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The Form of the Friendly Number of 10

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The Form of the Friendly Number of 10 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 divisors 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 square number 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 prime power 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. 푏 ′ Therefore 푛 is not a friend of 10. A friend of 10 must be of the form 5 푚, 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. 2 9 2 퐩퐫퐨퐩퐨퐬퐭퐢퐨퐧 ퟐ. If 퐼(푁 ) = , 푡ℎ푒푛 3 ∤ 푁 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). 2 2 Hence, 3 ∤ 푁 whenever 푁 is a friend of 10. 9 2푎 푘 푒푗 퐩퐫퐨퐩퐨퐬퐭퐢퐨퐧 ퟑ. If 퐼(푛) = , then 푛 = 5 ∏ 푝 where 푘 ≥ 4 5 푗=1 푗 Proof. Using the previous proposition and 퐸푙푒푚푒푛푡푎푟푦 푃푟표푝푒푟푡푖푒푠 표푓 퐴푏푢푛푑푎푛푐푦 퐼푛푑푒푥( 푝푟표푝푒푟푡푦 푛표. 6), 푒 we will construct the largest value of 퐼(푛 ) 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 2푎 푘 푒푗 퐩퐫퐨퐩퐨퐬퐭퐢퐨퐧 ퟒ. If 퐼(푛) = , then 푛 = 5 ∏ 푝 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 a b c d f 2 represented as 5 distinct primes. These are 푛 = (5 7 11 13 17 ) but this does not work because 9 2 the smallest it could be is: 퐼(5272112132172) > .
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