Idempotents and Congruence $\boldsymbol{ax}\boldsymbol{ \equiv b\pmod n}$

Abstract

Alomair et al. (J Math Cryptol 4(2):121–148, 2010, Lemma 3.1) noticed the following result which seems not to appear previously explicitly in the literature: Given a nonzero $a \in \mathbb{Z}_{n}$, the ring of residues modulo n, such that gcd(a, n) = d | b, not only there exists an element $x \in \mathbb{Z}_{n}$ such that $x \cdot a \equiv b\pmod n$, but that there even exists an invertible element $x \in \mathbb{Z}_{n}^{{\ast}}$ such that $x \cdot a \equiv b\pmod n$. Their sufficient and necessary condition for this says that gcd(b∕d, n∕d) = 1 with d as above.A typical structure result on finite commutative semigroup says that the multiplicative semigroup of $\mathbb{Z}_{n}$ decomposes into the so-called maximal subsemigroups belonging to the idempotents of $\mathbb{Z}_{n}$. Each such semigroup contains a maximal subgroup having for its identity the corresponding idempotent. In general this subgroup is a proper subset of the maximal subsemigroup containing it. However, the group of elements of $\mathbb{Z}_{n}$ coprime to n is an example of the case when this maximal subsemigroup and the maximal subgroup coincide (both evidently belonging to the idempotent 1).In what follows we prove that if a congruence $x \cdot a \equiv b\pmod n$ is solvable there always exists a solution in the maximal semigroup belonging to the idempotent given by the divisor δ = gcd(b∕d, n∕d) and if δ is a unitary divisor of n then there even exists a solution in the maximal subgroup belonging to the idempotent given by δ.

Dedicated to the memory of Professor Wolfgang Schwarz