/4λ1q1/2−ε=−x1/4q∑c=1∞N4lc3;q−ϕq/qc5/4+Ox1/4−5/4νq1/2−ε+34x1/4q∫1∞∑c=1∞Fcμc5/4μ−7/4dμ+Ox1/4−3/4λ1q1/2−ε+Ox1/4−5/4νq1/2−ε=x1/4q−∑c=1∞N4lc3;q−ϕq/qc5/4+34x1/4q∫1∞∑c=1∞Fcμc5/4μ−7/4dμ+Ox1/4−3/4λ1q1/2−ε+Ox1/4−5/4νq1/2−ε.And the second term is(47)ϕqq2x1/4−4x−1/4ν+5−54∫1∞t−tt9/4dt+Ox−1/4ν4x1/4λ1−3−34∫1∞t−tt7/4dt+Ox−3/4λ=4ϕqq2x1/4+1/4λ1∑c≤xν1c5/4+ϕqq2x1/4−1/4ν12+3∫1∞t−tt7/4dt+ϕqq2x1/45−54∫1∞t−tt9/4dt−3−34∫1∞t−tt7/4dt+Oϕqq2x1/4−5/4ν+Oϕqq2x1/4−3/4λ1.So, we can get(48)T32=x1/4qϕqq5−54∫1∞t−tt9/4dt−3−34∫1∞t−tt7/4dt−∑c=1∞N4lc3;q−ϕq/qc5/4+34x1/4q∫1∞∑c=1∞Fcμc5/4μ−7/4dμ+4ϕqq2x1/4+1/4λ1∑c≤xν1c5/4+ϕqq2x1/4−1/4ν12+3∫1∞t−tt7/4dt+Oϕqq2x1/4−5/4ν+Oϕqq2x1/4−3/4λ1+Ox1/4−3/4λ1q1/2−ε+Ox1/4−5/4νq1/2−ε+Oxν+λ1qε.And forT34, following closely Chan [21] as shown in Figure 2 instead of Dirichlet’s hyperbola method shown in Figure 1, we just need to divide the interval a,b:a>xn,b>xm,a3b4≤x in the same way. Note that the sum(49)∑a>xλ1a3b4≤x/c5∑b>xμ1a3b4≡lc5¯modq1can be estimated with the help of asymptotic formula given at the end of page 101 in [21]; then, using Lemma 5, we can get(50)T34=∑c≤xν1c5/3⋅3ϕqq2x1/3−1/3μ1−∑c≤xν1c5/4⋅3ϕqq2x1/4−1/4λ1+ϕqq2xλ1+μ1+ν+Ox1/3−1/3μ12J1⋅q∑c≤xν1c5/4+2J1q1/2+εxνlogx,where 2J1 satisfies Xμ1/2J1q1/2≪1.Pickingλ1 to have the same size as μ1 and combining the previous results, we have(51)T3=x1/3q∑b=1∞∑c=1∞N3lb2c;qb4/3c5/3−6−2∫1∞t−tt7/3dtϕqqx1/3−2/3ν+ϕqq2x1/4−1/4ν12+3∫1∞t−tt7/4dt+x1/4qϕqq5−54∫1∞t−tt9/4dt−3−34∫1∞t−tt7/4dt−∑c=1∞N4lc3;q−ϕq/qc5/4+34x1/4q∫1∞∑c=1∞Fcμc5/4μ−7/4dμ+Oxμ1+νqε+2J1q1/2+εxν+x1/3−5/3νq1/2−ε+x1/4−5/4νq1/2−ε+x1/4−3/4λ1q1/2−ε+x1/3−4/3μ1q1/2−ε.ForT2, in the same way as T3, we have(52)T2=∑b≤xμb,q=1∑a3c5≤x/b4a3c5≡lb4¯modq1,=∑b≤xμb,q=1∑c≤xμ2c,q=1∑a3≤x/b4c5a3≡lb4c5¯modq1+∑b≤xμb,q=1∑a≤xλ2a,q=1∑c5≤x/a3b4c5≡la3b4¯modq1−∑b≤xμb,q=1∑a≤xλ2a,q=1∑c≤xμ2b4≡la3c5¯modq1+∑b≤xμb,q=1∑a>xλ2a3c5≤x/b4∑c>xμ2a3c5≡lb4¯modq1≔T21+T22−T23+T24.And we can obtain(53)T21=∑b≤xμ1b4/3∑c=1∞N3lb2c;qc5/3x1/3q−32ϕqq2∑b≤xμ1b4/3x1/3−2/3μ2+Oxμ+μ2qε+x1/3−5/3μ2q1/2−ε∑b≤xμ1b4/3,T23=ϕqq2xλ2+μ2+μ+Oxλ2q1/2−ε+Oxμ2+μqε,T24=∑b≤xμ1b4/3⋅32ϕqq2x1/3−2/3μ2−∑b≤xμ1b4/5⋅52ϕqq2x1/5−2/5λ2+ϕqq2xλ2+μ2+μ+Ox1/3−2/3μ22J2⋅q∑b≤xμ1b4/5+2J2q1/2+εxμlogx.ForT22, again in the same way as the proof of T3, we have(54)T22=∑b≤xμb,q=1∑a≤xλ2a,q=1∑c5≤xa3b4c5≡la3b4¯modq1,=x1/5q∑b≤xμ1b4/5∑a≤xλ2N5la2b;qa3/5+Oxμ+λ2qε=x1/5q∑b≤xμ1b4/5∑a≤xλ2N5la2b;q−ϕq/qa3/5+∑a≤xλ21a3/5ϕqq+Oxμ+λ2qε=x1/5q∑b≤xμ1b4/5∫1xλ21μ3/5d∑a≤μN5la2b;q−ϕqqμ+x1/5q∑b≤xμ1b4/5ϕqq52x2/5λ2−32−35∫1∞t−tt8/5dt+Ox−3/5λ2+Oxμ+λ2qε.If we letFbμ=∑a≤μN5la2b;q−ϕq/qμ, then the first term in the above formula is(55)x1/5q∑b≤xμ1b4/5Oq1/2+εx3/5λ2−N5lb;q+ϕqq+35∫1∞Fbμμ8/5dμ=−x1/5q∑b≤xμN5lb;q−ϕq/qb4/5+35x1/5q∫1∞∑b≤xμFbμ/b4/5μ8/5+Ox1/5−3/5λ2q1/2−ε∑b≤xμ1b4/5=N5l;q−ϕqq−45∫1∞∑b≤μN5lb;q−ϕq/qμ9/5dμx1/5q+1225∫1∞∫1∞∑b≤λFbμμ8/5λ9/5dλdμx1/5q−35x1/5q∫1∞∑a≤μN5la2;q−ϕq/qμ8/5dμ+Ox1/5−3/5λ2q1/2−ε∑b≤xμ1b4/5+Ox1/5−4/5μq1/2−ε,and the second term is(56)x1/5q∑b≤xμ1b4/5ϕqq52x2/5λ2+x1/5q∑b≤xμ1b4/5ϕqq−32−35∫1∞t−tt8/5dt+Oϕqqx1/5−3/5λ2∑b≤xμ1b4/5=52x1/5+2/5λq∑b≤xμ1b4/5ϕqq−x1/5qϕqq5x1/5μ−4−45∫1∞t−tt9/5dt+Ox−4/5μ32+35∫1∞t−tt8/5dt+Oϕqqx1/5−3/5λ2∑b≤xμ1b4/5=52ϕqq2x1/5+2/5λ2∑b≤xμ1b4/5+−4−45∫1∞t−tt9/5dt−32−35∫1∞t−tt8/5dtϕqq2x1/5+−152−3∫1∞t−tt8/5dtϕqq2x1/5+1/5μ+Oϕqqx1/5−3/5λ2∑b≤xμ1b4/5+Oϕqq2x1/5−4/5μ.Then, we can obtain(57)T22=x1/5q−4−45∫1∞t−tt9/5dt−32−35∫1∞t−tt8/5dtϕqq−45∫1∞∑b≤μN5lb;q−ϕq/qμμ8/5dμ−35x1/5q∫1∞∑a≤μN5la2;q−ϕq/qμ8/5dμ+N5l;q−ϕqq+1225∫1∞∫1∞∑b≤λFbμμ8/5λ9/5dλdμ+52ϕqq2x1/5+2/5λ2∑b≤xμ1b4/5+−152−3∫1∞t−tt8/5dtϕqq2x1/5+1/5μ+Ox1/5−3/5λ2q1/2−ε∑b≤xμ1b4/5+x1/5−4/5μq1/2−ε+ϕqq2x1/5−4/5μ+xμ+λ2qε.If we pickλ1 to have the same size as μ1, we can get(58)T2=∑b≤xμ1b4/3∑c=1∞N3lb2c;qc5/3x1/3q+x1/5q−4−45∫1∞t−tt9/5dt−32−35∫1∞t−tt8/5dtϕqq−45∫1∞∑b≤μN5lb;q−ϕq/qμμ8/5dμ+N5l;q−ϕqq−35x1/5q∫1∞∑a≤μN5la2;q−ϕq/qμ8/5dμ+1225∫1∞∫1∞∑b≤λFbμμ8/5λ9/5dλdμ+−152−3∫1∞t−tt8/5dtϕqq2x1/5+1/5μ+Ox1/3−5/3μ2q1/2−ε+x1/5−3/5λ2q1/2−ε∑b≤xμ1b4/5+x1/5−4/5μq1/2−ε+xμ+μ2qε+2J2q1/2+εxμ,where 2J2 satisfies Xμ1/2J2q1/2≪1.Similar to the proofs ofT2 and T3, we can get the following:(59)T1=∑a≤xλ1a3/4∑c=1∞N4lac3;qc5/4x1/4q+∑a≤xλ1a3/5−4ϕqq−N5la2;q+45∫1∞∑l1≤μN5ll1a2;q−ϕq/qμμ9/5dμx1/5q+Ox1/4−5/4μ3q1/2−ε∑a≤xλ1a3/4+xλ+μ3qε+2J3q1/2+εxλ+x1/5−4/5λ3q1/2−ε∑a≤xλ1a3/5,T4=52x2/5λ−32−35∫1∞t−tt8/5dt+Ox−3/5λ5ϕqq2x1/5+1/5μ+Ox1/5−4/5μq1/2−ε∑a≤xλ1a3/5+∑a≤xλ1a3/5−4ϕqq−N5la2;q+45∫1∞∑l1≤μN5ll1a2;q−ϕq/qμμ9/5dμx1/5q=252ϕqq2x1/5+μ/5+2/5λ+−152−3∫1∞t−tt8/5dtϕqq2x1/5+1/5μ+∑a≤xλ1a3/5−4ϕqq−N5la2;q+45∫1∞∑l1≤μN5ll1a2;q−ϕq/qμμ9/5dμx1/5q+Oϕqq2x1/5+μ/5−3/5λ+Ox1/5−4/5μq1/2−ε∑a≤xλ1a3/5,T5=∑b≤xμ1b4/3∑c=1∞N3lb2c;qc5/3x1/3q−32ϕqq2x1/3−2/3ν∑b≤xμ1b4/3+Ox1/3−2/3μq1/2−ε∑b≤xμ1b4/3=∑b≤xμ1b4/3∑c=1∞N3lb2c;qc5/3x1/3q+32ϕqq2x1/3−2/3ν3x−1/3μ−4+43∫1∞t−tt7/3dt+Ox−4/3μ+Ox1/3−5/3μq1/2−ε∑b≤xμ1b4/3=∑b≤xμ1b4/3∑c=1∞N3lb2c;qc5/3x1/3q+92ϕqq2x1/3−2/3ν−1/3μ+−6+2∫1∞t−tt7/3dtϕqqx1/3−2/3ν+Oϕqq2x1/3−2/3ν−4/3μ+Ox1/3−5/3μq1/2−ε∑b≤xμ1b4/3,T6=∑a≤xλ1a3/4∑c=1∞N4lac3;qc5/4x1/4q−4ϕqq2x1/4−1/4ν∑a≤xλ1a3/4+Ox1/4−5/4νq1/2−ε∑a≤xλ1a3/4=∑a≤xλ1a3/4∑c=1∞N4lac3;qc5/4x1/4q−4ϕqq2x1/4−21/4ν4x1/4λ−3−34∫1∞t−tt7/4dt+Ox−3/4λ+Ox1/4−5/4νq1/2−ε∑a≤xλ1a3/4=∑a≤xλ1a3/4∑c=1∞N4lac3;qc5/4x1/4q−16ϕqq2x1/4−1/4ν+1/4λ+12+3∫1∞t−tt7/4dtϕqqx1/4−1/4ν+Oϕqq2x1/4−1/4ν−1/4λ+Ox1/4−5/4νq1/2−ε∑a≤xλ1b3/4,T7=∑a≤xλa,q=1∑b≤xμb,q=1∑c≤xνc,q=1a3b4c5≡lmodq1=ϕqqxλ+μ+ν+Oxλq1/2−εxμ+ν.Finally, we discussT8 as follows: dividing interval xν,x1−3λ−4μ/5 into m parts with the length of each one being x1−3λ−4μ/5−xν/m. The nth part is(60)xν+n−1x1−3λ−4μ/5−xνm,xν+nx1−3λ−4μ/5−xνm.Leta0 be the intersection of the plane b=xμ, c=xν+nx1−3λ−4μ/5−xν/m, and curved surface a3b4c5=x, which is(61)a03x4μxν+nx1−3λ−4μ/5−xνm5=x,and then we obtain(62)a0=x1−4μ/3xν+nx1−3λ−4μ/5−xν/m5/3.Similarly, we have(63)b0=x1−3λ/4xν+nx1−3λ−4μ/5−xν/m5/4.We define the intersection ofa3b4c5=x and xν+nx1−3λ−4μ/5−xν/m as(64)a3b4=xxν+nx1−3λ−4μ/5−xν/m5≔xn.Now we divide the interval in thenth part and follow the construction in [23]; firstly, we have rectangles(65)Ri=2i−1xλ,2ixλ×xμ,xn2ixλ31/4,1≤i≤I=log2a0xλ.In the remaining regions(66)Si≔a,b,2i−1xλ<a≤2ixλ,b>xn2ixλ31/4,a3b4≤xm,we place additional rectangles Rijk. If we let 1≤k≤2j−1, in the same way we place a further rectangles and so on, then we can get Rijk which is(67)Rijk=2i−11+2k−22jxλ,2i−11+2k−12jxλ×xn2i−11+2k/2jxλ31/4,xn2i−11+2k−1/2jxλ31/4.The remaining regions are(68)Sijk=a,b:2i−11+2k−22jxλ<a≤2i−11+2k−12jxλ,b>xn2i−11+2k−1/2jxλ31/4,a3b4≤xn,Sijk′=a,b:2i−11+2k−12jxλ<a≤2i−11+2k2jxλ,b>xn2i−11+2k/2jxλ31/4,a3b4≤xn.In this part, expandingRi and Rijk to a cube with height x1−3λ−4μ/5−xν/m, we can get Rni, Rnijk, Sni, SnijkSnijk′, and the remaining region Sn correspondingly. Then, we have the asymptotic formula(69)T8=∑n∑i=1I−1∑a,b,c∈Rnia3b4c5≡lmodq1+∑n∑j=1J∑i=1I∑k=12j−1∑a,b,c∈Rnijka3b4c5≡lmodq1+∑n∑i=1I∑k=12J−1∑a,b,c∈SniJka3b4c5≡lmodq1+∑n∑i=1I∑k=12J−1∑a,b,c∈SniJk′a3b4c5≡lmodq1+∑n∑a,b,c∈Sna3b4c5≡lmodq1.If we letNA,B,CK,L,M≔#a,b,c:a3b4c5≡lmodq,A<a≤A+K,B<b≤B+L,C<c≤C+M, by Lemma 5, we know(70)NA,B,CK,L,M=ϕqq2KLM+OMq1/2−ε+KLqε.Then, we further obtain(71)T8=ϕqq2Area ofa,b,c:a>xλ,b>xμ,c>xν,a3b4c5≤x+O∑n∑i=1In∑k=12J−1Area ofSniJkq+O∑n∑i=1In∑k=12J−1Area ofSniJk′q+O∑n∑i=1I∑k=12J−1∑a,b,c∈SniJka3b4c5≡lmodq1+O∑n∑i=1I∑k=12J−1∑a,b,c∈SniJk′a3b4c5≡lmodq1+O∑n∑i=1In1q1/2−εx1−3λ−4μ/5−xνm+∑n∑i=1In2i−1xλxn1/42ixλ3/4−xμ+∑n∑j=1J∑i=1In∑k=12j−1x1−3λ−4μ/5−xνm1q1/2+∑n∑j=1J∑i=1In∑k=12j−12i−1xλ122jxn1/42i−1xλ1/4+O∑n∑a,b,c∈Sna3b4c5≡lmodq1,in which the main term is(72)ϕqq2∫xνx1−3λ−4μ/5∫xλx1−4μ/3/c5/3xa3c51/4−xμdadc=ϕqq2252x1+2λ+μ/5+92x1−μ−2ν/3−16x1+λ−ν/4−xλ+μ+ν.Now we deal with the error term ofT8. Note that(73)SniJk⊆2i−11+2k−22Jxλ,2i−11+2k−12Jxλ×xn2i−11+2k−1/2jxλ31/4,xn2i−11+2k−2/2jxλ31/4×xν+n−1x1−3λ−4μ/5−xνm,xν+nx1−3λ−4μ/5−xνm.Thus, the area ofSniJk is at most(74)2i−1xλ2J⋅xn1/42i−1xλ3/4⋅12J⋅x1−3λ−4μ/5−xνm,which implies that the estimation of the first error term is(75)∑n∑i=1In∑k=12J−1Area ofSniJkq≪12J⋅1q⋅m2/3x1−μ/3x1−3λ−4μ/52/3.Similarly, we have the same error bound for the second, third, and the fourth error term. The error terms in the fifth are(76)∑n∑i=1In1q1/2−εx1−3λ−4μ/5−xνm≪1q1/2x1−3λ−4μ/5−xν,∑n∑i=1In2i−1xλxn1/42ixλ3/4−xμ≪m5/3x1−μ/3x1−3λ−4μ/55/3,∑n∑j=1J∑i=1In∑k=12j−1x1−3λ−4μ/5−xνm1q1/2≪2J1q1/2x1−3λ−4μ/5−xν,∑n∑j=1J∑i=1In∑k=12j−12i−1xλ122jxn1/42i−1xλ1/4≪m5/3x1−μ/3x1−3λ−4μ/55/3.For the last error term, we extendSn into a ring sector which volume is(77)x1−3λ−4μ/5−xνm∫xλa1xn−1a31/4−xμda−∫xλa0xna31/4−xμda,where(78)a13x4μxν+n−1x1−3λ−4μ/5−xνm5=x.Then, we can addn and get(79)∑n∑a,b,c∈Sna3b4c5≡lmodq1≪1mx1−3λ−4μ/5x1−μ−5ν/3=1mx8−9λ−17μ−25ν/15.Therefore, we have(80)T8=ϕqq2252x1+2λ+μ/5+92x1−μ−2ν/3−16x1+λ−ν/4−xλ+μ+ν+O12J⋅1q⋅m2/3x1−μ/3x1−3λ−4μ/52/3+1q1/2x1−3λ−4μ/5−xν+m5/3x1−μ/3x1−3λ−4μ/55/3+2J1q1/2x1−3λ−4μ/5−xν+1mx8−9λ−17μ−25ν/15.Now we come to simplify the above error terms. Firstly, we have(81)1q1/2x1−3λ−4μ/5−xν≪x1/5−2/5λq1/2−ε.Whenm=O1 and λ=μ, we know that(82)m5/3x1−μ/3x1−3λ−4μ/55/3≪x2λ.If2J≪xλ and λ=μ, then(83)2J1q1/2x1−3λ−4μ/5−xν≪x1/5−2/5λq1/2−ε.If we pick2J=xλ and λ=μ, we get(84)12J⋅1q⋅m2/3x1−μ/3x1−3λ−4μ/52/3≪x1/5−2/5λq1/2−ε,with the condition m=O1.Therefore, if we chooseλ=μ=ν, μi=λii=1,2,3 and m=O1, then we get all error terms from T1 to T8:(85)Ox1/3−4/3μ1q1/2−ε+xμ1+λqε+2J1q1/2+εxλ+x1/3−5/3μ2+1/5λq1/2−ε+xμ2+λqε+2J2q1/2+εxλ+x1/4−5/4μ3+2/5λq1/2−ε+xμ3+λqε+2J3q1/2+εxλ+x1/3−5/3λq1/2−ε+x1/5−2/5λq1/2−ε+x8/15−51/15λ,where μ1, μ2, and μ3 satisfy the conditions(86)110<μ1<15−λ,xμ12J1q1/2≪1,225+325λ<μ2<15−λ,xμ22J2q1/2≪1,125+825λ<μ3<15−λ,xμ32J3q1/2≪1.Further if we chooseη=μ1=μ2=μ3, then we obtain(87)O1/3−4/3ηq1/2−ε+xη+λqε+2J1q1/21/2+εxλ+x1/5−2/5λq1/2−ε+x8/15−51/15λ+x1/3−5/3λq1/2−ε,where xη/2J1q1/2≪1, 5/51<λ<1/10 and 1/10<η<1/5−λ. Continue to simplify the error term and get(88)Ox1/3−4/3ηq1/2−ε+xη+λqε+2J1q1/2+εxλ,where xη/2J1q1/2≪1 and 8/15−66/15λ<η<1/5−λ,5/51<λ<13/132, or 1/10<η<1/5−λ13/132<λ<1/10. If q<x2η, we can choose 2J1=xη/q1/2; otherwise, we can choose 2J1=2.Finally, we have(89)∑a3b4c5≤xa3b4c5≡lmodq1=Aq′lx1/3q+Bq′lx1/4q+Cq′lx1/5q+Ox1/3−4/3ηq1/2−ε+maxxη,q1/2qεxλ,where(90)Aq′l=∑b=1∞∑c=1∞N3lb2c;qb4/3c5/3,Bq′l=ϕqq5−54∫1∞t−tt9/4dt−3−34∫1∞t−tt7/4dt−∑c=1∞N4lc3;q−ϕq/qc5/4+34x1/4q∫1∞∑c=1∞∑a≤μN4lac3;q−ϕq/qμc5/4μ−7/4dμ,Cq′l=−4−45∫1∞t−tt9/5dt−32−35∫1∞t−tt8/5dtϕqq−45∫1∞∑b≤μN5lb;q−ϕq/qμμ8/5dμ+N5l;q−ϕqq−35x1/5q∫1∞∑a≤μN5la2;q−ϕq/qμ8/5dμ+1225∫1∞∫1∞∑b≤λ∑a≤μN5la2b;q−ϕq/qλμμ8/5λ9/5dλdμ.


## 4. Proof of Theorem1

First recall that any cube-full number can be written uniquely asn=a3b4c5 where b, c are square-free numbers and b,c=1. Following the proof of Theorem 2.3 in [20], we have(91)∑n≤x,n∈P3n≡lmodq1=∑a3b4c5≤xb,c=1,b,csquarefreea3b4c5≡lmodq1=∑a3b4c5≤xa3b4c5≡lmodqb,c=1μ2bμ2c.In view of the identityμ2b=∑d2|bμd, we get(92)∑n≤x,n∈P3n≡lmodq1=∑a3b4c5d8≤xa3b4c5d8≡lmodqbd,c=1μdμ2c.From the identity(93)∑l|bl|cμl=1ifb,c=1,0otherwise,and μ2cl=μ2cμ2l for c,l=1, otherwise μ2cl=0 and μ2lμl=μl, we get(94)∑n≤x,n∈P3n≡lmodq1=∑a3b4c5d8≤xa3b4c5d8≡lmodqc,d=1μdμ2c∑e|be|cμe=∑a3b4c5d8e9≤xa3b4c5d8e9≡lmodqc,e,d=1μdμ2cμe.Using againμ2c=∑f2|cμf and(95)∑g|cg|edμg=1,ifed,c=1,0,otherwise,we have(96)∑n≤x,n∈P3n≡lmodq1=∑a3b4c5d8e9≤xa3b4c5d8e9≡lmodqc,e,d=1μdμe∑f2|cμf,=∑a3b4c5d8e9f10≤xa3b4c5d8e9f10≡lmodqcf,e,d=1μdμfμe=∑a3b4c5d8e9f10≤xa3b4c5d8e9f10≡lmodqf,e,d=1μdμfμe∑g|cg|edμg=∑a3b4c5g5d8e9f10≤xa3b4c5g5d8e9f10≡lmodqf,e,d=1μdμfμe∑g|edμg=∑g5d8e9f10≤xd,e,f=1μdμfμe∑g|edμg∑a3b4c5≤x/g5d8e9f10a3b4c5≡lg5d8e9f10¯modq1.Fromxη/2J1q1/2≪1, we know that if q≤x2η, then we can pick 2J1 of size xη/q1/2, and if q>x2η, we simply pick 2J1=2. Then, we obtain(97)∑n≤x,n∈P3n≡lmodq1=∑g5d8e9f10≤xq−1/2ηd,e,f,q=1μdμfμe∑g|edg,q=1μg∑a3b4c5≤x/g5d8e9f10a3b4c5≡lg5d8e9f10¯modq1+∑xq−1/2η<g5d8e9f10≤xd,e,f,q=1μdμfμe∑g|edg,q=1μg∑a3b4c5≤x/g5d8e9f10a3b4c5≡lg5d8e9f10¯modq1.From the result of Theorem2, we have(98)∑n⩽x,n∈P3n≡lmodq1=Aq′l∑d8e9f10≥1d,e,f,q=1μdμfμed8/3e3f10/3∑g|edg,q=1μgg5/3⋅x1/3q+Bq′l∑d8e9f10≥1d,e,f,q=1μdμfμed2e9/4f5/2∑g|edg,q=1μgg5/4⋅x1/4q+Cq′l∑d8e9f10≥1d,e,f,q=1μdμfμed8/5e9/5f2∑g|edg,q=1μgg⋅x1/5q+O∑d8e9f10⩽xq−1/2ηx1/3−4/3ηq1/2−εd8e9f101/3−4/3η+∑d8e9f10⩽xq−1/2ηxη+λd8e9f10η+λqε+∑xq−1/2η≤d8e9f10≤xx1/3−4/3ηq1/2−εd8e9f101/3−4/3η+∑xq−1/2η≤d8e9f10≤xxλd8e9f10λq1/2+ε.For the final term of theO-term, we have(99)∑xq−1/2η≤d8e9f10≤xxλd8e9f10λq1/2+ε≪xλq1/2+ε∑d≤x1/8∑e≤x1/9∑xq−1/2ηd−8e−9≤f10≤xd8e91d8λe9λf10λ≪xλq1/2+ε∑d≤x1/81d8λ∑e≤x1/91e9λ∑xq−1/2ηd−8e−9≤f10≤xd8e91f10λ≪xλq1/2+ε∑d≤x1/81d8λ∑e≤x1/91e9λx1/10−λd8/10−8λe9/10−9λ≪x1/10q1/2+ε∑d≤x1/81d8/10∑e≤x1/91e9/10=x49/360q1/2+ε.In the same way, we have(100)∑xq−1/2η≤d8e9f10≤xx1/3−4/3ηq1/2−εd8e9f101/3−4/3η≪x49/360q7/60μ1−7/6≪x49/360q1/2+ε.The first two terms of the error term are(101)x1/3−4/3ηq1/2−ε+xη+λqε.If we take(102)x1/3−4/3ηq1/2−ε=xη+λqε,and assume q=xθ, then we have η=1/7−3/7λ−3/14θ and(103)xη+λqε=x1/7+4/7λq3/14qε.Now we pickλ=5/51+ε and get(104)x1/3−4/3ηq1/2−ε+xη+λqε≪x71/357q3/14qε.Consequently, we obtain(105)∑n≤x,n∈P3n≡lmodq1=Aq′l∑d8e9f10≥1d,e,f,q=1μdμfμed8/3e3f10/3∑g|edg,q=1μgg5/3⋅x1/3q+Bq′l∑d8e9f10≥1d,e,f,q=1μdμfμed2e9/4f5/2∑g|edg,q=1μgg5/4⋅x1/4q+Cq′l∑d8e9f10≥1d,e,f,q=1μdμfμed8/5e9/5f2∑g|edg,q=1μgg⋅x1/5q+Ox49/360q1/2+x71/357q3/14qε,which is Theorem 2.



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*Source: 5552120-2021-03-10.xml*