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On the number of integers which form perfect powers in the way of x(y21+y22+y23+y24)=zk

  • Let k2 be an integer. We studied the number of integers which form perfect k-th powers in the way of

    x(y21+y22+y23+y24)=zk.

    For k4, we established a unified asymptotic formula with a power-saving error term for the number of such integers of bounded size under Lindelöf hypothesis, and we also gave an unconditional result for k=2.

    Citation: Tingting Wen. On the number of integers which form perfect powers in the way of x(y21+y22+y23+y24)=zk[J]. AIMS Mathematics, 2024, 9(4): 8732-8748. doi: 10.3934/math.2024423

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  • Let k2 be an integer. We studied the number of integers which form perfect k-th powers in the way of

    x(y21+y22+y23+y24)=zk.

    For k4, we established a unified asymptotic formula with a power-saving error term for the number of such integers of bounded size under Lindelöf hypothesis, and we also gave an unconditional result for k=2.



    Counting the number of integral solutions of a certain Diophantine equation is an interesting project in number theory. Let Nk(H) be the number of pairs of positive integers x1, x2H whose product x1x2 is a perfect k-th power. Tolev [1] first established an asymptotic formula for Nk(H). The proof combines Perron's formula with elementary ideas from the work of Heath-Brown and Moroz [2]. It is proved that for any integer k2,

    Nk(H)=ckH2/k(logH)k1+O(H2/k(logH)k2),

    where ck>0 is an explicit constant depending on k. De la Bretèche et al. [3] improved this result based on the multiple Dirichlet series theory and complex analysis. They showed that there exists a constant θk>0 such that

    Nk(H)=H2/kQ(logH)+O(H2/kθk+ε),

    where Q is a polynomial of degree k1 with leading coefficient ck. In fact, they considered a more general case and counted the number of tuples of n2 integers whose product is a perfect k-th power. Precisely, they proved that there exists a constant θn,k>0, such that

    Nn,k=|{(x1,,xn)[1,H]nNn:x1,,xn=zk,zN}|=Hn/kQn,k(logH)+O(Hn/kθn,k),

    where Qn,k is a polynomial of degree (n+k1k)n.

    It is natural to consider the analogue of the above results for polynomials. Liu and Niu [4] counted pairs of polynomials whose product is a cube over finite field and obtained an asymptotic formula (see [4, Theorem 1.2]) by contour integration. It is indicated that there are some differences and challenges between polynomials and integers.

    Returning to the case of integers, this problem is closely related to the integral points on algebraic surface

    S:x1x2x3=x30.

    This is a split toric surface and can also be regarded as the product of three integers forming a perfect cube. It has been studied by many authors including de la Bretèche [5], Fouvry [6], Heath-Brown and Moroz [2], and Salberger [7]. Denote by NS(H) the number of primitive integral points (i.e., gcd(x0,x1,x2,x3)=1) on S satisfying x00 and max0i3|xi|H. The sharpest unconditional result is proved by de la Bretèche [5], which states

    NS(H)=HP(logH)+O(H7/8exp(c(logH)3/5)(loglogH)1/5)),

    where P is a polynomial of degree 6 and c is a positive constant. Now, we look at the corresponding non-split toric surface

    S:x(y21+y22)=z3.

    It can be observed that S and S are isomorphic over Q(i). Denote by NS(H) the number of primitive integral points on S satisfying z0 and

    max{|x|,y21+y22,|z|}H.

    De la Bretèche et al. [8] studied this surface and proved that

    NS(H)=HP(logH)+O(H8/9+ε),

    where P is a cubic polynomial. Liu et al. [9] further considered the following case

    S4:x(y21+y22+y23+y24)=z3. (1.1)

    For ease of presentation, we let

    y=(y1,y2,y3,y4),y=y21+y22+y23+y24,

    and denote by N4(H) the number of integral tuples (x,y1,,y4), satisfying max{|x|,y}H, which form nonzero perfect cubes in the way of (1.1) since the variable z is completely determined by x and y. They showed that

    N4(H)=c4H3(logH)2+O(H3logH).

    In fact, they dealt with a more general case that the number of squares is a multiple of 4 (see [9, Theorem 7.1]). Zhai [10] improved this result by obtaining a power-saving error term that

    N4(H)=H3P4(logH)+O(H31/4+ε),

    where P4 is a quadratic polynomial. All above results are obtained by studying the corresponding multiple Dirichlet series. Liu et al. stated in [9] that the same idea can be applied to investigate the number of integral solutions of some higher-degree Diophantine equation like

    xd=(y21+y22+y23+y24)zd2,d4.

    They obtained an asymptotic formula in [11] for d=4, and Wen [12] established the asymptotic formulae for any integer d4 with power-saving error terms.

    We remark here that the Diophantine equations S, S, and S4 mentioned above are homogeneous, so there is an equivalent relation between rational points in projective space and integral points in affine space up to a scalar multiplication. They are actually related to another project in number theory, which is Manin's conjecture (see [2,5,6,7,8,9] for more details).

    Motivated by the above work, in this paper, we mainly focus on the Diophantine equation

    Sk4:x(y21+y22+y23+y24)=zk (1.2)

    for k2. Similar to (1.1), we denote by Nk4(H) the number of integral tuples (x,y1,,y4) satisfying max{|x|,y}H, which form nonzero perfect k-th powers in the way of (1.2). Recall that the Lindelöf hypothesis (LH in brief, [13, §Ⅱ.3.4]) states that

    ζ(1/2+it)(|t|+1)ε

    for any ε>0. Our main result is as follows.

    Theorem 1.1. Let k4 be any integer. Assuming LH, then for any ε>0, there exists a constant ϑk, such that

    Nk4(H)=H2+3/kP(logH)+O(H2+3/kϑk+ε),

    where

    ϑk=32k(12k/31[2k/3])>0,

    and [α] is the integral part of α, the implied constant only depends on k and ε, P is a polynomial of degree k1 given by (3.14) with leading coefficient 16Ck, and Ck is a positive constant given by (3.13).

    Remark 1.2. The assumption of LH is just for simplifying the calculation. We claim that the above asymptotic formulae still holds for k=4 unconditionally, since we can apply the fourth moment estimate of the Riemann zeta function instead of LH as in (4.8).

    The case of k=3 has been solved in [9,10] unconditionally as mentioned above. In fact, following the proof of Theorem 1.1, one can easily obtain an asymptotic formula for k=3 with a power-saving error term O(H31/5+ε). We shall leave that as an example to justify our result. We next give an unconditional result for k=2.

    Theorem 1.3. Unconditionally, we have

    N24(H)=16H2+3/2P(logH)+O(H2+3/21/3+ε),

    where P is a linear polynomial given by (4.7) and the implied constant only depends on ε.

    We first introduce the bivariate Perron's formula [10,Lemma 2.2], which plays an important role in our proof.

    Lemma 2.1. Suppose that f(n1,n2) is a bivariate arithmetic function and its Dirichlet series

    F(s1,s2)=n1=1n2=1f(n1,n2)ns11ns22

    is absolutely convergent for (sj)>σj(j=1,2) with some σ1, σ2>0. Let x1, x2, T1, T210 be parameters such that xjN, and define

    bj=σj+1/logxj,j=1,2.

    We have

    n1x1n2x2f(n1,n2)=1(2πi)2b1+iT1b1iT1b2+iT2b2iT2F(s1,s2)xs11xs22s1s2ds2ds1+O(xσ11xσ22E),

    where

    E:=2j=1n1=1n2=1|f(n1,n2)|nb11nb22min{1,1Tj|logxjnj|}.

    Recalling the definition of Nk4(H) before, we see that

    Nk4(H)=|{(x,y,z)Z6Sk4:max{|x|,y}H,z0}|.

    Let r4(n) be the number of representations of a positive integer n as the sum of four squares

    n=y21+y22+y23+y24

    with

    (y1,y2,y3,y4)Z4.

    It is well known that

    r4(n)=8r4(n) with r4(n)=dnd0(mod4)d, (2.1)

    and r4(n) is a multiplicative arithmetic function. Let 1k denote an indicator function of perfect k-th power defined by

    1k(n)={1, if n is a perfect kth power,0, otherwise. (2.2)

    In view of the above problem, we can write

    Nk4(H)=21mH1nH2r4(n)1k(n)=161mH1nH2r4(n)1k(n). (2.3)

    In order to deal with this double sum, we define the corresponding Dirichlet series

    Fk(s,w)=m,n=1r4(n)1k(mn)msnw. (2.4)

    The following proposition gives the expression and convergence of Fk(s,w), which allows us to extend the double Dirichlet series to a suitable large region.

    Proposition 2.2. Let k2. If (s)>1/k and (w)>1+1/k, then

    Fk(s,w)=kj=0ζ((kj)s+j(w1))Hk(s,w), (2.5)

    where Hk(s,w) is an Euler product given by (2.16), which is absolutely convergent if s and w satisfy the conditions

    min1ik1((ki)s+i(w1))1/2+ε and (w)1+ε. (2.6)

    Furthermore, we have

    Hk(s,w)1 (2.7)

    in the above region, and the implied constant is absolute.

    Proof. Note that r4(n) is multiplicative, and (2.1) yields r4(1)=1 and

    r4(pν)={1pν+11p, if p3,3, if p=2, (2.8)

    for any integer ν1, then we can rewrite Fk(s,w) in (2.4) as the Euler product

    Fk(s,w)=pμ0ν0r4(pν)1k(pμ+ν)pμs+νw=pd0pkds0νkdr4(pν)pν(ws)=:pFk,p(s,w). (2.9)

    Here, we let μ+ν=kd for d0, according to the definition of 1k in (2.2). On the other hand, a simple formal calculation shows

    d0xd0νkdyν1zkd+11z=11zd0xd(1ykd+11yz1(yz)kd+11yz)=11z(11y(11xy1xyk)z1yz(11xyz1x(yz)k))=Gk(x,y,z)(1x)(1xyk)(1x(yz)k) (2.10)

    with

    G2(x,y,z)=1+xy(1+z)+xy2z

    and

    Gk(x,y,z)=1+xy(1+z)+xy2(1+z+z2)++xyk1(1+z++zk1)+xyk(z+z2++zk1)+x2yk+1(z2+z3++zk1)+x2yk+2(z3+z4++zk1)++x2y2k2zk1

    for k3. Similarly, we have

    1+d1xd(1+31νkdyν)=11x+31yd1xd(yykd+1)=11x+3(xyxyk+1)(1y)(1x)(1xyk)=1+3xy(1+y++yk2)+2xyk(1x)(1xyk). (2.11)

    When p3, in view of (2.8), we apply (2.10) with

    (x,y,z)=(pks,p(ws),p)

    to deduce that

    Fk,p(s,w)=0jk(11p(kj)s+j(w1))1Hk,p(s,w), (2.12)

    where

    Hk,p(s,w)=Gk(pks,p(ws),p)(11pkw)11jk1(11p(kj)s+j(w1)). (2.13)

    Meanwhile, for p=2, the formula (2.11) with

    (x,y)=(2ks,2(ws))

    gives us

    Fk,2(s,w)=0jk(112(kj)s+j(w1))1Hk,2(s,w), (2.14)

    where

    Hk,2(s,w)=(1+32(k1)s+w(1+12ws++12(k2)(ws))+12kw1)×(112kw)11jk(112(kj)s+j(w1)). (2.15)

    It can be observed that Fk,p(s,w)/Hk,p(s,w) will give the Euler product of Riemann zeta functions in (2.12) and (2.14). Combining (2.9) and (2.12)–(2.15), we get (2.5) with

    Hk(s,w)=Hk,2(s,w)p3Hk,p(s,w). (2.16)

    Next, we shall discuss the convergence. In view of the expression of Hk,p(s,w) in (2.13) and (2.15), when expanding Hk,p(s,w) into 1 plus some monomials about p, we see that the power of p in the denominator of each monomial is great than 1 if

    min1ik1((ki)s+i(w1))1/2+ε  and  (w)1+ε,

    so we have

    Hk,p(s,w)=1+O(p1ε),

    and Hk(s,w) is absolutely convergent in this region, which implies (2.7). This completes the proof.

    We next make use of bivariate Perron's formula. Suppose HN and let T[10,H1/2] be a parameter to be chosen later. The analytic property of Fk(s,w) allows us to set

    x1=H,  x2=H2,  s1=s=σ+it,  s2=w=u+iv,  b1=1/k+ε,  b2=1/k+1+ε

    in Lemma 2.1, then we get

    Nk4(H)=16(2πi)2b1+iTb1iTb2+10iTb210iTFk(s,w)Hs+2wwsdwds+O(H2+3k+ε(J1(H,T)+J2(H,T)))

    with

    J1(H,T):=m,n=1r4(n)1k(mn)mb1nb2min(1,1T|logHm|),J2(H,T):=m,n=1r4(n)1k(mn)mb1nb2min(1,1T|logH2n|).

    Noticing that

    r4(n)nτ(n),

    and following the arguments in [10], one can easily get

    J1(H,T)m,n=1τ(n)1k(mn)(mn)b1min(1,1T|logHm|)HεT,

    and the same result holds for J2(H,T). It follows that

    Nk4(H)=16(2πi)2b1+iTb1iTb2+10iTb210iTFk(s,w)Hs+2wwsdwds+O(H2+3/k+εT). (2.17)

    It suffices to evaluate the double integral on the righthand side of (2.17).

    In this section, we shall prove the main theorem. The proof is divided into three steps. We shall first turn the double integral into some usual single integrals by Cauchy's residue theorem and then deal with the single integrals. The final step is choosing the suitable parameters.

    In this subsection, we shall apply Cauchy's residue theorem to evaluate the inner integral over w in (2.17) and derive the following result.

    Lemma 3.1. Let k2 and Nk4(H) be defined as in (2.3). Assuming LH, we have

    Nk4(H)=16kj=1Ij(H,T)+O(H2+3/k+ε/T+H2+2/k+ε),

    where

    Ij(H,T):=12πib1+iTb1iTresw=wjFk(s,w)wsHs+2wjds (3.1)

    with

    wj=(1(kj)s)/j+1

    for 1jk.

    Proof. In terms of the inner integral in (2.17), we consider the domain formed by four points

    w=b2±10iT,w=1/2k+1±10iT.

    In this domain, from Proposition 2.2, we easily see that the integrand function Fk(s,w)/(ws) has k simple poles:

    wj=1j(1(kj)s)+1

    for 1jk.

    Using the residue theorem for the variable w, we get

    Nk4(H)=16kj=1Ij(H,T)+R1(H,T)+R2(H,T)R3(H,T)+O(H2+3/k+ε/T), (3.2)

    where Ij(H,T) is given by (3.1) and

    R1(H,T):=16(2πi)2b1+iTb1iTb2+10iT12k+1+10iTFk(s,w)Hs+2wwsdwds,R2(H,T):=16(2πi)2b1+iTb1iT12k+1+10iT12k+110iTFk(s,w)Hs+2wwsdwds,R3(H,T):=16(2πi)2b1+iTb1iTb210iT12k+110iTFk(s,w)Hs+2wwsdwds.

    Set s=b1+it with TtT and w=u+10iT with 1/2k+1ub2. By Proposition 2.2, one has Hk(s,w)1, since the conditions in (2.6) are satisfied clearly. On the other hand, recall that LH states

    ζ(1/2+it)(|t|+1)ε,

    then we deduce from the Phragmén-Lindelöf principle that the subconvexity bound of the Riemann zeta function under LH satisfies

    ζ(σ+it){1, if σ>1,(|t|+1)max{12σ,0}+ε, if 0<σ1. (3.3)

    This will be used several times below, and it follows that

    Fk(b1+it,u+10iT)Tε

    holds uniformly for 1/2k+1ub2. This implies

    R1(H,T)TTdt|t|+1b212k+1T1+εHb1+2uduH2+3/k+ε/T,

    and the same bound holds for R3(H,T). Similarly, for u=1/2k+1, we have

    Fk(b1+it,1/2k+1+iv)(|t|+|v|+1)ε,

    and R2(H,T) can be estimated as

    R2(H,T)Hb1+2(1/2k+1)TT(|t|+1)1+εdt10T10T(|v|+1)1+εdvH2+2/k+ε.

    Inserting the upper bounds of Ri(H,T) (i=1,2,3) into (3.2), we obtain the required formula.

    In this subsection, we shall evaluate Ij(H,T) for 1jk, and our main idea is Cauchy's residue theorem.

    Lemma 3.2. Let k4 and Ij(H,T) for 1jk be defined as in (3.1), then we have the following estimates under LH for each Ij(H,T):

    Ij(H,T)H2+2k+j22k(kj1)+ε+H2+3k+ε/T,for1j<2k/3,I2k/3(H,T)=C2k/3H2+3/k+O(H2+3k+ε/T),Ij(H,T)=H2+3/kPj(logH)+O(H2+32k+2k32k(j1)+ε+H2+3k+ε/T),for2k/3<j<k,Ik(H,T)=H2+3/kPk(logH)+O(H2+2k+k22k(k1)+εT12(k1)+H2+3k+ε/T),

    where C3k/2, Pj(logH) for 2k/3<j<k and Pk(logH) are defined in (3.6), (3.9), and (3.12), respectively.

    Remark 3.3. In fact, the above result still holds for k=3 apart from the third formula, since the third case vanishes if k=3.

    Proof. Recall the definition of Ij(H,T) in (3.1). We deduce from Proposition 2.2 that

    resw=wjFk(s,w)ws=Hk(s,wj)jwjs0kjζ(j+(1j)ks)=:Gj(s).

    It follows that

    Ij(H,T)=12πib1+iTb1iTGj(s)H2+2j+(32kj)sds.

    The proof of this lemma is divided into four parts based on whether 32k/j is positive, negative, or zero. Keep in mind that the letter j is always an integer.

    If 1j<2k/3, the condition (2.6) becomes

    min1ik1((ki)s+i(wj1))1/2+ε,

    which implies

    (s)<(2k2j)/2k(k1j).

    So, we shift the line of integration from b1 to

    (s)=σj=(2k2j)/2k(k1j)ε.

    By Proposition 2.2, the Euler product Hk(s,wj) is absolutely convergent in the region b1(s)σj, and we have Hk(s,wj)1. It follows from this and (3.3) that

    Gj(σ+it)|Hk(s,wj)|(|t|+1)20kj|ζ(j+(1j)k(σ+it))||ζ(kj+(1kj)k(σ+it))|/(|t|+1)2ε(|t|+1)1/2(k1j)2+ε(|t|+1)3/2+ε (3.4)

    for b1σσj, since the real part of each zeta function is

    (/j+(1/j)ks)>1/2

    for 0k1 and j. The above bound allows us to extend the integral of Ij(H,T) on [T,T] to (,), so we have

    Ij(H,T)=12πi(b1)Gj(s)H2+2j(2kj3)sds+O(H2+3k+ε/T),

    where (b1) means b1+ib1i. Note that there is no pole in the region b1(s)σj. The residue theorem tells us that

    Ij(H,T)=12πi(σj)Gj(s)H2+2j(2kj3)sds+O(H2+3k+ε/T). (3.5)

    By (3.4), the integral on the righthand side can be estimated as

    H2+2j(2kj3)σj|Gj(σj+it)|dtH2+2k+j22k(kj1)+εdt(|t|+1)32εH2+2k+j22k(kj1)+ε.

    Combining this with (3.5), we obtain the first assertion in Lemma 3.2.

    If j=2k/3 is an integer, we find that

    Ij(H,T)=12πib1+iTb1iTGj(s)H2+2jds.

    Otherwise, this case does not exist. Note that

    G2k/3(s)(|t|+1)(2ε)

    for s=b1+it, which shows that G2k/3(b1+it,χ) has a good convergence as a function in t. It follows that

    I2k/3(H,T)=12πi(b1)G2k/3(s)H2+3kds+O(H2+3/k+ε/T)=:C2k/3H2+3/k+O(H2+3/k+ε/T),

    where

    C2k/3=12πi(1k+ε)G2k/3(s)ds (3.6)

    is an absolute constant. This gives the second assertion.

    If 2k/3<j<k, we shift the line of integration from b1 to

    (s)=σj=(j2)/2k(j1)+ε.

    It is easy to check that (2.6) is exactly satisfied, so we have Hk(s,wj)1. It follows from (3.3) again that

    Gj(σ+it)|Hk(s,wj)||wjs|0kj|ζ(j+(1j)k(σ+it))||ζ(k(σ+it))|(|t|+1)2(|t|+1)max{1/2kσ,0}2+ε(|t|+1)3/2+ε (3.7)

    for σj<σb1, which yields

    Ij(H,T)=12πi(b1)Gj(s)H2+2j+(32kj)sds+O(H2+3k+ε/T).

    Note that

    0<(/j+(1/j)ks)<1

    for 0<j, so there is only a pole s=1/k of order j given by Gj(s) in the strip σjσb1. The residue theorem gives us

    Ij(H,T)=ress=1/kGj(s)H2+3/k+12πi(σj)Gj(s)H2+2j+(32kj)sds+O(H2+3k+ε/T). (3.8)

    We can compute the residue in the main term as

    ress=1/kGj(s)=1(j1)!lims1/kdj1dsj1((s1k)jGj(s)Hs1k)=:Pj(logH). (3.9)

    This is a polynomial of degree j1 with leading coefficient

    Hk(1k,1k+1)(j/k)j2(k+1)((j1)!)2.

    The integral in (3.8) can be bounded by

    H2+2j+(32k3)σj|Gj(σj+it)|dtH2+32k+2k32k(j1)+ε.

    Inserting (3.9) and the above estimate into (3.8), we get the third formula in Lemma 3.2.

    Finally, it remains to deal with the case j=k. Now, we shift the line of integration from b1 to

    (s)=σk=(k2)/2k(k1)+ε.

    Noticing that wk=1/k+1 is no longer dependent on s, it follows from (3.3) that

    Gk(σ+it)=Hk(s,wk)kwks0<kζ(k+(1k)k(σ+it))|ζ(k(σ+it))|(|t|+1)1ε{(|t|+1)(1/2kσ)1+ε,if σkσ1/2k(|t|+1)1+ε,if 1/2k<σb1 (3.10)

    since Hk(s,wk) is absolutely convergent for σk(s)b1 according to (2.6). Note that there is only one pole s=1/k of order k given by Gk(s) in the rectangle formed by σk±iT and b1±iT. It follows from the residue theorem that

    Ik(H,T)=ress=1/kGk(s)H2+3/k+(σkiTb1iT+σk+iTσkiT+b1+iTσk+iT)Gk(s)H2+2k+sds=H2+3/kPk(logH)+K1(H,T)+K2(H,T)+K3(H,T), (3.11)

    where Ki(H,T) for i=1,2,3 corresponds to the above three integrals, respectively, and Pk is a polynomial of degree k1 given by

    Pk(logH)=1(k1)!lims1/kdk1dsk1((s1k)kGk(s)Hs1k) (3.12)

    with leading coefficient

    Ck=Hk(1k,1k+1)(k+1)((k1)!)2. (3.13)

    This is a positive constant depending on k. By (3.10), using the same method as before, we can estimate K1(H,T) by

    K1(H,T)12kσkH2+2/k+σT1/2+kσεdσ+b112kH2+2/k+σT1εdσH2+2k+k22k(k1)+ε/T2k32(k1)+H2+3k+ε/T,

    so does K3(H,T). As for K2(H,T), noting that

    Gk(σk+it)(|t|+1)(2k3)/2(k1)+ε,

    we get

    K2(H,T)H2+2k+σkTT(|t|+1)(2k3)/2(k1)+εdtH2+2k+k22k(k1)+εT12(k1).

    The last formula in Lemma 3.2 is obtained followed from the above two estimates and (3.11). This completes the proof.

    Combining Lemmas 3.1 and 3.2, we get

    Nk4(H)H2=16H3k(1j=2k/3C2k/3+2k/3<jkPj(logH))+O(H2k+ε+H3k+ε/T)+O(1j<2k/3H2k+j22k(kj1)+ε+2k/3<j<kH32k+2k32k(j1)+ε+H2k+k22k(k1)+εT12(k1))

    for k4. It can be simplified to

    Nk4(H)/H2=H3kP(logH)+O(H3k+ε/T+H32k+2k32k[2k/3]+ε+H2k+k22k(k1)+εT12(k1)),

    where [α] is the integral part of α and

    P(logH)=16(1j=2k/3C2k/3+2k/3<jkPj(logH)). (3.14)

    This is a polynomial of degree k1 with leading coefficient 16Ck. It suffices to choose a suitable T to balance the error terms. It can be observed that the error term is actually dominated by H32k+2k32k[2k/3]+ε for k4. Just letting H9/4k2TH1/k, one can get the second formula in Theorem 1.1.

    Following the proof of Theorem 1.1, we give a sketch of the proof for k=2. According to Proposition 2.2, the corresponding Dirichlet series can be written as

    F2(s,w)=ζ(2s)ζ(s+w1)ζ(2(w1))H2(s,w)

    with

    H2(s,w)=(1122w2)p(1+1ps+w+1ps+w1+1p2w1)(11p2w)1(11ps+w1),

    which is absolutely convergent and satisfies H2(s,w)1 in the region

    (s+w)3/2+ε,(w)1+ε. (4.1)

    Applying Perron's formula, we get

    N24(H)=16(2πi)2b1+iTb1iTb2+10iTb210iTF2(s,w)Hs+2wwsdwds+O(H2+3/2+εT) (4.2)

    for b1=1/2+ε and b2=3/2+ε.

    Following the arguments in Section 3, we still shift the path of integration over w to (w)=5/4. Clearly, (4.1) is satisfied. Applying the residue theorem to evaluate the inner integral in (4.2), we can get (3.2) for k=2. Note that the assumption of LH is used to bound Fk(s,w) and Gj(s) in some region before. Unconditionally, we have the following well-known estimate for the Riemann zeta function (see [13, Theorem Ⅱ.3.8] for example):

    ζ(σ+it){1, if σ>1,(|t|+1)1σ3+ε, if 1/2σ1,(|t|+1)34σ6+ε, if 0<σ<1/2. (4.3)

    Just replacing (3.3) by (4.3) and using the same method as in §3.1, we can get

    N24(H)=16(I1(H,T)+I2(H,T))+O(H2+32+ε/T+H3+εT12) (4.4)

    with

    I1(H,T)=12πib1+iTb1iTζ(2s)ζ(22s)(2s)sH2(s,2s)H4sds,I2(H,T)=12πib1+iTb1iTζ(2s)ζ(s+1/2)3sH2(s,3/2)H3+sds.

    The treatments of I1(H,T) and I2(H,T) are a little different from before. There are only two Riemann zeta functions involved here, which can be treated more carefully. As for I1(H,T), we shift the line of integration from b1 to σ1=1ε due to (4.1). It follows from (4.3) that

    ζ(2s)ζ(22s)H2(s,2s)|ζ(22s)|(|t|+1)max{(2σ1)/3,(8σ5)/6}+ε

    for s=σ+it and b1σσ1. Noticing that there is no pole for the integrand function in this strip, similar to (3.5), we deduce from the residue theorem that

    I1(H,T)=12πi(σ1)ζ(2s)ζ(22s)(2s)sH2(s,2s)H4sds+O(H2+32+ε/T).

    For s=σ1+it, the integral above can be bounded by

    H4σ1ζ(2ε2it)(|t|+1)2dtH3+εdt(|t|+1)3/2εH3+ε.

    It follows that

    I1(H,T)H3+ε+H2+3/2+ε/T. (4.5)

    It suffices to deal with I2(H,T). In view of (4.1), we shift the path of integration from b1±iT to ε±iT. Note that there is a pole s=1/2 of order 2 in this region. It follows from the residue theorem that

    I2(H,T)=H2+32P(logH)+(εiTb1iT+ε+iTεiT+b1+iTε+iT)ζ(2s)ζ(s+1/2)3sH2(s,3/2)H3+sds, (4.6)

    where P is a linear polynomial given by

    P(logH)=lims1/2dds(13s(s1/2)2ζ(2s)ζ(s+1/2)H2(s,3/2)Hs12)=13H2(1/2,3/2)logH+(γ2/3)H2(1/2,3/2)+13H2(1/2,3/2) (4.7)

    and γ is Euler's constant. We derive from (4.3) that the first and third integrals in (4.6) can be bounded by

    1/4εT25σ31+εH3+σdσ+b11/4T1/2σ1+εH3+σdσH3+ε/T13+H3+14+ε/T34+H3+12+ε/T.

    Recall the fourth moment estimate of the Riemann zeta function (see [14, Theorem 5.1])

    TT|ζ(σ+it)|4dtTlog4T

    for 1/2σ<1 and the functional equation, which states

    ζ(s)=χ(s)ζ(1s)withχ(s)(|t|+1)1/2σ

    for σ1/2. By Hölder's inequality and partial integration, we can estimate the second integral in (4.6) as follows

    H3+εTT|ζ(2ε+2it)ζ(1/2+ε+it)||t|+1dtH3+ε(TT|ζ(2ε+2it)|4|t|+1dt)1/4(TT|ζ(1/2+ε+it)|4|t|+1dt)1/4(TTdt|t|+1)1/2H3+ε(TT|(|t|+1)1/22εζ(12ε2it)|4|t|+1dt)1/4H3+εT1/2. (4.8)

    Combining all the above, we get

    I2(H,T)=H2+32P(logH)+O(H3+εT1/2+H3+1/4+ε/T3/4+H3+1/2+ε/T). (4.9)

    Finally, inserting (4.5) and (4.9) into (4.2), we get

    N24(H)=16H2+32P(logH)+O(H2+32+ε/T+H3+εT12+H3+14+ε/T34).

    Choosing T=H1/3, the error terms are balanced to H2+3/21/3+ε. This gives the required formula in Theorem 1.3.

    In this paper, we study the number of integers which form perfect powers in the way of

    x(y21+y22+y23+y24)=zk

    and the proof relies on techniques coming from complex analysis. We point out that this problem was never done before except for the case k=3. It is not easy to establish a unified asymptotic formula with power-saving error terms for large k, so we assume the Lindelöf hypothesis for k4. Theorem 1.1 gives an asymptotic formula with a power-saving error term for the number of such integers of bounded size under LH. This is the novelty of this paper. Moreover, Theorem 1.3 gives an unconditional result for k=2.

    The author declares she 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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