Let be a finite multiplicatively written commutative semigroup with identity . By a sequence over , we mean a finite unordered sequence of terms from where repetition is allowed. For a sequence T over we denote by the product of its terms and we say that T is a product-one sequence if . If is a finite abelian group, the Davenport constant of is the smallest positive integer such that every sequence T over of length has a nonempty product-one subsequence. The Davenport constant has mainly been studied for finite abelian groups but also in more general settings (we refer to      for work in the setting of abelian groups, to   for work in case of non-abelian groups, and to      for work in commutative semigroups).
In the present paper we study the Erdös-Burgess constant of which is defined as the smallest positive integer such that every sequence T over of length has a non-empty subsequence whose product is an idempotent of . Clearly, if happens to be a finite abelian group, then the unique idempotent of is the identity , whence . The study of for general semigroups is initiated by a question of Erdös and has found renewed attention in recent years (e.g.,      ). For a commutative unitary ring R, let be the multiplicative semigroup of the ring R, and the group of units of R, noticing that the group is a subsemigroup of the semigoup . We state our main result.
Theorem 1.1. Let be an integer, and let be the ring of integers modulon. Then
where is the number of primes occurring in the prime-power decomposition of n counted with multiplicity, and is the number of distinct primes. Moreover, if n is a prime power or a product of pairwise distinct primes, then equality holds.
Let be a finite multiplicatively written commutative semigroup with the binary operation *. An element of is said to be idempotent if . Let be the set of idempotents of . We introduce sequences over semigroups and follow the notation and terminology of Grynkiewicz and others (cf.  , Chapter 10] or   ). Sequences over are considered as elements in the free abelian monoid with basis . In order to avoid confusion between the multiplication in and multiplication in , we denote multiplication in by the boldsymbol and we use brackets for all exponentiation in . In particular, a sequence has the form
where are the terms of T, and is the multiplicity of the term a in T. We call the length of T. Moreover, if and , then has length has length , is a sequence of length 2. If and , then . Any sequence is called a subsequence of T if for every element , denoted . In particular, if , we call a proper subsequence of T, and let denote the resulting sequence by removing the terms of from T.
Let T be a sequence as in (1). Then
・ is the product of all terms of T, and
・ is the set of subsequence products of T.
We say that T is
・ a product-one sequence if ,
・ an idempotent-product sequence if ,
・ product-one free if ,
・ idempotent-product free if .
Let be an integer. For any integer , we denote the congruence class of modulo n. Any integer is said to be idempotent modulo n if , i.e., in . A sequence T of integers is said to be idempotent-product free modulo n provided that T contains no nonempty subsequence with being idempotent modulo n. We remark that saying a sequence T of integers is idempotent-product free modulo n is equivalent to saying the sequence is idempotent-product free in the multiplicative semigroup of the ring .
3. Proof of Theorem 1.1
Lemma 3.1. Let be a positive integer where , , and are distinct primes. For any integer , the congruence holds if and only if or for every .
Proof. Noted that if and only if divides for all , since , it follows that holds if and only if divides or , i.e., or for every , completing the proof.
Proof of Theorem 1. 1. Say
where are distinct primes and for all . It is observed that
taking a sequence V of integers of length such that
Now we show that the sequence is idempotent-product free modulo n, supposing to the contrary that contains a nonempty subsequence W, say , such that is idempotent modulo n, where is a subsequence of V and
It follows that
If , then is a nonempty subsequence of V. By (5) and (6), there exists some such that and . By Lemma 3.1, is not idempotent modulo n, a contradiction. Otherwise, for some , say
Since , it follows from (7) that . Combined with (8), we have that and . By Lemma 3.1, we conclude that is not idempotent modulo n, a contradiction. This proves that the sequence is idempotent-product free modulo n. Combined with (3) and (4), we have that
Now we assume that n is a prime power or a product of pairwise distinct primes, i.e., either or in (2). It remains to show the equality holds. We distinguish two cases.
Case 1. in (2), i.e., .
Taking an arbitrary sequence T of integers of length , let and . By the Pigeonhole Principle, we see that either or . It follows either , or . By Lemma 3.1, the sequence T is not idempotent-product free modulo n, which implies that . Combined with (9), we have that
Case 2. in (2), i.e., .
Taking an arbitrary sequence T of integers of length , by the Chinese Remainder Theorem, for any term of T we can take an integer such that for each ,
Note that and thus . Since , it follows that , and so there exists a nonempty subsequence W of T such that for each . Combined with (11), we derive that or , where . By Lemma 3.1, we conclude that is idempotent modulo n. Combined with (10), we have that . It follows from (9) that , completing the proof.
We close this paper with the following conjecture.
Conjecture 3.2. Let be an integer, and let be the ring of integers modulo n. Then .
This work is supported by NSFC (Grant No. 61303023, 11301381, 11501561).
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