Open Access
Issue
Wuhan Univ. J. Nat. Sci.
Volume 27, Number 1, March 2022
Page(s) 49 - 52
DOI https://doi.org/10.1051/wujns/2022271049
Published online 16 March 2022

© Wuhan University 2022

Licence Creative CommonsThis is an Open Access article distributed under the terms of the Creative Commons Attribution License (https://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

0 Introduction

Let p be an odd prime. Let Fq be a finite field of q elements with q=ps,s1 and Fq* denote the set of all the nonzero elements of Fq. Let N(f = b) denote the number of solutions of the equation f(x1,x2,,xn)=b in Fqn=Fq××Fq, where f(x1,x2,,xn) is a polynomial in Fq[x1,,xn] and bFq. That is,N(f=b)=#{(x1,x2,,xn)Fqn}| f(x1,x2,,xn)=0}Studying the value of N(f = b) is one of the main topics in finite fields. Generally speaking, it is nontrivial to give the formula for N(f = b). Finding the explicit formula for N(f = b) under certain condition has attracted lots of authors for many years.

Markoff-Hurwitz-type equations belong to the following type of the Diophantine equationsx12+x22++xn2=ax1x2xnwhere n, a are positive integers and n3. This type of equations was first studied by Markoff[1] for the case n=3,​ a=3. More generally, these equations were studied by Hurwitz[2].

Recently, Baoulina[3-5] studied the generalized Markoff-Hurwitz-type equationsa1x1m1+a2x2m2++anxnmn=ax1x2xnwhere ai,aFq* and mi are positive integers satisfying mi|(q1) for i=1,,n and n2. Baoulina[4-6] and Pan et al[7] considered the further generalized Markoff- Hurwitz-type equations of the form:(a1x1m1+a2x2m2++anxnmn)k=ax1k1x2k2xnkn(1)where n2, mi,ki,k are positive integers, a,aiFq*, for i=1,,n. The special case (1) of k=1 is investigated by Cao[8]. Song and Chen[9] presented the formulas for the number of solutions of the following equationsx1m1+x2m2++xnmn=ax1x2xtover the finite field Fq under some certain restrictions, where n2,t>n, mi|(q1) for i=1,,n and aFq*

Hu and Li[10] consider the rational points of the further generalized Markoff-Hurwitz-type equations of the form(a1x1m1+a2x2m2++anxnmn)k=ax1k1x2k2xtktover the finite field Fq under some certain cases, where n2, mi, kj, k and t>n are positive integers, ai, aFq*, for 1in, 1jt.

In this paper, we consider the number of solutions of the following equations(x1m1+x2m2++xnmn)k=x1x2xnxn+1kn+1xtkt(2)over the finite field Fq under some other restrictions, where n2, t>n, k, kj(n+1jt),mi(1in) are positive integers. In what follows, we always letdj=gcd(mj,q1), 1jn, M=lcm[m1,,mn],D=lcm[d1,,dn],d0=gcd(d,k),d=gcd(i=1nM/mikM,(q1)/D).

For any positive integers ν1,ν2,,νr, we let I(ν1,ν2,,νr) denote the number of r-tuples(j1,j2,,jr) of integers with 1jiνi1(1ir), such that j1/v1+j2/v2++jr/vr is an integer. Denote by Nq the number of solutions of (2) in Fqn. Our main result is the following theorem.

Theorem 1   Suppose that gcd(kn+1,,kt=d,dD>2 and there is a positive integer l such that dD|(pl+1), with l as chosen minimal. Then 2l|s andNq=qt1+(1)((s/2l)1)nqtn/21(q1)I(d1,,dn)+qtn((1)n1+(1)n1r=2n((1)rs/2lqr/2)1j1jrnI(dj1,,djr)+(1)((s/2l)1)(n1)d1dnDq(n1)/2(dd0)(1)((s/2l)1)nd1dnDq(n2)/2 (d01))

This paper is organized as follows. In Section 1, we review some useful known lemmas which will be needed later. Subsequently, in Section 2, we prove Theorem 1. Some interesting applications of Theorem 1 will be provided as corollaries at the end of this paper.

1 Preliminary Lemmas

In this section, we present some useful lemmas that are needed in the proof of Theorem 1 as follows.

Lemma 1  [9,11] For any positive integer m, the number of elements of m-th power in Fq* is q1m.

Lemma 2  [10] Let t1,t2,,tr be positive integers and t=gcd(t1,t2,,tr,q1). Then for any elements a,αFq*,  we haveN(ax1t1xrtr=α) =N(a(x1xr)t=α) ={t(q1)r1,  if  a1α is a t-th power in Fq*0,                     otherwise

The following two lemmas are the main results in Ref.[6] and fundamental for our results.

Lemma 3  [6] Let n>2. Suppose that there is a positive integer l such that 2l|s and dD|(pl+1). ThenN(x1m1++xnmn=0)=qn1+(1)((s/2l)1)nq(n2)/2(q1)I(d1,,dn)

Lemma 4  [6] Suppose that dD>2 and there is a positive integer l such that dD|(pl+1), with l chosen minimal. Then 2l|s andN((x1m1++xnmn)k=ax1xn)=qn1+(1)n1+(1)((s/2l)1)nq(n2)/2(q1)I(d1,,dn)       +(1)n1r=2n(1)rs/2lqr/21j1jrnI(dj1,,djr)       +(1)((s/2l)1)(n1)d1dnDq(n1)/2T1       (1)((s/2l)1)nd1dnDq(n2)/2T2whereT1={dd0, if a is a d-th power in Fq  d0,  if a is a d0-th power but not             a d-th power in Fq     0,    if a is not a d0-th power in Fq andT2={d01, if a is a d0-th power in Fq 1,    if a is not a d0-th power in Fq 

2 Proof of Theorem 1

In this section, we give the proof of Theorem 1.

Proof of Theorem 1   Let N¯q (resp. N˜q) denote the number of the solutions of the equations (x1m1+x2m2++xnmn)k=x1x2xnxn+1kn+1xtkt with xn+1kn+1xtkt=0 (resp. xn+1kn+1xtkt0). Clearly, one hasNq=N¯q+N˜q(3)

Then we can solve the problem in two cases. One is xn+1kn+1xtkt=0 and the other one is xn+1kn+1xtkt0.

Case (i) xn+1kn+1xtkt=0. Then one hasN(xn+1kn+1xtkt=0)=N(xn+1xt=0)                             =j=1tn(tnj)(q1)tnj                             =qtn(q1)tn  (4)

Using the assumption there is a positive integer l such that 2l|s and dD|(pl+1). Thus, by (4) and Lemma 3 ,N¯q=(qtn(q1)tn)N((x1m1++xnmn)k=0)     =(qtn(q1)tn)N(x1m1++xnmn=0)     =(qtn(q1)tn)(qn1+(1)((s/2l)1)nq(n2)/2         ×(q1)I(d1,,dn))(5)

Case (ii) If xn+1kn+1xtkt0, we let δ=xn+1kn+1xtkt. DefineU:={βFq*:β be a d-th power in Fq*}

Note that gcd(kn+1,,kt,q1) =d. Then from Lemma 2, we can deduce thatN˜q=N((x1m1+x2m2++xnmn)k=δx1x2xn)     =d(q1)tn1    δUN((x1m1+x2m2++xnmn)k=δx1x2xn)(6)

Noting that integer l such that dD|(pl+1), with l chosen minimal. Thus for any given δU, from Lemma 1 and Lemma 3, one hasδUN((x1m1++xnmn)k=δx1xn)= ​q1d​ (qn1+​ (1)n1+(1)((s/2l)1)nq(n2)/2(q​ ​ 1)I(d1,,dn)       +(1)n1r=2n(1)rs/2lqr/21j1jrnI(dj1,,djr)       +(1)((s/2l)1)(n1)d1dnDq(n1)/2(dd0)       (1)((s/2l)1)nd1dnDq(n2)/2(d01))(7)

Then by (6) together with (7), we haveN˜q=(q1)tn(qn1+(1)n1+(1)((s/2l)1)n       q(n2)/2(q1)I(d1,,dn)+(1)n1       r=2n(1)rs/2lqr/21j1jrnI(dj1,,djr)       +(1)((s/2l)1)(n1)d1dnDq(n1)/2(dd0)       (1)((s/2l)1)nd1dnDq(n2)/2(d01))(8)

The desired result can follow immediately from (3), (5) and (8). This ends the proof of Theorem 1.

To conclude this section, we present some corollaries. It is clear that I(d1,,dn) plays a central role in Theorem 1. For any positive integers ν1,ν2,,νr, Sun and Wan[12] showed that I(ν1,ν2,,νr)=0 if and only if either gcd(νj,ν2,,νr/νj)=1 for some j or t is odd, ω1/2,,ωt/2 are pairwise coprime, and each ωj is coprime with any odd number in {ν1,ν2,,νr}, where {ω1,,ωt} is the set of even integers among ν1,ν2,,νr. In Ref.[13], they also showed that I(ν1,ν2,,νr)=1 if and only if 2|r,ν1/2,,νr/2 are pairwise coprime, and at least (r-1) of the νj/2 are odd.

Therefore we can easily deduce the following corollary.

Corollary 1   Suppose that d1,,dm are odd, dm+1,,dn are even, d1,,dm,dm+1/2,,dn/2 are pairwise coprime, 0mn. Under the conditions of Theorem 1, we haveNq=qt1+T3+(q1)tn((1)n1    +(1)n1r=2,2|rnm(nmr)qr/2​ +(1)((s/2l)1)(n1)q(n1)/2(dd0)    (1)((s/2l)1)nq(n2)/2(d01))T4whereT3={qtn/21(q1), if m=0 and n is even                0,     otherwiseandT4={1,          if m=n2nm1 ,   if m<n 

Let v be a positive integer. It is also known (Ref. [14], Proposition 6.17) thatI(v,,v)r=(v1)r+(1)r(v1)v

Then we have the second corollary.

Corollary 2   Suppose that d1==dn=D. Under the conditions of Theorem 1, we haveNq=qt1+(1)((s/2l)1)nqtn/21(q1)       (D1)n+(1)n(D1)D+(q1)tn       ((1)n1m=0n(1)ms/2lqm/2(D1)m+(1)m(D1)D       +(1)((s/2l)1)(n1)Dn1q(n1)/2(dd0)       (1)((s/2l)1)nDn1q(n2)/2(d01))

Clearly, Corollaries 1-2 are some special cases of Theorem 1. For example, consider the further generalized Markoff-Hurwitz-type equation over F32(x15+x22+x36)2=x1x2x3x42x54(9)

Clearly m1=5,m2=2,m3=6,k=2,k4=2,k5=4. Then we get d1=gcd(5,8)=1,d2=gcd(2,8)=2,d3=gcd(6,8)=2,D=lcm(d1,d2,d3)=2,M=lcm(5, 2, 6)=30, d0=gcd(d,k)=2 and d=gcd(i=13M/mikM,(q1)/D)=2. One can immediately conclude that (9) has 6 817 solutions in F32 by Corollary 1.

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