A Platonic basis of Integers

Transcription

1 A Platonic basis of Integers Maya Mohsin Ahmed arxiv: v1 [mathnt] 3 Apr 019 Abstract In this article, we prove that every integer can be written as an integer combination of exactly 4 tetrahedral numbers Moreover, we compute the modular periodicity of platonic numbers 1 Introduction Figurate numbers related to the five platonic solids are called platonic numbers In [5], Sir Frederick Pollock conjectured that every positive integer is the sum of at most five, seven, nine, thirteen, and twenty one tetrahedral numbers, octahedral numbers, cubes, icosahedral numbers, and dodecahedral numbers, respectively See [1] for a detailed study of Platonic numbers In this article, we discuss integer combinations of platonic numbers instead of sums Thus we also include the operation of subtraction Let t n,o n,c n,i n, andd n represent thetetrahedral numbers, octahedralnumbers, cubes, icosahedral numbers, and the dodecahedral numbers, respectively The following identities are discussed in [1] and [4]: We list a few numbers in the sequences: t n = n(n+1)(n+) o n = n(n +1) 3 c n = n 3 i n = n(5n 5n+) d n = n(9n 9n+) (1) t n : 1,4,10,0,35,5,84,10,15,0, o n : 1,,19,44,85,14,31,344,489,70, c n : 1,8,7,4,15,1,343,51,79,1000, i n : 1,1,48,14,55,45,74,118,19,0, d n : 1,0,84,0,455,81,1330,04,95,400, 1

2 In Section, we prove that every integer can be written as an integer combination of exactly 4 tetrahedral numbers Moreover, we prove that integers that are divisible by four, six, forty five, and fifty four, can be written as integer combinations of exactly four octahedral numbers, cubes, icosahedral numbers, and dodecahedral numbers, respectively We also conjecture that every integer can be written as a sum of at most five different platonic numbers Wesayasequence s n isperiodicmoddifthereexistsaninteger msuchthat, s n+m s n mod d for every n The smallest such integer m is called the period of s n mod d Example 11 Let r n denote the residues of the the tetrahedral numbers t n modulo : t n : r n : Observe that the reduced sequence r n repeats after every four terms Consequently, t n+4 t n mod Hence t n has period 4 mod In Section, we show that the platonic numbers satisfy the following linear recurrence relation: y n = 4y n 1 y n +4y n 3 y n 4 () It is well known that linear recurrence sequences are eventually periodic See [3] and the references therein for a discussion of periodic sequences defined by linear recurrences In Section 3, we derive the periods of the platonic numbers Forward Differences In this section, we use forward differences of tetrahedral numbers to prove that every integer can be written as an integer combination of exactly 4 tetrahedral numbers Let y n denote a sequence of integers, then the first forward difference, denoted by y n, is defined as y n = y n+1 y n is called the forward difference operator The differences of the first forward differences are called the second forward differences and are denoted by y i Continuing thus, the n-th forward difference is defined as n y n = n 1 y n+1 n 1 y n We provide the forward differences tables of the five platonic numbers below

3 t n t n t n 3 t n 4 t n o n o n o n 3 o n 4 o n c n c n c n 3 c n 4 c n i n i n i n 3 i n 4 i n d n d n d n 3 d n 4 d n We see the fourth forward differences of all the platonic numbers are zero The following identities are derived from the first forward differences of the Platonic numbers t n = t n+1 t n = 1 n + 3 n+1 (3) o n = o n+1 o n = n +n+1 c n = c n+1 c n = 3n +3n+1 i n = i n+1 i n = 15 n + 5 n+1 d n = d n+1 d n = 7 n + 9 n+1 Since, y n = y n+ y n+1 +y n, we get the following identities 3

4 t n t n+1 +t n+ = n+ (4) o n o n+1 +o n+ = 4n+4 c n c n+1 +c n+ = n+ i n i n+1 +i n+ = 15n+10 d n d n+1 +d n+ = 7n+18 Since 3 y n = y n+3 3y n+ +3y n+1 y n, we get the following identities t n+3 3t n+ +3t n+1 t n = 1 (5) o n+3 3o n+ +3o n+1 o n = 4 c n+3 3c n+ +3c n+1 c n = i n+3 3i n+ +3i n+1 i n = 15 d n+3 3d n+ +3d n+1 d n = 7 The fourth forward difference is given by 4 y n = y n+4 4y n+3 +y n+ 4y n+1 +y n Consequently, we get the following identities t n+4 4t n+3 +t n+ 4t n+1 +t n = 0 () o n+4 4o n+3 +o n+ 4o n+1 +o n = 0 c n+4 4c n+3 +c n+ 4c n+1 +c n = 0 i n+4 4i n+3 +i n+ 4i n+1 +i n = 0 d n+4 4d n+3 +d n+ 4d n+1 +d n = 0 Consequently, it follows that all the platonic numbers satisfy the linear recurrence relation given by Equation From Equations 4 and 5, we get t n 3 t n = n Consequently, 3t n 8t n+1 +7t n+ t n+3 = n (7) Equation 7 implies that every integer can be written as an integer combination of four tetrahedral numbers Again, from Equations 4 and 5, we get o n 3 o n = 4n Therefore, o n 5o n+1 +4o n+ o n+3 = 4n Therefore, we conclude that integers divisible by 4 can be written as integer combinations of four octahedral numbers Similarly, since c n 3 c n = n, we get c n 5c n+1 +4c n+ c n+3 = n, 4

5 which implies that the integers which are divisible by are integer combinations of four cubes From Equations 4 and 5, we get 3 i n 3 i n = 45n, therefore 5i n 1i n+1 +9i n+ i n+3 = 45n Hence integers that are divisible by 45 can be written as integer combinations of four icosahedral numbers Finally, from Equations 4 and 5, we get 3 d n 3 d n = 54n, therefore 5d n 1d n+1 +9d n+ d n+3 = 54n Hence we conclude that integers that are divisible by 54 are integer combinations of four dodecahedral numbers Note that since the second forward differences are linear in n, some combination of these differences might lead to a proof of the Pollock s conjecture We leave that to the reader to explore In our study of the Pollock s conjecture, we observed that we could write the numbers between two platonic numbers as sums of at most five Platonic numbers Example 1 Integers between o 5 = 85 to t 8 = 10 written as sums of at most five different Platonic numbers 8 = 85+1, 87 = , 88 = , 89 = 85+4, 90 = , 91 = , 9 = , 93 = 85+8, 94 = , 95 = 85+10, 9 = , 97 = 85+1, 98 = , 99 = , 100 = , 101 = , 10 = , 103 = , 104 = , 105 = 85+0, 10 = , 107 = , 108 = , 109 = , 110 = , 111 = , 11 = 85+7, 113 = , 114 = , 115 = , 11 = , 117 = , 118 = , 119 = Thus, we add one more conjecture to the collection of the many beautiful unproved conjectures on platonic numbers Conjecture 1 Every integer can be written as a sum of at most five different Platonic numbers 3 Modular Periods of Platonic numbers In this section, we derive the modular periods of platonic numbers We use the formulas (1) in Section 1 for the proof of the following proposition 5

6 Proposition 31 Let d > 1 be an integer 1 Consider the sequence of tetrahedral numbers t n Let p t denote the period of t n mod d Case 1: d is an even integer Case : d is an odd integer { d if d is divisible by 3; p t = d otherwise { 3d if d is divisible by 3; p t = d otherwise Let p o denote of the period of the sequence of octahedral numbers o n mod d { 3d if d is divisible by 3; p o = d otherwise 3 If p c denotes the period of the sequence of cubes mod d, then p c = d 4 Let p i denote the period of the sequence of icosahedral numbers mod d { d if d is even; p i = d otherwise 5 Let p d denote the sequence of dodecahedral numbers, then { d if d is even; p d = d otherwise Proof 1 We derive the periods of the sequence of tetrahedral numbers Case 1: d is even Let d be not divisible by 3 Since d is even, we write d = k for some k For n 1, t n+4k t n = (n+4k)((n+4k)+1)((n+4k)+) (n(n+1)(n+) = k 3 ( 3n +1nk +n+1k +1k + ) Since d is not divisible by 3, and t n+4k t n is an integer, 3 divides (3n +1nk+ n+1k +1k+) Hence t n+4k t n = ks for some integer s Consequently, t n+d t n mod d, which implies p t = d On the other hand, if d is divisible by 3, then d = k for some integer k 1 For any n 1, t n+3k t n = k(3n +108nk +n+19k +108k +) Hence, t n+3k t n mod k Consequently, p t = d

7 Case : d is an odd integer Let d be not divisible by 3 Since d > 1 is odd, d = k + 1 for some integer k 1 For n 1, (k +1) ( t n+(k+1) t n = 3n +nk +9n+4k +10k+ ) Nowdisnotdivisiblebyor3 Hencek+1isnotdivisibleby Consequently, divides (3n +nk+9n+4k +10k+) because t n+(k+1) t n is an integer Therefore t n+k+1 t n = (k +1)s for some integer s Thus, p t = d Now consider the case when d is divisible by 3 Then d = 3(k +1), for some k 0 For n 1, t n+3d t n = 3(k +1) (3n +54nk +33n+34k +378k+110) Since t n+3d t n is an integer, divides (3n +54nk+33n+34k +378k+110) Therefore t n+9(k+1) t n = 3(k+1)s for some integer s Consequently, p t = 3d in this case Next, we consider the sequence of octahedral numbers o n Let d be not divisible by 3 For n 1, o n+d o d = d 3 (n +nd+d +1) Since o n+d o d is an integer, and d is not divisible by 3, we get 3 divides (n + nd+d +1) Consequently, o n+d o d mod d Hence p o = d in this case On the other hand, if d is divisible by 3, then d = 3k for some k 1 For n 1, o n+3d o n = 3k(n +54nk +1k +1) Consequently, o n+3d o d mod d Hence p o = 3d 3 p c = d because for n 1, c n+d c n = (n+d) 3 n 3 = 3n d+3nd +d 3 = d(3n +3nd+d ) 4 We now compute p i Let d be an odd integer For n 1, i n+d i n = d (15n +15nd 10n+5d 5d+) Since d is odd, and i n+d i n is aninteger, divides (15n +15nd 10n+5d 5d+) Consequently, p i = d in this case On the other hand, when d is even, d = k for some k 1 For n 1, i n+d i n = (n+4k)(5(n+4k) 5(n+4k)+) n(5n 5n+) Thus p i = d = k(15n +0nk 10n+80k 0k +) 7

8 5 Finally, we compute the periods of the sequence of dodecahedral numbers d n Let d be an odd integer For n 1, d n+d d n = (n+d)(9(n+d) 9(n+d)+) n(9n 9n+) = d (7n +7nd 18n+9d 9d+) Since d is odd and d n+d d n is an integer, we get divides (7n +7nd 18n+ 9d 9d+) Consequently, we get p d = d in this case On the other hand, when d is even, d = k for some k 1 For n 1, Thus, p d = d d n+d d n = k(7n +108nk 18n+144k 3k +) References [1] John H Conway and Richard K Guy, The Book of Numbers, Springer-Verlag, Copernicus, New York, 199 [] Dickson, L E, History of the Theory of Numbers, Vol : Diophantine Analysis, Dover, New York, 005 [3] Engstrom, H T Periodicity in sequences defined by linear recurrence relations, Proceedings of the National Academy of Sciences of the United States of America, 1 (10) (1930), 3-5 [4] Hyun Kwang Kim, On regular polytope numbers, Proceedings of the American Mathematical Society, 131 (003), 5-75 [5] Pollock, F On the Extension of the Principle of Fermat s Theorem of the Polygonal Numbers to the Higher Orders of Series Whose Ultimate Differences Are Constant With a New Theorem Proposed, Applicable to All the Orders, Abs Papers Commun Roy Soc London 5, 9-94,