The Form Of The Friendly Number Of 10

Sourav Mandal

Abstract

9 Any positive integer 푛 other than 10 with abundancy index ⁄5 must be a square with at least 6 distinct prime factors, the smallest being 5, and my new argument about the form of the friendly 25(52푐+1−1)(8푛+1−2푐) number of 10 is ,if exist. Futher,at least one of the prime factors must be 4 congruent to 1 modulo 3 and appear with an exponent congruent to 2 modulo 6 in the prime factorization of 푛.

Introduction

휎(푛) For a positive integer 푛, the sum of the positive of 푛 is denoted by 휎(푛); the ratio is 푛 known as the abundancy index of 푛 or sometimes the ratio denoted as 휎̂(푛) or 퐼(푛). A pair (푎, 푏) is called a friendly pair if휎̂(푎) = 휎̂(푏) in this case,it is also common to say that 푏 is 푎 friend of a or simply that 푏 and 푎 are friend. Perfect numbers have abundancy index 2,and thus are all friendly numbers with abundancy index less than 2 are often called deficient, while numbers whose abundancy index are greater than 2 are called abundant. The original problem was to show that the density of friendly integers is ℕ is unity and the density of solitary numbers (numbers with no friends) is zero.

휎(푛) We used two main approaches: one was an analysis of the function. While the other used 푛 number theoretic arguments to find a representation for a friend of 10. In [1], it was shown that 10 is the smallest number where it is unknown whether there are any friends of it. We will assume a basic understanding of the function 휎(푛).This can be found in [2] . For more on number theoretic techniques, see [3] . And the question “Is 10 a solitary number” is still unanswered. If 10 does have a friend, the following may be of use in finding it.

1. Elementary Properties of Abundancy Index :

Let 푎 and 푏 be positive integers. In what follows, all primes are positive.

1. 퐼(푎) ≥ 1 with equality only if 푎 = 1 2. If 푎 divides 푏 then 퐼(푎) ≤ 퐼(푏) with equality only if 푎 = 푏

16 May 2019 14:32:36 BST Version 1 - Submitted to J. London Math. Soc. 3. If 푝1 , 푝2, ⋯ ⋯ , 푝푘 are distinct primes and 푒1, 푒2, ⋯ ⋯ , 푒푘 are positive integers then

푘 푘 푒푗 푒푗 −푖 퐼 (∏ 푝푗 ) = ∏ (∑ 푝푗 ) 푗=1 푗=1 푖=0

푘 푒푗+1 푝푗 − 1 = ∏ 푒푗 푗=1 푝푗 (푝푗 − 1)

These formulae follow from well-known analogues for 휎:

푘 푘 푒푗 푘 푒푗+1 푒 푝푗 − 1 휎 (∏ 푝 푗 ) = ∏ (∑ 푝푖) = ∏ 푗 푗 (푝 − 1) 푗=1 푗=1 푖=0 푗=1 푗

Property 3 directly implies a property of 퐼 shared by 휎 .

4. 퐼 is weakly multiplicative (meaning, if gcd(푎, 푏) = 1, then 퐼(푎푏) = 퐼(푎)퐼(푏)).

5. Suppose that 푝1 , 푝2, ⋯ ⋯ , 푝푘 are distinct primes , 푞1 , 푞2, ⋯ ⋯ , 푞푘 are distinct primes 푒1, 푒2, ⋯ ⋯ , 푒푘 are positive integers and 푝푗 ≤ 푞푗 , 푗 = 1,2, ⋯ ⋯ , 푘. Then

푘 푘 푒푗 푒푗 퐼 (∏ 푝푗 ) ≥ 퐼 (∏ 푞푗 ) 푗=1 푗=1

With equality only if 푝푗 = 푞푗 , 푗 = 1,2, ⋯ ⋯ , 푘. This follows from 3 and the observation that if 푥푒+1−1 푒 ≥ 1, then is a decreasing function of 푥 on (1, ∞). 푥푒(푥−1)

푘 푝푗 6. If the distinct prime factors of 푎 are 푝1 , 푝2, ⋯ ⋯ , 푝푘, then 퐼(푎) < ∏푗=1 . Although 푝푗−1 related to 5, 7 is most easily seen by applying 3 and the observation that for 푝 > 1,

1 푒+1 푝 − 푝 − 1 푝푒 = 푝푒(푝 − 1) 푝 − 1 푝 Increases to as 푒 → ∞. 푝−1

2. Theorem 1:

16 May 2019 14:32:36 BST Version 1 - Submitted to J. London Math. Soc. If 푛 is a friend of 10 then 푛 is a odd with at least 6 distinct prime factors, the smallest being 5. Further, at least one of 푛’s prime factors must be congruent to 1 modulo 3, and appear in the factorization of 푛 to a power congruent to 2 modulo 6. If there is only one such prime dividing 푛, then it appears to a power congruent to 8 modulo 18 in the factorization of 푛.

퐩퐫퐨퐩퐨퐬퐭퐢퐨퐧 ퟏ. 퐼(푎푛) > 퐼(푛) 푓표푟 푎 > 1

Proof. In general 푎 can share prime factors with 푛 . Let 푎 = 푢푣 where gcd(푎, 푛) = 푢, gcd(푣, 푛) = 1 . We thus have 퐼(푎푛) = 퐼(푢푛)퐼(푣) by 퐸푙푒푚푒푛푡푎푟푦 푃푟표푝푒푟푡푖푒푠 표푓 퐴푏푢푛푑푎푛푐푦 퐼푛푑푒푥 .

퐼(푎푛) = 퐼(푢푛)퐼(푣) > 퐼(푢푛)

퐼(푢푛) > 퐼(푛)

푇ℎ푢푠,

퐼(푎푛) > 퐼(푛) 푓표푟 푎 > 1

휎(푛) 휎(푎푛) . 퐴 푓푟푖푒푛푑 표푓 푛 푐푎푛푛표푡 푏푒 푎 푚푢푙푡푖푝푙푒 표푓 푛. 푇ℎ푎푡 푖푠 ≠ 푓표푟 푎 > 1. Lemma 1 푛 푎푛

휎(푎푛) 휎(푛) . This follows directly from 푝푟표푝표푠푡푖표푛 1 . Since 푎푛 is a multiple of 푛. > , Proof 푎푛 푛

휎(푎푛) 휎(푛) So ≠ 푎푛 푛

Corollary 1. 퐴 푓푟푖푒푛푑 표푓 10 푐푎푛푛표푡 푏푒 푡ℎ푒 푓표푟푚 푛 = 2푎5푏푚. 푇ℎ푢푠 푎 푓푟푖푒푛푑 표푓 10 푐푎푛푛표푡 푏푒 푎푛 푒푣푒푛 푖푛푡푒푔푒푟.

Proof. For a, b > 1. 2푎5푏 is a multiple of 10, and thus so is 푛 = 2푎5푏푚.

Therefore 푛 is not a friend of 10. A friend of 10 must be of the form 5푏푚, So a friend of 10 can′t be an even integer.

2푒 퐴 푓푟푖푒푛푑 표푓 10 푚푢푠푡 푏푒 푡ℎ푒 푠푞푢푎푟푒 표푓 푠표푚푒 푛푢푚푏푒푟푠: 푛 = 52푏 ∏ 푘 푝 푗 Corollary 2. 푗=1 푗

휎(푛) 9 푒 Proof. Suppose = 푎푛푑 푛 = 5푏푑, 푤ℎ푒푟푒 푑 = ∏ 푘 푝 푗 푡ℎ푒푛 푛 5 푗=1 푗

휎(푛) 9 = 푛 5

푘 푘 푏 푒푗 푏 푒푗 5 ∙ 휎(5 )휎 (∏ 푝푗 ) = (9) ∙ 5 ∏ 푝푗 푗=1 푗=1

푏 푒1 푒2 푒3 푒푘 푏 푒1 푒2 푒3 푒푘 5 ∙ 휎(5 ) ∙ 휎(푝1 ) ∙ 휎(푝2 ) ∙ 휎(푝3 ) ⋯ 휎(푝푘 ) = (9) ∙ 5 ∙ 푝1 푝2 푝3 ⋯ 푝푘

16 May 2019 14:32:36 BST Version 1 - Submitted to J. London Math. Soc. e b e1 ek b e1 e2 e3 k 5(1 + 5 + ⋯ 5 )(1 + p1 + ⋯ + p1 ) ⋯ (1 + pk + ⋯ + pk ) = (9) ∙ 5 p1 p2 p3 ⋯ pk

From the 퐶표푟표푙푙푎푟푦 1, we have 푝푖 > 2 for any 푖 ≤ k. So the right side must be odd. 푥 To obtain this, we must have that every sum on the left side is also odd. Since each 푝푖 is odd, we must have an odd number of terms in the sum for the whole sum to be odd. To obtain this, each 푒푖 and 푏 must be even. Thus 푛 must 푏푒 푡ℎ푒 푠푞푢푎푟푒 표푓 푠표푚푒 푛푢푚푏푒푟푠.

9 퐩퐫퐨퐩퐨퐬퐭퐢퐨퐧 ퟐ. If 퐼(푁2) = , 푡ℎ푒푛 3 ∤ 푁2 5

9 . From the 푐표푟표푙푙푎푟푦 2, we have that If 퐼(푁2) = 푎푛푑 3 ∣ 푁2 . 푡ℎ푒푛 Proof 5 푘 2푒 푁2 = 32푎52푏 ∏ 푝 푗 푗 푗=1 푘 2푒 9 where 2,3,5 ∤ ∏ 푝 푗 . It is easy to varify that 퐼(3254) and 퐼(3452) > . Combining this 푗 5 푗=1

푘 푘 2푒 2푒 9 with 퐿푒푚푚푎 1, we have that 퐼 (3254 ∏ 푝 푗) and 퐼 (3452 ∏ 푝 푗) > 푗 푗 5 푗=1 푗=1

푘 2푒 9 Thus, the problm reduces to a single case: 퐼 (3252 ∏ 푝 푗) = 푗 5 푗=1

푘 푘 2푒 2푒 5 ∙ 휎 (32) ∙ 휎 (52) ∙ 휎 (∏ 푝 푗) = 9 ∙ (32) ∙ (52) ∙ ∏ 푝 푗 푗 푗 푗=1 푗=1

푘 푘 2푒푗 4 2푒푗 13 ∙ (31)휎 (∏ 푝푗 ) = (3 ) ∙ (5) ∙ ∏ 푝푗 푗=1 푗=1

푘 푘 ℎ 2푒푗 2푒푗 2푐 2푑 2푒푖 we can see that 13,31 ∣ ∏ 푝푗 , Thus ∏ 푝푗 = 13 31 ∏ 푝푖 푗=1 푗=1 푖=1

ℎ 2푒푖 where 2,3,5,13,31 ∤ ∏ 푝푖 . We will divide by 13,31 immediately. 푖=1

ℎ ℎ 2푐 2푑 2푒푖 4 2푐−1 2푑−1 2푒푖 휎(13 ) 휎(31 )휎 (∏ 푝푖 ) = 3 (5)(13 )(31 ) ∏ 푝푖 푖=1 푖=1

16 May 2019 14:32:36 BST Version 1 - Submitted to J. London Math. Soc. ℎ 405 405 퐼 (∏ 푝2푒푖) = < < 1 푖 2푐 2푑 448 푖=1 휎 (13 ) 휎 (31 ) 2푐−1 ∙ 2푑−1 ⏟(13 ) ⏟(31 ) ≥14 ≥32

This is a contradiction with 퐸푙푒푚푒푛푡푎푟푦 푃푟표푝푒푟푡푖푒푠 표푓 퐴푏푢푛푑푎푛푐푦 퐼푛푑푒푥( 푝푟표푝푒푟푡푦 푛표. 1).

Hence, 3 ∤ 푁2 whenever 푁2 is a friend of 10.

9 푒 퐩퐫퐨퐩퐨퐬퐭퐢퐨퐧 ퟑ. If 퐼(푛) = , then 푛 = 52푎 ∏ 푘 푝 푗 where 푘 ≥ 4 5 푗=1 푗 Proof. Using the previous proposition and 퐸푙푒푚푒푛푡푎푟푦 푃푟표푝푒푟푡푖푒푠 표푓 퐴푏푢푛푑푎푛푐푦 퐼푛푑푒푥( 푝푟표푝푒푟푡푦 푛표. 6),

we will construct the largest value of 퐼(푛푒) with 4 distinct primes. Let 푛 = (5)(7)(11)(13).

Here is a largest value with 4 distinct primes because from 푝푟표푝표푠푖푡푖표푛 2 . We have that 푝푖 ≥ 7.

To maximize the value of 퐼(푛푒) we let 푒 → ∞. Hence ,

5 ∙ 7 ∙ 11 ∙ 13 1001 9 lim 퐼(푛푒) = = < 푒 →∞. 4 ∙ 6 ∙ 10 ∙ 12 576 5

So there must be at least 5 distinct primes in the factorization of 푛.

9 푒 퐩퐫퐨퐩퐨퐬퐭퐢퐨퐧 ퟒ. If 퐼(푛) = , then 푛 = 52푎 ∏ 푘 푝 푗 where 푘 ≥ 5 5 푗=1 푗 Proof. We can use the same technique as in the last proposition to show that only 3 cases could work if 푛 were 2 represented as 5 distinct primes. These are 푛 = (5a7b11c13d17f) but this does not work because

9 2 the smallest it could be is: 퐼(5272112132172) > . Case2 gives 푛 = (5a7b11c13d19f) , but this 5 9 does not work because the smallest it could be is: 퐼(5272112132192) > . The final case takes 5

2 9 a little more work: 푛 = (5a7b11c13d23f) . We can see that if a > 1, then 퐼(푛) > , 푠표 푎 = 1. Let us 5 9 examine when 퐼(푛) = , then 5

16 May 2019 14:32:36 BST Version 1 - Submitted to J. London Math. Soc. 2 5휎 (52) 휎 (72푏) 휎 (112푐) 휎 (132푑) 휎 (232푓) = 9(52)(7b11c13d23f)

2 31휎 (52) 휎 (72푏) 휎 (112푐) 휎 (132푑) 휎 (232푓) = 9(52)(7b11c13d23f) = 푙

Clearly, 31 ∤ 푙. Since the left hand side is some integer , this results in a contradiction.

Hence a friend of 10 must be be composed of at least 6 distinct primes.

Finally, since

푘 2푒 If 퐼(푁2) = 52푎 ∏ 푝 푗 , then 푗 푗=1

푘 2푒푗 2 2푎 푖 2 5휎(푁 ) = 5(1 + 5 ⋯ + 5 ) ∏ (∑ 푝푗) = 9푁 , 푗=1 푖=0

We have that 9 ∣ 휎(푁2). if 푝 ≡ 2 (mod 3), then 휎(푝2푒) ≡ 1 (푚표푑 3)for any positive integer 푒.

2푒 Consequently, some 푝 ≡ 1 (푚표푑 3), and 휎 (푝 푗) ≡ 0 (푚표푑 3) implies 2푒 + 1 ≡ 0 (푚표푑 3). 푗 푗 푗

Thus 푒푗 = 3푡 + 1 for some integer 푡, 푠표 2푒푗 = 6푡 + 2

2푒 If 푝 is only such prime dividing 푁2 , then 휎 (푝 푗) ≡ 0 (푚표푑 9). Checking the possibilities 푗 푗

푝푗 ≡ 1,4 표푟 7 (푚표푑 9), one finds that 2푒푗 ≡ 8 (mod 18)

3. Theorem 2:

25(52푐+1−1)(8푛+1−2푐) If 푁2 is a friend of 10 then 푁2 is must be in the form of . 4 Proof.

Let, the friendly number of 10 푖푠 푁2

푘 2 2푐 2푒푗 푁 = 5 ∏ 푝푗 푗=1

16 May 2019 14:32:36 BST Version 1 - Submitted to J. London Math. Soc. 휎(푁2) 9 = 푁2 5

퐴푠 푁2 is a square number from 퐶표푟표푙푙푎푟푦 2. , so it is in the form of (4푎 + 1)

∵ 5|(4푎 + 1)

∴ 푁2 is in the form of (20푎 − 15)

휎(20푎 − 15) 9 = 20푎 − 15 5

⇒ 휎(20푎 − 15) = −27 + 36푎

휎(20푎 − 15) ≡ −27 (푚표푑 36)

&

휎(20푎 − 15) ≡ −27 (푚표푑 푎)

Again (20푎 − 15) is a square number so 푎 = 5푎0(푎0 − 1) + 2

2 After putting 푎 = 5푎0(푎0 − 1) + 2, 푤푒 푔푒푡 (20푎 − 15) = (10푎0 − 5)

Last digit of 5푎0(푎0 − 1) + 2 is always 2 so

휎(20푎 − 15) + 27 휎[(10푎 − 5)2 ] + 27 = 0 = 푎 here the last digit of 푎 is must be 2 36 36 1 1

2 휎[(10푎0 − 5) ] ≡ −27 (푚표푑 36)

푘 2푐 2푒푗 휎 (5 ∏ 푝푗 ) ≡ −27 (푚표푑 36) 푗=1

Here the last digit of the quotient is 2

hence,

푘 2푒푗 휎 (∏ 푝푗 ) ≡ 45 − 90푐 (푚표푑 360) 푗=1

푘 2푒푗 ⇒ 휎 (∏ 푝푗 ) = 360푛 + 45 − 90푐 푗=1

16 May 2019 14:32:36 BST Version 1 - Submitted to J. London Math. Soc. 푘 2푐+1 2푒 (5 − 1)(360푛 + 45 − 90푐) ⇒ 휎(52푐 )휎 (∏ 푝 푗) = 푗 4 푗=1

푘 2푐+1 2푒 (5 − 1)(360푛 + 45 − 90푐) ⇒ 휎 (52푐 ∏ 푝 푗) = 푗 4 푗=1

휎(푁2) 9 ∵ = 푁2 5

2푐 푘 2푒푗 휎 (5 ∏푗=1 푝푗 ) 9 ⇒ = 2푐 푘 2푒푗 5 5 ∏푗=1 푝푗

(52푐+1 − 1)(360푛 + 45 − 90푐) 9 ⇒ = 2푐 푘 2푒푗 5 4 (5 ∏푗=1 푝푗 )

푘 2푐+1 2푒 25(5 − 1)(8푛 + 1 − 2푐) ⇒ 52푐 ∏ 푝 푗 = 푗 4 푗=1

25(52푐+1 − 1)(8푛 + 1 − 2푐) ⇒ 푁2 = 4

So we have come to the conclusion that the friendly number of 10 is must be in the form of

25(52푐+1 − 1)(8푛 + 1 − 2푐) if exist. 4

25(52푐+1 − 1)(8푛 + 1 − 2푐) 푁2 = 4

From 푝푟표푝표푠푡푖표푛 2. 3 ∤ 푁2.

3 ∤ (8푛 + 1 − 2푐)

∴ 푛 ≠ 3푛0 + (푐 − 2)

Just for fun, we introduce a new definition.

Definition 1 : (Theoretical Friend). A sequence 푛푒 is a theoretical Friend of 푚 if:

16 May 2019 14:32:36 BST Version 1 - Submitted to J. London Math. Soc.

lim 퐼( 푛푒) = 퐼(푚 ) 푒→∞

푒 퐩퐫퐨퐩퐨퐬퐭퐢퐨퐧 ퟓ. 10 has at least one theoretical Friend, namely 푛푒 = 3 ∙ 5

Proof.

휎(3푒) ∙ 휎(5) lim 퐼( 푛푒) = lim 푒→∞ 푒→∞ 3푒 ∙ 5

3 6 = ( ) ∙ ( ) = 퐼(10) 2 5

For further reading on the topic of 퐼(푛) and 휎(푛), see [5] and [4]. See [5] for information concerning when 휎(푛)= k has exactly 푚 solutions (Sierpi´nski Conjecture). See [4] for a more indepth study of 휎(푛) and on the distribution and density of numbers of this form.

4. References:

[1] C.W. Anderson and D. Hickerson. Friendly integers, (partial) solution 6020 of problem 307 (1975). American Mathematical Monthly, pages 65–66, January 1977. in Advanced Problems and Solutions.

[2] , chapter 6. Dover Publications, 1971.

[3] Introduction to Analytic Number Theory. Springer-Verlag, 1968.

[4] P. Erdos. On the distribution of numbers of the form 휎 (n)/n and on some related questions. Pacific J. Math, 1974.

[5] K. Ford and S. Konyagin. On two conjectures of sierpinski concerning the arithmetic functions 휎 and 휑. http://www.math.uiuc.edu/ford/wwwpapers/sigma.pdf.

[6] J. A. Holdener. Conditions equivalent to the existence of odd perfect numbers. Math Magazine, 79(5):389–391, December 2006.

[7] R. Laatsch. Measuring the abundancy of integers. Math Magazine, 59:84–92,

16 May 2019 14:32:36 BST Version 1 - Submitted to J. London Math. Soc. 1986.

[8] P.A. Weiner. The abundancy ratio, a measure of perfection. Math Magazine, 73:307–310, 2000.

[9] E.W. Weisstein. Abundance. http://mathworld.wolfram.com/Abundancy.html, 2002.

16 May 2019 14:32:36 BST Version 1 - Submitted to J. London Math. Soc.