A Note on Multiplicatively $E$-Perfect Numbers

Journal of Inequalities in Pure and Applied Mathematics http://jipam.vu.edu.au/

Volume 7, Issue 3, Article 99, 2006

A NOTE ON MULTIPLICATIVELY e-PERFECT NUMBERS LE ANH VINH AND DANG PHUONG DUNG SCHOOLOF MATHEMATICS UNIVERSITYOF NEW SOUTH WALES SYDNEY 2052 NSW [email protected]

SCHOOLOF FINANCEAND BANKING UNIVERSITYOF NEW SOUTH WALES SYDNEY 2052 NSW [email protected]

Received 17 October, 2005; accepted 26 October, 2005 Communicated by J. Sándor

ABSTRACT. Let Te(n) denote the product of all exponential divisors of n. An integer n is called 2 2 multiplicatively e-perfect if Te(n) = n and multiplicatively e-superperfect if Te(Te(n)) = n . In this note, we give an alternative proof for characterization of multiplicatively e-perfect and multiplicatively e-superperfect numbers.

Key words and phrases: Perfect number, Exponential divisor, Multiplicatively perfect, Sum of divisors, Number of divisors.

2000 Mathematics Subject Classiﬁcation. 11A25, 11A99.

1. INTRODUCTION Let σ(n) be the sum of all divisors of n. An integer n is called perfect if σ(n) = 2n and α1 αk superperfect if σ(σ(n)) = 2n. If n = p1 ··· pk is the prime factorization of n > 1, a divisor β1 βk d | n, called an exponential divisor (e-divisor) of n is d = p1 ··· pk with βi | αi for all 1 ≤ i ≤ k. Let Te(n) denote the product of all exponential divisors of n. The concepts of multiplicatively e-perfect and multiplicatively e-superperfect numbers were ﬁrst introduced by Sándor in [1]. 2 Deﬁnition 1.1. An integer n is called multiplicatively e-perfect if Te(n) = n and multiplica- 2 tively e-superperfect if Te(Te(n)) = n . In [1], Sándor completely characterizes multiplicatively e-perfect and multiplicatively e- superperfect numbers.

Theorem 1.1 ([1]). a) An integer n is multiplicatively e-perfect if and only if n = pα, where p is prime and α is a perfect number. b) An interger n is multiplicatively e-superperfect if and only if n = pα, where p is a prime, and α is a superperfect number.

Sándor’s proof is based on an explicit expression of Te(n). In this note, we offer an alternative proof of Theorem 1.1.

2. PROOFOF THEOREM 1.1 2 a) Suppose that n is multiplicatively e-perfect; that is Te(n) = n . If n has more than one prime α1 αk factor then n = p1 ··· pk for some k ≥ 2, αi ≥ 1 and p1, . . . , pk are k distinct primes. We have three separate cases.

(1) Suppose that α1 = ··· = αk = 1. Then d is an exponential divisor of n if and only if d = p1 ··· pk = n. Hence Te(n) = n, which is a contradiction. (2) Suppose that two of α1, . . . , αk are greater 1. Without loss of generality, we may assume α2 αk α1 α3 αk that α1, α2 > 1. Then d1 = p1p2 ··· pk , d2 = p1 p2p3 ··· pk , d3 = n are three 2α1+1 distinct exponential divisors of n. Hence d1d2d3 | Te(n). However, p1 | d1d2d3 so 2 Te(n) 6= n , which is a contradiction. (3) Suppose that there is exactly one of α1, . . . , αk which is greater than 1. Without loss of generality, we may assume that α1 > 1 and α2 = ··· = αk = 1. We have that if β1 d is an exponential divisor of n then d = p1 p2 ··· pk for some β1 | α1. Hence if n 3 2α1 2 2 has more than two distinct exponential divisors then p2 | Te(n) = p1 p2 ··· pk, which α1 is a contradiction. However, d1 = p1p2 ··· pk, d2 = p1 p2p3 ··· pk are two distinct exponential divisors of n so d1, d2 are all exponential divisors of n. Hence Te(n) = α1+1 2 2 2α1 2 2 p1 p2 ··· pk = p1 p2 ··· pk. This implies that α1 = 1, which is a contradiction. Thus n has only one prime factor; that is, n = pα for some prime p. In this case then σ(α) 2 2α Te(n) = p . Hence Te(n) = n = p if and only if σ(α) = 2α. This concludes the proof. 2 b) Suppose that n is multiplicatively e-superperfect; that is Te(Te(n)) = n . If n has more than α1 αk one prime factor then n = p1 ··· pk for some k ≥ 2, αi ≥ 1 and p1, . . . , pk are k distinct primes. We have two separate cases.

(1) Suppose that α1 = ··· = αk = 1. Then d is an exponential divisor of n if and only if d = p1 ··· pk = n. Hence Te(n) = n and Te(Te(n)) = Te(n) = n which is a contradiction. (2) Suppose that there is at least one of α1, . . . , αk which is greater 1. Without loss of gener- α2 αk α1 α2 α3 αk ality, we may assume that α1 > 1. Then d1 = p1p2 ··· pk , d2 = n = p1 p2 p3 ··· pk , are two distinct exponential divisors of n. Hence d1d2 | Te(n). However, d1d2 = α1+1 2α2 2αk γ1 γk p1 p2 ··· pk so Te(n) = p1 ··· pk where γ1 ≥ α1 + 1, γi ≥ 2αi ≥ 2 for γ1 γ3 γk γ1 γ2 γ3 γk i = 2, . . . , k. Thus, t1 = p1 p2p3 ··· pk and t2 = Te(n) = p1 p2 p3 ··· pk are 2γ1 two distinct exponential divisors of Te(n). Hence t1t2 | Te(Te(n)). However, p1 | t1t2 and γ1 > α1, which is a contradiction. α Thus n has only one prime factor; that is n = p for some prime p. In this case then Te(n) = σ(α) σ(σ(n)) 2 2α p and Te(Te(n)) = p . Hence Te(Te(n)) = n = p if and only if σ(σ(α)) = 2α. This concludes the proof. Remark 2.1. In an e-mail message, Professor Sándor has provided the authors some more recent references related to the arithmetic function Te(n), as well as connected notions on e- perfect numbers and generalizations. These are [2], [3], and [4].

J. Inequal. Pure and Appl. Math., 7(3) Art. 99, 2006 http://jipam.vu.edu.au/ MULTIPLICATIVELY E-PERFECT 3

REFERENCES [1] J. SÁNDOR, On multiplicatively e-perfect numbers, J. Inequal. Pure Appl. Math., 5(4) (2004), Art. 114. [ONLINE: http://jipam.vu.edu.au/article.php?sid=469] [2] J. SÁNDOR, A note on exponential divisors and related arithmetic functions, Scientia Magna, 1 (2005), 97–101. [ONLINE http://www.gallup.unm.edu/~smarandache/ ScientiaMagna1.pdf] [3] J. SÁNDOR, On exponentially harmonic numbers, to appear in Indian J. Math.

[4] J. SÁNDOR AND M. BENCZE, On modiﬁed hyperperfect numbers, RGMIA Research Report Collection, 8(2) (2005), Art. 5. [ONLINE: http://eureka.vu.edu.au/~rgmia/v8n2/ mhpn.pdf]

J. Inequal. Pure and Appl. Math., 7(3) Art. 99, 2006 http://jipam.vu.edu.au/