Formation of Triples Consist Some Special Numbers with Interesting Property

Total Page:16

File Type:pdf, Size:1020Kb

Formation of Triples Consist Some Special Numbers with Interesting Property International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056 Volume: 04 Issue: 07 | July -2017 www.irjet.net p-ISSN: 2395-0072 FORMATION OF TRIPLES CONSIST SOME SPECIAL NUMBERS WITH INTERESTING PROPERTY V. Pandichelvi1, P. Sivakamasundari2 1Assistant Professor, Department of Mathematics, UDC, Trichy. 2Guest Lecturer, Department of Mathematics, BDUCC,Lalgudi, Trichy. ---------------------------------------------------------------------***---------------------------------------------------------------- Abstract- In this communication, we discover the triple them is a perfect square is given in the following two a,b,c involving some figurate numbers such that the sum sections. of any two of them is a perfect square. Also, we find some interesting relations among the triples. Section- A Key words Diophantine m -tuples, polygonal numbers. : Let an 2CP2n1 bn Dec Gno INTRODUCTION: Let n be an integer. A set of positive 2n2 2n2 which are equivalent to the following two equations integers a1,a2,a3,......amis said to have the an 20n2 30n 12,bn 16n2 30n 13 the property D(n) if ai a j n is a perfect square for all Now, we assume that 1 i j m such a sets is called a Diophantine m-tuple. such an bn 2 sets were studied by Diophantus [1].The set of numbers 1,2,7is Diophantine triple with property D(2).For an Let cn be any non-zero integer such that extensive review of various articles one may refer [2-12]. In b(n) cn 2 (1) this communication, we find the triple a,b,c involving centered pentagonal numbers, decagonal numbers, a(n) cn 2 (2) Gnomonic numbers, Kynea numbers and Jacobsthal-lucas Subtracting (2) from (1), we get numbers such that the sum of any two of them is a perfect 2 2 b(n) an (3) square. Also, a few interesting relations among the triples are presented. Put A 1, A in (3), we get A 2n2 (4) Notations: Substituting (4) in (2), the values of c are represented by 4 2 5n2 5n 2 cn 4n 20n 30n 12 (5) Let CP be a centered pentagonal number of n 2 Hence, rank . 20n2 30n 12,16n2 30n 13,4n4 20n2 30n 12 is a 2 t10,n 4n 3n be a decagonal number of rank n . triple in which the sum of any two of them is a perfect square. Gnon 2n 1 be a Gnomonic number of rank n . 2n n1 Kn 2 2 1 be a Kynea number of rank n . Table-1: Some numerical examples are illustrated n n below: jn 2 1 be a Jacobshthal-lucas number of rank n . n an bn cn a(n) bn a(n) cn b(n) cn Method of analysis: 1 62 59 58 112 22 12 The procedure for finding the triple involving 152 137 88 2 2 2 some interesting numbers such that the sum of any two of 2 17 8 7 © 2017, IRJET | Impact Factor value: 5.181 | ISO 9001:2008 Certified Journal | Page 2372 International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056 Volume: 04 Issue: 07 | July -2017 www.irjet.net p-ISSN: 2395-0072 3 282 247 42 232 182 172 Table-2: Some numerical examples are illustrated below: 4 452 389 572 292 322 312 n an bn cn a(n) bn a(n) cn b(n) cn 5 662 563 1838 352 502 492 1 23 58 266 92 172 182 A few interesting relations among the numbers are given 287 154 4202 2 2 2 below: 2 21 67 66 4223 538 3392426 2 2 2 1. anbn 36 Pron 12Gnon1 13 69 1843 1842 66047 2074 572 2 2 2 2. (1/ 4)an cn is a bi-quadratic integer 261 31987 31986 1050623 8218 2 2 3. (1/ 4)[bncn36Pron 12Gnon 37] is a 2.7165151661011 1029 521203 5212022 bi-quadratic integer 4. (1/ 4)[ancn 40Pron 10Gnon 34]is a CONCLUSION: bi-quadratic integer In this communication, we discover the triple involving SECTION-B various special numbers in such a way that the sum of any two of them is a perfect square. In this manner, one may seek out other triples and quadruples satisfying some other Let an K 2n properties. bn 8j2n j1 j4 which are equivalent to the following two equations REFERENCES: 4n 2n 2n cn an 2 2.2 1,bn 8.2 26 [1]. Baker A.,Davenport H., The equations 3x2 2 y 2 Now, we assume that 2 2 an bn 2 and 8x 7 z , J.Math.Oxford Ser. 20 (1969),129-137. Let be any non-zero integer such that [2]. I.G Bashmakova(ed), Diophantus of Alexandria, Arithmetic and the Book of polygonal numbers, Nauka, b(n) cn 2 (6) Moscow, 1974. a(n) cn 2 (7) [3]. Dickson L.E; History of theory of numbers ,Vol.2, Chelsea , New York 1966; 513-520. Subtracting (7) from (6), we get 2 2 b(n) an (8) [4]. Thamotherampillai N; The set of numbers {1, 2, 7}.Bull of Calcutta math soc.,1980; 72:195-197. The choices A 1, A lead (8) to [5]. Brown E; sets in which xy k is always a square, Math A 3.22n 24n1 13 (9) comp., 1985; 45: 613-620. Substituting (9) in (7), the values of c are represented by 4n 8n2 6n1 2n 4n1 [6]. Gupta H, singh K; On K-triad sequences. Internet J Math cn 8.2 2 6.2 76.2 26.2 170 Sci., 1985; 5: 799 - 804. Hence, [7]. Dujella A., Complete solution of a family of simultaneous 24n 2.22n 1,8.22n 26,8.24n 28n2 6.26n1 76.22n 26.24n1 170 pellian equations, Acta Math. Inform. Univ. Ostraviensis is a triple in which the sum of any two of them is a perfect 6 (1998), 59-67. square. [8]. Beardon AF, Deshpande MN; Diophantine triples. The athematical Gazette, 2002, 86:258-260 [9]. Deshpande MN; Families of Diophantine triplets,Bulletin of the Marathwada Mathematical Society, 2003; 4: 19-21. © 2017, IRJET | Impact Factor value: 5.181 | ISO 9001:2008 Certified Journal | Page 2373 International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056 Volume: 04 Issue: 07 | July -2017 www.irjet.net p-ISSN: 2395-0072 [10].Bugeaud Y, Dujella A, Mignotte M; On the family of Diophantine triples { k 1,k 1,16k 3 4k }. Glasgo Math J. 49(2007),333-334. [11].Gopalan M.A., Pandichelvi V; On the extendibility of the Diophantine triple involving Jacobsthal numbers J2n1, J2n1 3,2J2n J2n1 J2n1 3, International journal of Mathematics and Applications, 2009;2(1):1-3. [12]Pandichelvi V, Construction of the Diophantine triple involving polygonal numbers. Impact J SciTech., 2011; 5(1): 7-11. © 2017, IRJET | Impact Factor value: 5.181 | ISO 9001:2008 Certified Journal | Page 2374 .
Recommended publications
  • Use Style: Paper Title
    Volume 4, Issue 11, November – 2019 International Journal of Innovative Science and Research Technology ISSN No:-2456-2165 On Three Figurate Numbers with Same Values A. Vijayasankar1, Assistant Professor, Department of Mathematics, National College, Affiliated to Bharathidasan University Trichy-620 001, Tamil Nadu, India Sharadha Kumar2, M. A. Gopalan3, Research Scholar, Professor, Department of Mathematics, Department of Mathematics, Shrimati Indira Gandhi College, National College, Affiliated to Bharathidasan University, Affiliated Bharathidasan University, Trichy-620 001, Tamil Nadu, India Trichy-620 002, Tamil Nadu, India Abstract:- Explicit formulas for the ranks of Triangular Centered Octagonal Number of rank M numbers, Hexagonal numbers, Centered Hexagonal ct8,M 4MM 11 numbers, Centered Octagonal numbers, Centered Decagonal numbers and Centered Dodecagonal numbers Centered Decagonal Number of rank M satisfying the relations; ct10,M 5MM 11 t 3, N t 6,h ct 6, H , t3,N t6,h ct 8,M , t 3,N t6,h ct 10,M , t t ct 3,N 6,h 12,D are obtained. Centered Dodecagonal Number of rank D ct12,D 6DD 11 Keywords:- Equality of polygonal numbers, Centered Hexagonal numbers, Centered Octagonal numbers, Triangular Number of rank N Centered Decagonal numbers, Centered Dodecagonal NN 1 numbers, Hexagonal numbers, Triangular numbers. t3, N 2 Mathematics Subject Classification: 11D09, 11D99 Hexagonal Number of rank h 2 I. INTRODUCTION t6,h 2h h The theory of numbers has occupied a remarkable II. METHOD OF ANALYSIS position in the world of mathematics and it is unique among the mathematical sciences in its appeal to natural human 1. Equality of t t ct curiosity.
    [Show full text]
  • Notations Used 1
    NOTATIONS USED 1 NOTATIONS ⎡ (n −1)(m − 2)⎤ Tm,n = n 1+ - Gonal number of rank n with sides m . ⎣⎢ 2 ⎦⎥ n(n +1) T = - Triangular number of rank n . n 2 1 Pen = (3n2 − n) - Pentagonal number of rank n . n 2 2 Hexn = 2n − n - Hexagonal number of rank n . 1 Hep = (5n2 − 3n) - Heptagonal number of rank n . n 2 2 Octn = 3n − 2n - Octagonal number of rank n . 1 Nan = (7n2 − 5n) - Nanogonal number of rank n . n 2 2 Decn = 4n − 3n - Decagonal number of rank n . 1 HD = (9n 2 − 7n) - Hendecagonal number of rank n . n 2 1 2 DDn = (10n − 8n) - Dodecagonal number of rank n . 2 1 TD = (11n2 − 9n) - Tridecagonal number of rank n . n 2 1 TED = (12n 2 −10n) - Tetra decagonal number of rank n . n 2 1 PD = (13n2 −11n) - Pentadecagonal number of rank n . n 2 1 HXD = (14n2 −12n) - Hexadecagonal number of rank n . n 2 1 HPD = (15n2 −13n) - Heptadecagonal number of rank n . n 2 NOTATIONS USED 2 1 OD = (16n 2 −14n) - Octadecagonal number of rank n . n 2 1 ND = (17n 2 −15n) - Nonadecagonal number of rank n . n 2 1 IC = (18n 2 −16n) - Icosagonal number of rank n . n 2 1 ICH = (19n2 −17n) - Icosihenagonal number of rank n . n 2 1 ID = (20n 2 −18n) - Icosidigonal number of rank n . n 2 1 IT = (21n2 −19n) - Icositriogonal number of rank n . n 2 1 ICT = (22n2 − 20n) - Icositetragonal number of rank n . n 2 1 IP = (23n 2 − 21n) - Icosipentagonal number of rank n .
    [Show full text]
  • The Secret Code of God
    The Secret Code of God Abul Hassan Copy Rights and 19 19 – The Secret Code of God By: Abul Hassan 1st Edition: 2010 No Part of this book may be reproduced in any form or by any means electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher or author. However any one can download the electronic version of this book, i.e. free eBook from any website on the internet without our permission for the purposes of reading and research or from our website – www.ali-pi.com. Abul Hassan www.ali-pi.com [email protected] [email protected] Price: US$ 19.00/CAN$ 19.00 19 – The Secret Code of God Page 2 Contents and 19 Topics Page 1. Copy Rights and 19 2 2. Abul Hassan’s Desk and 19 5 3. Dedication and 19 7 4. Coding and 19 8 5. Religions of the world and 19 10 6. Bahai’s and 19 11 7. Hinduism and 19 12 8. Symbolism and 19 12 9. Judaism and 19 13 10. Other faiths and 19 16 11. Christianity and 19 17 12. Islam and 19 25 13. Security Code and 19 29 14. Holy Quran and 19 30 15. 99 Names of Allah and 19 73 16. Arabic Alphabets and 19 87 17. Everything and 19 90 18. Mathematics and 19 94 19. Mathematical Properties of 19 97 20. Prime Numbers and 19 102 21. Perfect Numbers and 19 111 22. Powerful Numbers and 19 115 23. Enneadecagon and 19 116 24.
    [Show full text]
  • Numbers 1 to 100
    Numbers 1 to 100 PDF generated using the open source mwlib toolkit. See http://code.pediapress.com/ for more information. PDF generated at: Tue, 30 Nov 2010 02:36:24 UTC Contents Articles −1 (number) 1 0 (number) 3 1 (number) 12 2 (number) 17 3 (number) 23 4 (number) 32 5 (number) 42 6 (number) 50 7 (number) 58 8 (number) 73 9 (number) 77 10 (number) 82 11 (number) 88 12 (number) 94 13 (number) 102 14 (number) 107 15 (number) 111 16 (number) 114 17 (number) 118 18 (number) 124 19 (number) 127 20 (number) 132 21 (number) 136 22 (number) 140 23 (number) 144 24 (number) 148 25 (number) 152 26 (number) 155 27 (number) 158 28 (number) 162 29 (number) 165 30 (number) 168 31 (number) 172 32 (number) 175 33 (number) 179 34 (number) 182 35 (number) 185 36 (number) 188 37 (number) 191 38 (number) 193 39 (number) 196 40 (number) 199 41 (number) 204 42 (number) 207 43 (number) 214 44 (number) 217 45 (number) 220 46 (number) 222 47 (number) 225 48 (number) 229 49 (number) 232 50 (number) 235 51 (number) 238 52 (number) 241 53 (number) 243 54 (number) 246 55 (number) 248 56 (number) 251 57 (number) 255 58 (number) 258 59 (number) 260 60 (number) 263 61 (number) 267 62 (number) 270 63 (number) 272 64 (number) 274 66 (number) 277 67 (number) 280 68 (number) 282 69 (number) 284 70 (number) 286 71 (number) 289 72 (number) 292 73 (number) 296 74 (number) 298 75 (number) 301 77 (number) 302 78 (number) 305 79 (number) 307 80 (number) 309 81 (number) 311 82 (number) 313 83 (number) 315 84 (number) 318 85 (number) 320 86 (number) 323 87 (number) 326 88 (number)
    [Show full text]
  • Issue 32 of the Secondary Magazine
    Welcome to Issue 32 of the Secondary Magazine. As the summer term starts and the exam season rapidly approaches, now is the time to plan what you will do with some of that time gained when Year 11 have departed. In the meantime, there is a lot of revision to do! Contents From the editor – Gauss I can remember my primary school teacher teaching me a ‘trick’ to quickly add up a set of consecutive numbers to ‘amaze my friends’. It wasn’t until much later that I discovered that this ‘trick’ had been used by Gauss whilst a schoolboy. Perhaps your pupils would like to perform this ‘trick’ to honour Gauss’ birthday? Up2d8 Maths The fortnightly Up2d8 maths resources explore a range of mathematical themes in a topical context. What makes a rollercoaster scary? SAW - The Ride carried its first passengers earlier this year and has claimed to be the scariest roller coaster in the world. In this resource, students use this claim as the basis for a statistical investigation into what makes a roller coaster scary. The Interview – Peter Ashwin Peter is a professor of mathematics at the University of Exeter. Here he shares a snapshot of his work and an insight into his personality. Focus on…10 Is it true that some numbers have greater prominence than others? As the basis for the metric system, 10 is certainly an important number. Here are some facts to help you form an opinion. An idea for the classroom – 3D co-ordinates A colleague recently went for an interview and was asked to teach a lesson on 3D co-ordinates.
    [Show full text]
  • On the Solution of Diophantine Problems in Whole Numbers∗
    On the solution of Diophantine problems in whole numbers∗ Leonhard Euler [175] §1. It is often the case in solving Diophantine problems that one comes across a formula in which no more than one indeterminate is involved; typically, whole numbers are required to be put in place of the indeterminate in order to satisfy the formula. Indeed, whenever this cannot be done, it may be necessary to settle for fractional number values. However, it is observed that if in this formula the highest degree of the indeterminate is a square and the same expression is also equal to a square number, then a good many of these problems are satisfied by infinitely many whole numbers, which are linked with each other through a certain law, and form a particular series. But if the formula is to be equal to either a cube or some other higher power, or if the indeterminate has more than two dimensions, then one may not be able to produce much more than can be generated by fractional numbers. §2. Moreover, in this manner a method for all problems is established whereby one must find by guessing one number that solves the problem, from which infinitely many others can be found in turn. In fact, the rule may not be able to reveal anything at first, as the case may occur in which there does not admit of a solution at all, as for 3x2 + 2, a formula which can never be made equal to a square. Therefore, we will always set forth in what follows a single instance to be found by which the condition of the problem is satisfied, and we will give the rule whereby from this one countless others can be generated.
    [Show full text]
  • 12 Number Patternsmep Pupil Text 12
    12 Number PatternsMEP Pupil Text 12 12.1 Simple Number Patterns A list of numbers which form a pattern is called a sequence. In this section, straightfor- ward sequences are continued. Worked Example 1 Write down the next three numbers in each sequence. (a) 2, 4, 6, 8, 10, . (b) 3, 6, 9, 12, 15, . Solution (a) This sequence is a list of even numbers, so the next three numbers will be 12, 14, 16. (b) This sequence is made up of the multiples of 3, so the next three numbers will be 18, 21, 24. Worked Example 2 Find the next two numbers in each sequence. (a) 6, 10, 14, 18, 22, . (b) 3, 8, 13, 18, 23, . Solution (a) For this sequence the difference between each term and the next term is 4. Sequence 6, 10, 14, 18, 22, . Difference 4 4 4 4 So 4 must be added to obtain the next term in the sequence. The next two terms are 22+= 4 26 and 26+= 4 30 , giving 6, 10, 14, 18, 22, 26, 30, . (b) For this sequence, the difference between each term and the next is 5. Sequence 3, 8, 13, 18, 23, . Difference 5 5 5 5 Adding 5 gives the next two terms as 23+= 5 28 and 28+= 5 33, giving 3, 8, 13, 18, 23, 28, 33, . 262 MEP Pupil Text 12 Exercises 1. Write down the next four numbers in each list. (a) 1, 3, 5, 7, 9, . (b) 4, 8, 12, 16, 20, . (c) 5, 10, 15, 20, 25, .
    [Show full text]
  • Relations Between Special Polygonal Numbers Generated Through the Solutions of Pythagorean Equation
    International Journal of Innovation in Science and Mathematics Volume 2, Issue 2, ISSN (Online): 2347–9051 Relations between Special Polygonal Numbers Generated through the Solutions of Pythagorean Equation K. Meena S. Vidhyalakshmi B. Geetha Former VC, Professor, M.Phil. Student, Bharathidasan University, Department of Mathematics, SIGC, Department of Mathematics, SIGC, Trichy-24, Tamilnadu , India Trichy-620002, Tamilnadu, India Trichy-620002, Tamilnadu, India Email: [email protected] Email: [email protected] Email: [email protected] A. Vijayasankar M. A. Gopalan Assistant Professor, Department of Mathematics, Professor, Department of Mathematics, National College, Trichy-620001, Tamilnadu, India SIGC, Trichy-620002, Tamilnadu, India Email: [email protected] Email: [email protected] Abstract – Employing the solutions of the Pythagorean We illustrate below the process of obtaining relations equation, we obtain the relations between the pairs of special between the pairs of special polygonal numbers such that polygonal numbers such that the difference in each pair is a each relation is a perfect square. perfect square. The choices 8N-3= r 2 s2 , 2M+1= r 2 s2 (3) Keywords – Ternary Quadratic Equation, Integer in (1) leads to the relation 16t10,N 8t3,M 8 is a square Solutions, Pythagorean Equation, Polygonal Numbers integer Centered Polygonal Numbers. From (3), the values of ranks of the Decagonal and triangular numbers are respectively given by I. INTRODUCTION r 2 s2 3 r 2 s2 1 N= , M= In [1,2,4-8,10], employing the integral solutions of 8 2 special binary quadratic Diophantine equation, special It is seen that N and M are integers when patterns of Pythagorean triangles are generated.
    [Show full text]
  • Pythagorean Equation and Special M-Gonal Numbers
    IOSR Journal of Mathematics (IOSR-JM) e-ISSN: 2278-5728,p-ISSN: 2319-765X, Volume 8, Issue 1 (Sep. - Oct. 2013), PP 31-34 www.iosrjournals.org Pythagorean Equation and Special M-Gonal Numbers M. A. Gopalan1, V. Geetha2 1(Department of Mathematics, Shrimathi Indira Gandhi College,Trichirappalli-620 002.) 2(Department of Mathematics, Cauvery College For Women,Trichirappalli-620 018.) Abstract: Employing the solutions of the Pythagorean equation, we obtain the relations between the pairs of special polygonal numbers such that the difference in each pair is a perfect square. Key Words: Pythagorean equation,Polygonal numbers MSC Classification Number: 11D09. NOTATIONS PenP = Pentagonal number of rank P HexQ = Hexagonal number of rank Q HepH = Heptagonal number of rank H DodecD = Dodecagonal number of rank D I. Introduction In [1,2,4-8,10],employing the integral solutions of special binary quadratic Diophantine equation, special patterns of Pythagorean triangles are generated. In [3],the relations among the pairs of special m-gonal numbers generated through the solutions of the binary quadratic equation yx2221 are determined. In [9], the relations among special figurate numbers through the equation yx2210 1 are obtained. In[11],employing the solutions of the Pythagorean equation, and obtain the relations between Triangular number and Pentagonal number, Octagonal number, Hexagonal number, Heptagonal number, Decagonal number, Dodecagonal number , Pentagonal number and Hexagonal number, Octagonal number such that the difference in each pair is a perfect square. In this communication, employing the solutions of the Pythagorean equation, we obtain the relations between the pairs of special polygonal numbers which are not mentioned in [11] such that the difference in each pair is a perfect square.
    [Show full text]
  • IJESM Volume 2, Issue 2 ISSN: 2320-0294 2013 ______
    June IJESM Volume 2, Issue 2 ISSN: 2320-0294 2013 ___________________________________________________ EQUALITY OF CENTERED DECAGONAL NUMBER WITH SPECIAL M-GONAL NUMBERS Manju Somanath V.Sangeetha* M.A.Gopalan Abstract Explicit formulas for the ranks of Centered Decagonal numbers which are simultaneously equal to Triangular number,Square number,Pentagonal number,Hexagonal number,Octagonal number and Decagonal number in turn are presented. Keywords Centered Decagonal number, Triangular number,Square number,Pentagonal number,Hexagonal number,Octagonal number,Decagonal number Assistant Professor, Department of Mathematics, National College,Trichy-1 Professor,Department of Mathematics, Srimathi Indira Gandhi College,Trichy-2 A Quarterly Double-Blind Peer Reviewed Refereed Open Access International e-Journal - Included in the International Serial Directories Indexed & Listed at: Ulrich's Periodicals Directory ©, U.S.A., Open J-Gage, India as well as in Cabell’s Directories of Publishing Opportunities, U.S.A. International Journal of Engineering, Science and Mathematics http://www.ijmra.us 208 June IJESM Volume 2, Issue 2 ISSN: 2320-0294 2013 ___________________________________________________ Introduction In [1],the equality of Triangular numbers which are simultaneously equal to Pentagonal numbers and Hexagonal numbers are illustrated through examples. In [2], explicit formulas for the ranks of Triangular numbers which are simultaneously equal to Pentagonal,Octagonal,Decagonal and Dodecagonal numbers in turn are presented. In [3],explicit formula for the ranks of centered Hexagonal numbers which are simultaneously equal to Triangular number,Pentagonal number,Hexagonal number,Heptagonal number,Decagonal number,Dodecagonal number in turn are presented. In [4],explicit formula for finding the ranks n of Hex-numbers which are simultaneously equal to Centered m-gonal numbers such as Centered Triangular, Centered Square, Centered Pentagonal, Centered Heptagonal, Centered Octagonal, Centered Nonagonal, Centered Decagonal numbers of rank m are presented.
    [Show full text]
  • A Unique Perfect Power Decagonal Number
    Introduction Descent The Modular Method and FLT Eliminating Newforms A Unique Perfect Power Decagonal Number Philippe Michaud-Rodgers University of Warwick Young Researchers in Algebraic Number Theory III 20.08.2021 Introduction Descent The Modular Method and FLT Eliminating Newforms Polygonal Numbers The nth s-gonal number is (s − 2)n2 − (s − 4)n P (n)= : s 2 n2+n Triangular numbers: P3(n) = 2 . 2 Square numbers: P4(n) = n . Decagonal numbers: P10(n) = D(n) = n(4n − 3). Introduction Descent The Modular Method and FLT Eliminating Newforms Perfect Powers in Sequences Perfect Power m N > 1 is a perfect power if N = y for some y 2 Z and m > 1. Perfect powers in Fibonacci and Lucas sequences. Perfect powers in arithmetic progressions. Perfect power polygonal numbers (Kim, Park, Pint´er,2013): m All solutions to Ps (n) = y when m > 2, Ps (n) > 1, and s 2 f3; 5; 6; 8; 20g are 3 9 3 P8(2) = 2 and P20(8) = 2 = 8 : A unique perfect power decagonal number (M. 2021) m The only solution to P10(n) = D(n) = y when m > 1 and D(n) > 1 is D(3) = 33. Introduction Descent The Modular Method and FLT Eliminating Newforms Descent We have D(n) = n(4n − 3) = y p, with p prime. Case 1: 3 - n. Then n and 4n − 3 are coprime. So n = ap and 4n − 3 = bp; with (a; b) = 1. So 4ap − bp = 3: (1) Case 2: 3 k n. Then n = 3tp and 4n − 3 = 3p−1up; with (t; u) = 1 and 3 - t.
    [Show full text]
  • The Natural Logarithm E 4
    Generated by Foxit PDF Creator © Foxit Software http://www.foxitsoftware.com For evaluation only. The Aikaantha State of "Black Hole" Matter and Base 60 By John Frederick Sweeney Abstract Space, the fundamental reality, can only exist in one state of Aikantha, the coherent state of three - dimensional, cubic resonances. The Universe remains in the Aikaantha state, fluctuating between the two values of the e logarithm and π /10. The latter value implies a circle with many trigonometric equations related to the Square Root of Five, or the Golden Section. In addition, this forms the basis for Base 60 Math in the Qi Men Dun Jia Model, and thus the icosahedron with its 60 stellated permutations. The value of π /10 thus limits the bounds of visible matter, and gives rise to Pisano Periodicity of the icosahedron. At the same time, the author dispels the current "comic book" notions of black holes: the Aikantha state is the state of black holes, yet misunderstood by modern science. 1 Generated by Foxit PDF Creator © Foxit Software http://www.foxitsoftware.com For evaluation only. Table of Contents Introduction 3 Three States of Matter 4 The Natural Logarithm e 4 Pi Divided by Ten 7 Instant Karma 9 Conclusion 12 Appendix I Decagonal Number 14 Appendix II Trigonometry Angles--Pi/10 17 Appendix III Decagram 21 Appendix IV Golden Ratio and Pi 23 Appendix V Natural logarithm e 25 2 Generated by Foxit PDF Creator © Foxit Software http://www.foxitsoftware.com For evaluation only. Introduction We all have an image of what a black hole is, thanks to TV documentaries and the boyish, feverish imaginations of today's scientists.
    [Show full text]