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Let In be the set of all idempotent elements of Zn

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Problem 11591

(American Mathematical Monthly, Vol.118, August-September 2011) Proposed by Dan White and Lenny Jones (USA).

Let In be the set of all idempotent elements of Zn. That is, e ∈ In if and only if e2= e (mod n).

Let In1 = In, and for k ≥ 2, let Ink be the set of all sums of the form u + v where u ∈ In, v ∈ Ink−1 , and the addition is done modulo n. Determine, in terms of n, the least k such that Ink= Zn. Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

We first show that if n is a power of a prime p then the only idempotents in Znare 0 and 1. Suppose that x 6= 0 is a solution of x2 ≡ x (mod pe). Let x = pab, where 0 ≤ a < e and gcd(b, p) = 1.

Then b(pab − 1) ≡ 0 (mod pe−a) which yields that pab ≡ 1 (mod pe−a). Thus a = 0 and x = b ≡ 1 (mod pe). It follows at once that for such n, Ink= {0, 1, . . . , k} for k = 1, . . . , n − 1 and Ink= Zn as soon as k ≥ n − 1.

In the general case, where the factorization of n is pe11pe22· · · perr, by the Chinese Remainder Theorem Ink = Ipke1

1 × Ipke2

2 × · · · × Ipker r . So, Ink= Znwhen Ipkei

i = Zpeii for all i = 1, . . . , r, which implies that the least k such that the desired condition is fulfilled is

min{peii : i = 1, . . . , r} − 1.



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