Let R be a commutative ring with the identity 1R, and let R∗ be the multiplicative group of units in R. An element a∈R∗ is called an exceptional unit if there exists a b∈R∗ such that a+b=1R. We set R∗∗ to be the set of all exceptional units in R. In this paper, we consider the residue-class ring Zn. For any positive integers n,s, and c∈Zn, let Ns(n,c):=♯{(x1,...,xs)∈(Z∗∗n)s:x1+...+xs≡c(modn)}. In 2016, Sander (J.Number Theory 159 (2016)) got a formula for N2(n,c). Later on, Yang and Zhao (Monatsh. Math. 182 (2017)) extended Sander's theorem to finite terms by using exponential sum theory. In this paper, using matrix theory, we present an explicit formula for Ns(n,c). This extends and improves earlier results.
Citation: Junyong Zhao. Counting sums of exceptional units in Zn[J]. AIMS Mathematics, 2024, 9(9): 24546-24554. doi: 10.3934/math.20241195
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Let R be a commutative ring with the identity 1R, and let R∗ be the multiplicative group of units in R. An element a∈R∗ is called an exceptional unit if there exists a b∈R∗ such that a+b=1R. We set R∗∗ to be the set of all exceptional units in R. In this paper, we consider the residue-class ring Zn. For any positive integers n,s, and c∈Zn, let Ns(n,c):=♯{(x1,...,xs)∈(Z∗∗n)s:x1+...+xs≡c(modn)}. In 2016, Sander (J.Number Theory 159 (2016)) got a formula for N2(n,c). Later on, Yang and Zhao (Monatsh. Math. 182 (2017)) extended Sander's theorem to finite terms by using exponential sum theory. In this paper, using matrix theory, we present an explicit formula for Ns(n,c). This extends and improves earlier results.
Let R be a commutative ring with the identity 1R, and let R∗ be the multiplicative group consisting of all the units in R. An element a∈R∗ is said to be an exceptional unit if 1R−a∈R∗, i.e., if a−1R∈R∗, or, in other words, if there exists a b∈R∗ satisfying a+b=1R. In 1969, exceptional units were first introduced by Nagell [7] to study certain cubic diophantine equations. From then on, many types of diophantine equations have been studied by means of exceptional units, for example, Thue equations [15], Thue-Mahler equations [16], and discriminant form equations [12].
Exceptional units also became a useful tool in number theory. For example, in 1977, Lenstra [5] introduced a new method to find Euclidean number fields by using exceptional units. Later on, many new Euclidean number fields were found with this method (see [4,6]). Furthermore, exceptional units also have connections with cyclic resultants [13,14] and Lehmer's conjecture related to Mahler's measure[10,11].
Let Z, Z+, and P be the sets of integers, positive integers, and primes, respectively. For n∈Z+, let Zn={0,1,…,n−1} be the ring of residue classes modulo n. By definition, one has Z∗n={a∈Zn:gcd(a,n)=1}. In this note, we set Z∗∗n to be the set of all exceptional units in Zn, i.e., Z∗∗n:={a∈Zn:gcd(a,n)=gcd(a−1,n)=1}. Given p∈P, we denote by νp(n) the p-adic valuation of n, i.e., νp(n) is the unique nonnegative integer r satisfying pr|n and pr+1∤n. Moreover, we let ξn stand for the primitive n-th root of unity, i.e., ξn:=e2πi/n.
In 2010, Harrington and Jones [3] obtained the following identity:
♯Z∗∗n=n∏p|n,p∈P(1−2p). |
This result can also be deduced immediately from the theorems of Deaconescu [2] or Sander [8]. By the definition of an exceptional unit, we can see that
♯Z∗∗n=♯{(u,v)∈(Z∗n)2:u+v≡1(modn)}. |
For c∈Zn, in 2009, it was proved by Sander [8] that
♯{(u,v)∈(Z∗n)2:u+v≡c(modn)}=n∏p∈Pp|n,p|c(1−1p)∏p∈Pp|n,p∤c(1−2p). |
In this paper, we shall describe the elements in Zn, which could be written as the sum of one or more expected units. In addition, for these elements, we will derive the number of representations as such a sum. More specifically, for n,s∈Z+, and c∈Zn, we set
Ns(n,c):=♯{(x1,...,xs)∈(Z∗∗n)s:x1+...+xs≡c(modn)}. |
In 2016, Sander [9] presented an explicit formula for N2(n,c). Now, we state Sander's theorem as follows:
Theorem 1.1. (Sander [8]) Given n,k∈Z+ and c∈Zn. The number N2(n,c) satisfies the following relations:
N2(2k,c)=0,N2(3k,c)={3k−1if c≡1(mod3),0otherwise, |
while for all primes p⩾5,
N2(pk,c)={pk−1(p−2)ifc≡1(modp),pk−1(p−3)if c≡0(modp)orc≡2(modp),pk−1(p−4)otherwise. |
Let ω(n):=∑p|n,p∈P1 be the number of distinct prime divisors of n. In 2017, Yang and Zhao [17] extended Sander's theorem to finite terms by means of exponential sums, as below.
Theorem 1.2. (Yang and Zhao [17]) For n,s∈Z+⩾2 and c∈Zn, we have
Ns(n,c)=(−1)sω(n)∏p|n,p∈Ppνp(n)(s−1)−s(ps∑j=0j≡c(modp)(sj)+(2−p)s−2s). |
In this paper, by using matrix theory, we give the following two results:
Theorem 1.3. Let p∈P,s∈Z+, and let ξj:=e2πi/j. Then
(Ns(p,0)Ns(p,1)⋮Ns(p,p−1))=1p((p−2)s+p−1∑j=1(−1−ξ−1j)s(p−2)s+p−1∑j=1ξj(−1−ξ−1j)s⋮(p−2)s+p−1∑j=1ξ(p−1)j(−1−ξ−1j)s). |
The second main result of this paper is the following corollary:
Corollary 1.1. Let n,s∈Z+⩾2 and c∈Zn. We have
Ns(n,c)=∏p|n,p∈Pp(vp(n)−1)(s−1)Ns(p,c), |
where Ns(p,c) is determined by Theorem 1.3.
This paper is organized as follows: Section 2 provides several lemmas that are needed in the proof of Theorem 1.3 and Corollary 1.1. Then we give the proofs of Theorem 1.3 and Corollary 1.1 in Section 3.
In this section, we supply several lemmas that will be needed in the proof of Theorem 1.3 and Corollary 1.1. We begin with the following result, which can be proved by using the Chinese remainder theorem:
Lemma 2.1. [1] Let k,s∈Z+, f(x1,...,xs)∈Z[x1,...,xs], and let m1,...,mk be pairwise relatively prime positive integers. For any integer j with 1≤j≤k, let Nj be the number of zeros of f(x1,...,xs)≡0(modmj), and let N denote the number of zeros of f(x1,...,xs)≡0(mod∏kj=1mj). Then N=∏kj=1Nj.
Lemma 2.2. Let k∈Z+,p∈P. For any integer c, we have Ns(pk+1,c)=ps−1Ns(pk,c).
Proof. Let (b1,⋯,bs) be a solution of x1+⋯+xs≡c(modpk), with bj (1≤j≤s) being exceptional units. One has gcd(bj,p)=1. Let b1+⋯+bs−c=apk for some a∈Z. For k1,⋯,ks∈Zpk, the congruence
(b1+k1pk)+⋯+(bs+kspk)≡c(modpk+1) |
holds if and only if
a+k1+⋯+ks≡0(modp). | (2.1) |
Clearly, the number of solutions to (2.1) is ps−1.
Thus, one get Ns(pk+1,c)=ps−1Ns(pk,c).
In this paper, we view vector v as a column vector and vT as the transpose of v. For a∈Z, we let <a>m denote the unique integer r such that r≡a(modm) with 0⩽r⩽m−1.
Definition 2.1. Let v=(a0,⋯,am−1)T be a complex vector. The circulant matrix Av associated with v is a m×m complex matrix having the form
Av=(a0a1⋯am−1am−1a0⋯am−2⋮⋮⋮a1a2⋯a0). |
In other words, if we let Av=(Ai,j), then Ai,j=a<j−i>m.
Lemma 2.3. Let Av be a circulant matrix associated to the vector v=(a0,⋯,am−1)T, and let f(x)=∑m−1i=0aixi. Then, for each j=0,1,⋯,m−1, f(ξjm) is an eigenvalue of Av and vj=(1,ξjm,ξ2jm,⋯,ξj(m−1)m)T is an eigenvector corresponding to f(ξjm).
Proof. Let ω be any m-th root of unity. Set
α=(1ω⋮ωm−1). |
Consider
Avα=(a0a1⋯am−1am−1a0⋯am−2⋮⋮⋮a1a2⋯a0)(1ω⋮ωm−1):=(b1b2⋮bm). |
Clearly,
b1=a0+a1ω+a2ω2+⋯+am−2ωm−2+am−1ωm−1=f(ω). |
For any k≥2, one has
bk=am−k+1+am−k+2ω+⋯+am−1ωk−2+a0ωk−1+a1ωk−2+⋯+am−kωm−1=(am−k+1ωm−k+1+am−k+2ωm−k+2+⋯+am−kωm−k)ωk−1=f(ω)ωk−1. |
Therefore, we obtain that
Avα=(f(ω)f(ω)ω⋮f(ω)ωm−1)=f(ω)α. |
In particular, take ω=ξjm, where j runs from 0 to m−1. It then follows that f(ξjm) is an eigenvalue and
(1ξjm⋮ξj(m−1)m) |
is an eigenvector corresponding to f(ξjm) for each j=0,1,⋯,m−1.
This completes the proof of Lemma 2.3.
Lemma 2.4. Let k be a nonnegative integer and m be a positive integer. Then
m−1∑j=0ξkjm={m, if m∣k,0, if m∤k. |
Proof. First, if m∣k, then ξkjm=1 for any integer j. So
m−1∑j=0ξkjm=m−1∑j=01=m. |
Next, we let m∤k. Then k=qm+r for 0<r<m. Then one has
ξkm=ξqm+rm=(ξmm)qξrm=ξrm≠1. |
It follows that
m−1∑j=0ξkjm=m−1∑j=0ξrjm=ξmrm−1ξrm−1=1−1ξrm−1=0. |
The proof of Lemma 2.4 is complete.
We also need the following result, which can be found in any standard linear algebra textbook.
Lemma 2.5. Let A be a m×m matrix. Let λ1,λ2,⋯,λm be all the eigenvalues of A, and αj be an eigenvector corresponding to λj for every 1⩽j⩽m. If α1, α2,⋯,αm are linearly independent, then Q−1AQ=diag(λ1,λ2,⋯,λm) with Q=(α1,α2,⋯,αm).
Lemma 2.6. Let V be a Vandermonde matrix of the form
(111⋯11ξmξ2m⋯ξm−1m1ξ2mξ4m⋯ξ2(m−1)m⋮⋮⋮ ⋮1ξm−1mξ2(m−1)m⋯ξ(m−1)(m−1)m). |
Then V is invertible, and
V−1=1m(111⋯11ξm−1mξ2(m−1)m⋯ξ(m−1)(m−1)m1ξm−2mξ2(m−2)m⋯ξ(m−2)(m−1)m⋮⋮⋮ ⋮1ξmξ2m⋯ξm−1m). |
Proof. The proof follows from a direct calculation.
Proof of Theorem 1.3. Let (x1,...,xs)∈(Z∗∗n)s. It then follows that xs≠0 and xs≠1. Since
Ns(n,c):=♯{(x1,...,xs)∈(Z∗∗n)s:x1+...+xs≡c(modn)}, |
it is easy to see that for any integer i with 0≤k≤p−1, one has
Ns(p,k)=p−1∑j=0j≠k,<k−1>pNs−1(p,j). |
That is,
(Ns(p,0)Ns(p,1)⋮Ns(p,p−1))=(01⋯1000⋯11⋮⋮⋮⋮11⋯00)(Ns−1(p,0)Ns−1(p,1)⋮Ns−1(p,p−1)):=Av(Ns−1(p,0)Ns−1(p,1)⋮Ns−1(p,p−1)). |
It is clear that Av is a circulant matrix associated with the vector v=(0,1,⋯,1,0)T. For simplicity, we set ξ:=ξp in the following. Then ξ1:=ξ, ξ2=ξ2,⋯, ξp−1=ξp−1 are all the primitive p-th roots of unity. Let f(x)=x+x2+⋯+xp−2. By Lemma 2.3, for each j=0,1,⋯,p−1, f(ξj) is an eigenvalue of Av and vj=(1,ξj1,ξj2,⋯,ξjp−1)T is an eigenvector corresponding to the eigenvalue f(ξj).
Let
B=(v0,v1,⋯,vp−1)=(111⋯11ξ1ξ21⋯ξp−111ξ2ξ22⋯ξp−12⋮⋮⋮ ⋮1ξp−1ξ2p−1⋯ξp−1p−1). |
Since det(B)=∏0⩽i<j⩽p−1(ξj−ξi)≠0, one has v0,v1,⋯,vp−1 are linearly independent. By Lemmas 2.5 and 2.6, we have
B−1=1p(111⋯11ξp−11ξp−12⋯ξp−1p−11ξp−21ξp−22⋯ξp−2p−1⋮⋮⋮ ⋮1ξ1ξ2⋯ξp−1) |
and
Av=B diag(f(ξ0),f(ξ1),⋯,f(ξp−1)) B−1. |
Notice that N1(p,0)=N1(p,1)=0, N1(p,j)=1 for 2⩽j⩽p−1, and for 1⩽j⩽p−1,
f(ξj)=−1−ξ−j |
by Lemma 2.4. Therefore, one has
(Ns(p,0)Ns(p,1)⋮Ns(p,p−1))=B diag(f(ξ0),f(ξ1),⋯,f(ξp−1)) B−1(Ns−1(p,0)Ns−1(p,1)⋮Ns−1(p,p−1))=B diag(fs−1(ξ0),fs−1(ξ1),⋯,fs−1(ξp−1)) B−1(N1(p,0)N1(p,1)⋮N1(p,p−1))=B diag((p−2)s−1,(−1−ξ−1)s−1,⋯,(−1−ξ−p+1)s−1) B−1(00⋮1)=B diag((p−2)s−1,(−1−ξ−11)s−1,⋯,(−1−ξ−1p−1)s−1) B−1(00⋮1)=((p−2)s−1 (−1−ξ−11)s−1 ⋯ (−1−ξ−1p−1)s−1(p−2)s−1 ξ1(−1−ξ−11)s−1 ⋯ ξp−1(−1−ξ−1p−1)s−1⋮ ⋮ ⋮(p−2)s−1 ξp−11(−1−ξ−11)s−1 ⋯ ξp−1p−1(−1−ξ−1p−1)s−1) B−1(00⋮1)=1p((p−2)s−1 (−1−ξ−11)s−1 ⋯ (−1−ξ−1p−1)s−1(p−2)s−1 ξ1(−1−ξ−11)s−1 ⋯ ξp−1(−1−ξ−1p−1)s−1⋮ ⋮ ⋮(p−2)s−1 ξp−11(−1−ξ−11)s−1 ⋯ ξp−1p−1(−1−ξ−1p−1)s−1)(p−2−1−ξp−11⋮−1−ξ1)=1p((p−2)s−1 (−1−ξ−11)s−1 ⋯ (−1−ξ−1p−1)s−1(p−2)s−1 ξ1(−1−ξ−11)s−1 ⋯ ξp−1(−1−ξ−1p−1)s−1⋮ ⋮ ⋮(p−2)s−1 ξp−11(−1−ξ−11)s−1 ⋯ ξp−1p−1(−1−ξ−1p−1)s−1)(p−2−1−ξ−11⋮−1−ξ−1p−1)=1p((p−2)s+p−1∑j=1(−1−ξ−1j)s(p−2)s+p−1∑j=1ξj(−1−ξ−1j)s⋮(p−2)s+p−1∑j=1ξ(p−1)j(−1−ξ−1j)s). |
This completes the proof of Theorem 1.3.
Proof of Corollary 1.1. Let n=∏p|npvp(n) be the canonical decomposition of n. By Lemmas 2.1 and 2.2, we get
Ns(n,c)=∏p|nNs(pvp(n),c)=∏p|np(vp(n)−1)(s−1)Ns(p,c). |
This finishes the proof of Corollary 1.1.
In the current study, by means of matrix theory, we present an explicit expression for ♯{(x1,...,xs)∈(Z∗∗n)s:xk1+...+xks≡c(modn)} with k=1. Naturally, one will ask for the formula for ♯{(x1,...,xs)∈(Z∗∗n)s:xk1+...+xks≡c(modn)} with k>1. Moreover, exceptional units are interesting and deserve further research.
The author declares he has not used Artificial Intelligence (AI) tools in the creation of this article.
The author woulds like to thank the anonymous referees for their careful reading of the manuscript and helpful suggestions that improve the presentation of the paper.
The author declares that he have no conflict of interest.
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