O ct 2 00 7 Regular integers modulo n László Tóth ( Pécs , Hungary ) October

Let n = p ν 1 1 · · · p νr r > 1 be an integer. An integer a is called regular (mod n) if there is an integer x such that a 2 x ≡ a (mod n). Let ̺(n) denote the number of regular integers a (mod n) such that 1 ≤ a ≤ n. Here ̺(n) = (φ(p ν 1 1) + 1) · · · (φ(p νr r) + 1), where φ(n) is the Euler function. In this paper we first summarize some basic properties of regular integers (mod n). Then in order to compare the rates of growth of the functions ̺(n) and φ(n) we investigate the average orders and the extremal orders of the functions ̺(n)/φ(n), φ(n)/̺(n) and 1/̺(n). 1. Introduction Let n > 1 be an integer. Consider the integers a for which there exists an integer x such that a 2 x ≡ a (mod n). In the background of this property is that an element a of a ring R is said to be regular (following J. von Neumann) if there is an x ∈ R such that a = axa. In case of the ring Z n this is exactly the condition of above. Properties of these integers were investigated by J. Morgado [7], [8], who called them regular (mod n). In a recent paper O. Alkam and E. A. Osba [1] using ring theoretic considerations rediscovered some of the statements proved elementarly by J. Morgado. It was observed in [7], [8] that a > 1 is regular (mod n) if and only if the gcd (a, n) is a unitary divisor of n. We recall that d is said to be a unitary divisor of n if d | n and gcd (d, n/d) = 1, notation d || n. These integers occur in the literature also in an other context. It is said that an integer a possesses a weak order (mod n) if there exists an integer k ≥ 1 such that a k+1 ≡ a (mod n). Then the weak order of a is the smallest k with this property, see [4], [2]. It turns out that a is regular (mod n) if and only if a possesses a weak order (mod n). Let Reg n = {a : 1 ≤ a ≤ n, a is regular (mod n)} and let ̺(n) = # …