On Perfect Numbers and Their Relations 1 Introduction

Home , Euclid number, Hyperperfect number, Superperfect number

Int. J. Contemp. Math. Sciences, Vol. 5, 2010, no. 27, 1337 - 1346

On Perfect Numbers and their Relations

Recep G¨ur

2000 Evler Mahallesi Boykent Sitesi E Blok 26/7 Nevsehir, Turkey [email protected]

Nihal Bircan

Technische Universitaet Berlin Fakultaet II Institut f¨ur Mathematik MA 8 − 1 Strasse des 17. Juni 136 D-10623 Berlin, Germany [email protected]

Abstract

In this paper, we get some new formulas for generalized perfect numbers and their relationship between arithmetical functions φ, σ concerning Ore’s harmonic numbers and by using these formulas we present some examples.

Mathematics Subject Classiﬁcation: 11A25, 11A41, 11Y70

Keywords: perfect number, 2-hyperperfect number, Euler’s totient function, Ore harmonic number

1 Introduction

In this section, we aimed to provide general information on perfect numbers shortly. Here and throughout the paper we assume m, n, k, d, b are positive integers and p is prime. N is called perfect number, if it is equal to the sum of its proper divisors. The ﬁrst few perfect numbers are 6, 28, 496, 8128,... since,

6 = 1+2+3 28 = 1+2+4+7+14 496 = 1+2+4+8+16+31+62+124+248

Euclid discovered the first four perfect numbers which are generated by the formula 2n−1(2n−1), called Euclid number. In his book ’Elements’ he presented the proof of the formula which gives an even perfect number whenever 2n −1is 1338 Recep Gür and Nihal Bircan prime. In order for 2n −1 to be a prime n must itself be a prime. Moreover, the perfect numbers are all even and end with 6 or 8. In [5] the author provided detailed information on perfect numbers. In general, we can say that if ap − 1 is prime for some numbers a ≥ 2 and p ≥ 2, then a must equal to 2 and p must be a prime. This means that if we are interested in primes of the form ap − 1 we only need to consider the case that a = 2 and p is prime. Primes of the form 2p − 1 are called Mersenne primes which are available at [15], [16]. If N is an even perfect number, then N looks like N =2p−1(2p − 1) where 2p − 1 is a Mersenne prime. Up to know, only 47 Mersenne primes are known which means that there are 47 known even perfect numbers. There is a conjecture that there are infinitely many perfect numbers. To search for new ones is to keep on going. But to this day no one has been able to discover any odd perfect number. However, no one has yet been able to prove conclusively that none exist. In [8] the author proved any odd perfect number must have the form n =12m +1orn =36q + 9 shortly. In [6] the author proved that no odd perfect number exist in the interval [1, 10300] ⊂ N. We can also define perfect numbers as σ(N)=2N,(N ∈ N) where σ is a divisor function. If p is a prime number and k ≥ 1 than its only divisors are 1 and p. So, σ(p)=p + 1 and

σ pk p p2 ... pk pk+1−1 . ( )=1+ + + + = p−1

Furthermore, if 2n+1 − 1=p, N =2np is perfect, then we have,

σ(N)=(2n+1 − 1)(p +1)=2n+1p.

For any prime p, we have φ(p)=p − 1 where φ is Euler’s totient function deﬁned as the number of invertible elements in an complete residue system. Here are some properties of Euler’s totient function which we use to prove our results;

φ(m) (i) If a and m are integers such that (a, m) = 1 then a ≡ 1 (mod m) (ii) If n ≥ 1 we have φ(d)=n d|n n ≥ φ n n. − 1 (iii) For 1 we have ( )= 1 p p|n k n , ..., n k . The harmonic mean of numbers 1 k is 1 +...+ 1 For example, the n1 nk harmonic mean of 1 and 2 is 2/(1 + 1/2) = 4/3. Such an integer is called Ore harmonic number or harmonic divisor number. According the theorem of Ore; every perfect number is harmonic. n is called superperfect number if σ(σ(n)) = 2n and every even superperfect number n must be a power of 2, that is, 2p−1 such that 2p − 1 is a Mersenne prime. On perfect numbers and their relations 1339

A unitary perfect number is an integer which is the sum of its positive proper unitary divisors, not including the number itself. (A divisor d of a number n is d n unitary divisor if and d share no common factors). T. Yamada [13] showed that 9 and 165 are all of the odd unitary superperfect numbers. A. Bege gave a table of k-hyperperfect numbers in [1] which deﬁned n as; k− n k σ n − n − σ n k+1 n k−1 . hyperfect number if =1+ [ ( ) 1] and so ( )= k + k Also, they proved a conjecture which states that all 2-hyperferfect numbers are of the form n =(3k − 1).(3k − 2) where 3k − 2 is prime. In [7] the author computed all hyperperfect numbers less than 1011. σ σ n 3 n n If ( ( )) = 2 ( + 1), then is called super-hyperperfect number. In [1] the authors studied generalized perfect numbers, presented some conjectures and n p−1 p 3p−1 numerical results. They conjectured that, if =3 where and 2 are primes, then n is a super-hyperperfect number. In [11] the author studied on perfect numbers and their composition with arithmetical functions. 2 n is called multiplicatively e-perfect if Te(n)=n where Te(n) denote the prod- uct of exponential divisors of n. In [10] a characterization of multiplicatively e-perfect and similar numbers is given. An alternative proof for characterization of e-perfect and e-superperfect numbers can be found in [3]. For further information on perfect numbers see also [2], [4], [9], [12], [14].

2 Main Results

Theorem 2.1. If k ≥ 1, 1+2+4+... + k =2k − 1 where (2k − 1) is prime and k(2k − 1) is a perfect number, then

φ(k[2k − 1]) = k(k − 1). (1)

Moreover, if k[2k − 1] = 2n(2n+1 − 1) is an Euclid number, then

φ(2n[2n+1 − 1]) = 2n(2n − 1). (2)

Proof. Let k ≥ 1 , 1+2+4+... + k =2k − 1 and 2k − 1 is prime. Since φ(k[2k − 1]) is a multiplicative function and (k, 2k − 1) = 1, we can write φ(k[2k − 1]) = φ(k).φ(2k − 1). From hypothesis k is even and is of the form 2n (n ≥ 1). By using a property of Euler’s totient function, we can write, 1 k φ(k)=k.(1 − )= (3) 2 2 Since 2k − 1 is prime,

φ(2k − 1) = 2k − 2 (4)

From (3) and (4), we obtain 1340 Recep G¨ur and Nihal Bircan

k φ(k[2k − 1]) = φ(k).φ(2k − 1) = (2k − 2) 2 φ(k[2k − 1]) = k(k − 1).

Any even perfect number is an Euclid number, that is, it is of the form 2n(2n+1 − 1) where 2n+1 − 1 is prime. So, if k =2n and 2k − 1=2n+1 − 1is prime, then k[2k−1] is an Euclid number. So, φ(2n[2n+1 −1]) = 2n(2n −1). Proposition 2.2. If N>6 is a perfect number, then φ(N) is even and φ(N) is not a prime. Proof. Let N is a perfect number. From the result of Euclid theorem and theorem 2.1, N = k.(2k − 1) and φ(k[2k − 1]) = k(k − 1). Also, 2|k.(k − 1) so φ(N) is an even number and φ(6) = 2 and from hypothesis N>6soφ(N) > 2. Moreover, k|φ(N) and (k − 1)|φ(N)soφ(N) is not a prime. Corollary 2.3. If N is a perfect number and N = k.(2k − 1), then there is at least a natural number n which satisﬁes n|φ(N) and (n +1)|φ(N). Proof. It is clear from theorem 2.1.

√ Proposition 2.4. If φ(N)=b and N is a perfect number, then 4b +1is a Mersenne prime. Proof. From the result of theorem 2.1, we can write b = k.(k − 1),

b = k2 − k 1 1 b =(k − )2 − 2 4 2 1 1 b + = k − 4 2 1 1 ± b + = k − 4 2 (1 + 4b) 1 k = ± + √ 4 2 2k − 1=± 4b +1

From the√ result of theorem 2.1, (2k −1) is a Mersenne prime. So, it is positive. Thus, 4b + 1 is a Mersenne prime. Proposition 2.5. If N =2n(2n+1 − 1) is an Euclid number, then φ(N 2)=N.φ(N). On perfect numbers and their relations 1341

Proof. By using a property of Euler’s totient function and theorem 2.1, we can write 1 1 φ(N 2)=φ(22n(2n+1 − 1)2)=22n(2n+1 − 1)(2n+1 − 1). .(1 − ) 2 2n+1 − 1 =22n(2n+1 − 1)(2n − 1) =2n(2n+1 − 1)[2n(2n − 1)] = N.φ(N).

Theorem 2.6. Let p is a Mersenne prime and p =2n+1 − 1. If N =2n(2n+1 − 1) is a perfect number, then 4p σ(N)= φ(N). p − 1 or 2n+2 − 2 σ(N)= φ(N). 2n − 1 Proof. From hypothesis and theorem 2.1, we can write φ N φ np n n − . n p+1 ( )= (2 )=2 (2 1) So, 2 = 2 , then p +1 p − 1 2n(2n − 1) = ( )( ) 2 2 p2 − 1 φ(N)= 4 σ(N)=σ(2np)=σ(2n).σ(p) σ n 2n+1−1 σ p p where (2 )= 2−1 and ( )= + 1. So,

σ(N)=p(p +1)

From φ(N) and σ(N) we obtain,

4p σ(N)= φ(N). p − 1 or p =2n+1 − 1. So,

2n+2 − 2 σ(N)= φ(N). 2n − 1 1342 Recep G¨ur and Nihal Bircan

See example 3.5 for the use of this theorem.

Theorem 2.7. Let N is a perfect number and σ(N) is harmonic. If N is of the form 2n(2n+1 − 1) where 2n+1 − 1 is a Mersenne prime, then

(n + 2)(2n+2 − 2) H(σ(N)) = . (2n+2 − 1)

Proof. If N is a perfect number; this means σ(N)=2N or σ(2n(2n+1 − 1)) = 2n+1(2n+1 − 1).

H(σ(N)) = H(2n+1[2n+1 − 1]) −1 1 1 1 1 1 1 = .(1 + + ... + + + + ... + ) 2n +4 2 2n+1 2n+1 − 1 2.(2n+1 − 1) 2n+1.(2n+1 − 1) −1 1 2n+2 − 1 2n+2 − 1 = .( + ) 2.(n +2) 2n+1 2n+1.(2n+1 − 1) −1 1 (2n+2 − 1)2n+1 = . 2.(n +2) 2n+1.(2n+1 − 1) (n +2).(2n+2 − 2) = . (2n+2 − 1) This completes the proof.

Corollary 2.8. If N is a perfect number, then σ(N) is harmonic.

Proof. Ore’s harmonic number theorem tells us that every perfect number is harmonic. From Ore’s theorem and theorem 2.7, N is harmonic. Also, σ(N)=2N. So, σ(N) is harmonic.

Theorem 2.9. Let 3k − 2 is prime. If n is a 2-hyperperfect number, then

φ(n)=n − 32k−2.

Proof. From hypothesis, we can take n =3k−1.(3k − 2). Also, φ(n)isa multiplicative function, and (3k−1, 3k − 2) = 1. So,

φ(n)=φ(3k−1.(3k − 2)) = φ(3k−1).φ(3k − 2) =(3k−1 − 3k−2).(3k − 3) =3k−1.[(3k − 2) − 3k−1] = n − 32k−2.

Theorem 2.10. If n is a super-hyperperfect number, then On perfect numbers and their relations 1343

φ φ n 2 n ( ( )) = 9 . Proof. From the deﬁnition of super-hyperperfect number we know that n p−1 p 3p−1 p−1, =3 where and 2 are primes and (3 2) = 1. So,

1 φ(φ(3p−1)) = φ((3p−1).(1 − )) 3 = φ(3p−2.2) = φ(3p−2).φ(2) 1 =3p−2.(1 − ) 3 2.3p−1 2 = = n 9 9

Theorem 2.11. If n is an even superperfect number, then

n φ(φ(n)) = . 4 Proof. If n is an even superperfect number, then n is of the form 2p−1 where 2p − 1 is a Mersenne prime. So, 1 φ(n)=φ(2p−1)=2p−1.(1 − ) 2 =2p−2 1 φ(φ(n)) = φ(2p−2)=2p−2.(1 − ) 2 2p−1 = 4 n φ(φ(n)) = 4

3 Examples

In this section, we introduced some examples related to our theorems. Example 3.1. N p+1 . φ p σ p p n+1 − If = 4 ( ( )+ ( )) where =2 1 is a Mersenne prime, then N is a perfect number.

Let p =2n+1−1 is a Mersenne prime. Also, by using properties σ(p)=p+1 φ p p − σ p φ p p N p(p+1) n n+1 − and ( )= 1, ( )+ ( )=2 . Therefore, = 2 =2(2 1). So, according to theorem 2.1, N is a perfect number. 1344 Recep G¨ur and Nihal Bircan

Example 3.2. Let p is a Mersenne prime and N is a perfect number. By . N 2p φ 2N using proposition 2 4,If b = p−1 , then ﬁnd ( p+1 ). √ p b b p2−1 Since = 4 +1 so, = 4 . Then,

N(p − 1) = 2pb p2 − 1 N(p − 1) = 2p( ) 4 4N(p − 1) = 2p(p2 − 1) 2N = p p +1 2N φ( )=p − 1 p +1 So, the result holds.

Example 3.3. d φ(N) N Every = 2 is a triangular number where is an even perfect number.

This follows directly from theorem 2.1.If m =2n then φ(N)=m(m − 1). φ(N) So, 2 is a triangular number.

Example 3.4. Let >0,Nis a perfect number, n ∈ N. Is there any perfect φ(N) < 1 numbers which satisﬁes N 2 ?

Assume p =2n+1 − 1 is a Mersenne prime, N =2np is a perfect number,

φ(N)=2n(2n − 1) p2 − 1 = 4 p(p +1) N = 2 φ(N) p − 1 1 1 = = − N 2p 2 2p

1 φ(N) < 1 Since = 2p , so, N 2 .

Example 3.5. N φ N > N If is an even perfect number then, ( ) 4 and σ(N) < φ(N) 8. On perfect numbers and their relations 1345

We can verify these inequalities in two ways, Firstly, we write N =2n.p, p =2n+1 − 1. So,

N φ N N. − 1 >N. − 1 N. 1 N. −2 > ( )= (1 p) (1 )= = 2 p|N p|N 2 p|N 2 4 σ(N)=2N N φ(N) > 4 σ N N ( ) < 2 < φ N N 8 ( ) 4

The second way is shorter. By using theorem 2.6,

σ(N) 2n+2 − 2 = < 8 φ(N) 2n − 1

The inequality is true because 3 < 2n+1.

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Received: April, 2010