Lecture Amiens

Information about Lecture Amiens

Published on June 19, 2007

Author: Charlie

Source: authorstream.com

Content

Resort to computingon a problem of Euler and of Lucas:  Resort to computing on a problem of Euler and of Lucas Christian Boyer Lecture of March 31st 2005, Amiens (Picardy, France) Jules Verne, le Centenaire (1905 – 2005) ASSOCIATION pour le DEVELOPPEMENT de la CULTURE SCIENTIFIQUE Union Régionale des Ingénieurs et Scientifiques de Picardie Jules Verne - Edouard Lucas:  Jules Verne - Edouard Lucas (Nantes 1828 – Amiens 1905) (Amiens 1842 – Paris 1891) Edouard Lucas:  E. Lucas Amiens Paris Edouard Lucas 1842 : Born in Amiens, studied at the imperial school of Amiens (current Louis Thuillier school) 1859 : Scientific diploma (« Bac ») at Amiens, then math studies at Douai 1861 : Passed at Polytechnique and at Normale Sup, Chose Normale Sup, and left Amiens 1864 : Paris Observatory 1870 : Artillery lieutenant, took part in the battles of the Loire (Orléans, Blois, Le Mans, …) 1872 : Math teacher at Mougins 1876 : Math teacher at the Charlemagne school and at the Saint-Louis school, in Paris 1891 : Accidental death, following a banquet J. Verne Nantes Paris Le Crotoy / Amiens Unsolved problemof Martin Gardner:  Unsolved problem of Martin Gardner Is it possible to construct a 3x3 magic square using 9 distinct square integers? Gardner asks this question in Quantum, 1996 then again in Scientific American, 1998 He attributes the initial problem to Martin LaBar, Southern Wesleyan University, USA problem 270 published on 2 lines in The College Mathematics Journal, 1984 He has been offering a $100 prize since 1996 Two ways to see the problem:  Two ways to see the problem Impose 8 magic sums (3 rows, 3 columns, and 2 diagonals ► sums = 3∙a) And try to use the maximum of square integers among the 9 distinct integers used Impose 9 distinct square integers And try to get the maximum of magic sums among the 8 sums Near solution with 9 squares:  Near solution with 9 squares OK for the 9 square integers, but… only 7 correct sums out of 8 S2 = 21 609 = 147² for 3 rows, 3 columns, 1 diagonal Unfortunately S2 = 38 307 for the other diagonal By computing, and independently 1996: Lee Sallows, University of Nijmegen, Netherlands 1996: Michaël Schweitzer, Göttingen, Germany A lot of other known solutions of this type S2 is often a square, as it is in this solution Near solution with 8 sums:  Near solution with 8 sums OK for the 8 sums (3 rows, 3 columns, 2 diagonals), but… only 7 square integers out of 9 S2 = 3 ∙ Center = 3 ∙ 425² = 541 875 By computing, and independently 1997: Lee Sallows, University of Nijmegen, Netherlands 1997: Andrew Bremner, Arizona State University, USA Only known solution of this type Link with 3 othermathematical problems:  Link with 3 other mathematical problems John Robertson, USA, Mathematics Magazine: Arithmetic progressions Right triangles with the same area Congruent numbers and elliptic curves y2 = x3 – n2x No possible solutionfor powers ≥ 3:  No possible solution for powers ≥ 3 In each 3x3 magic square, we must have xn + yn = 2zn Because xn + zn + yn = Magic sum = 3 ∙ Center = 3zn x2 + y2 = 2z2 possible (i.e.: 12 + 72 = 2 ∙ 52) No usable conclusion x3 + y3 = 2z3 impossible with x≠y≠z (Euler) Implies no 3x3 magic square of cubes x4 + y4 = 2z2 impossible (Legendre) ► of course 2z4 impossible Implies no 3x3 magic square of 4th powers xn + yn = 2zn impossible for n ≥ 3 (Noam Elkies, using the Andrew Wiles proof of the last theorem of Fermat) Implies no 3x3 magic square of powers ≥3 Computing research:  Computing research Duncan Buell Department of Computer Science and Engineering, University of South Carolina, USA Background code through most of calendar year 1998 Multiple processors, SGI Challenge computer of his university Result… No « magic hourglass » having 7 square integers for all a andlt; 2.5 ∙ 1025 Computing research:  Computing research Andrew Bremner, Acta Arithmetica, 2001 Simple remark In order that a 3x3 magic square could have 9 square integers, it is necessary that all the possible combinations of 6 square integers, among the 9, have solutions Result… Numerous solutions for each of these 16 possible combinations So there is no « locking » at this level 4x4 solution:  4x4 solution Andrew Bremner, 2001 S2 = 2823 for the 4 rows, 4 columns and 2 diagonals Much research and many publications from 1996 to 2004:  Much research and many publications from 1996 to 2004 Andrew Bremner Acta Arithmetica (2 articles) Duncan Buell Martin Gardner Quantum then Scientific American Richard Guy, problem D15 Unsolved Problems in Number Theory, 3rd edition American Math Monthly Landon Rabern Rose-Hulman Institute Math Journal John Robertson Mathematics Magazine Lee Sallows The Mathematical Intelligencer For all the authors, Martin LaBar is always introduced as the original submitter of the problem My own researchof 2004 - 2005:  My own research of 2004 - 2005 3x3 case: alas no progress 4x4 case: first simple parametric solutions 5x5 case: first known solution First solutions with prime numbers (^²) First solutions with cubes Discovery that Euler, then Lucas, had already worked on the subject (far before LaBar and Gardner) ∑ = Publication in 2005 of an article in The Mathematical Intelligencer 3x3 case: alas no progress:  3x3 case: alas no progress Among the 8 possible configurations of 7 square integers, only one config. allowed the construction of an example (but already known) Numerous attempts with centers reaching 1020 or 1030 (however non exhaustive) for nothing… So, still very far from a supposed example of 8 square integers (Magic hourglass) 4x4 case: first simpleparametric solutions:  S2 = 85(k² + 29) With k = 3 S2 = 85(3² + 29) = 3230 4x4 case: first simple parametric solutions 5x5 case: first known solution:  5x5 case: first known solution S2 = 1375 Distinct integers from 1 to 31 (only 4, 18, 26, 28, 29, 30 are missing) First solutions withprime numbers only (^²):  First solutions with prime numbers only (^²) Why? Only to strengthen the difficulty… 3x3: impossible 4x4: OK ! S2 = 509 020 5x5: OK ! S2 = 34 229 See www.primepuzzles.net First solutions with cubes:  First solutions with cubes 3x3 proved impossible (Reminder: x3 + y3 = 2z3 impossible with x≠y≠z) 4x4 first solution… hmmm… S3 = 0… 5x5 first solution… hmmm… S3 = 0… x is « distinct » from –x,and#xB;and x3 + (-x)3 = 0 Edouard Lucas was the first to propose the 3x3 problem!:  Edouard Lucas was the first to propose the 3x3 problem! Edouard Lucas in 1876, in Nouvelle Correspondance Mathématique (magazine of the Belgian mathematician Eugène Catalan), more than one century before LaBar Jules Verne in Amiens: 1872 Around the World in 80 Days, 1874 The Mysterious Island, 1876 Michael Strogoff Parametric solution of a semi-magic square 6 correct sums S2 = (p²+q²+r²+s²)² Example with (p, q, r, s) = (6, 5, 4, 2), so S2 = 812 = 38 then moving rows and columns Smallest possible squarewith the Lucas method:  Smallest possible square with the Lucas method With 6 sums (p, q, r, s) = (1, 2, 4, 6) S2 = (1²+2²+4²+6²)² = 57² With 8 sums, Lucas mathematically proves that his method does not allow them But with 7 sums, Lucas had not seen that his method did allow them (p, q, r, s) = (1, 3, 4, 11), we retrieve exactly the Sallows and Schweitzer square! And it explains why S2 was a square S2 = (1²+3²+4²+11²)² = 147² Leonhard Euler was the first to construct a square of squares!:  Leonhard Euler was the first to construct a square of squares! Letter sent to Lagrange in 1770, without the method « Permettez-moi, Monsieur, que je vous parle d’un problème fort curieux et digne de toute attention » S2 = 8515 Euler’s letter to Lagrange, of 1770:  Euler’s letter to Lagrange, of 1770 Original found in the Bibliothèque de l’Institut de France, within the correspondence of Lagrange Bibliothèque de l’Institut de France Photo C. Boyer Euler’s 4x4 method:  Euler’s 4x4 method Published in 1770 Method linked to his works attempting to demonstrate that each positive integer is the sum of at most 4 square integers old conjecture of Diophantus, then Bachet and Fermat conjecture which will be completely demonstrated by Lagrange from the partial results of Euler Precursor to Hamilton’s quaternions theory S2 = (a²+b²+c²+d²)(p²+q²+r²+s²) Square sent to Lagrange (a, b, c, d, p, q, r, s) = (5, 5, 9, 0, 6, 4, 2, -3) S2 = 131∙65 = 8515 Euler’s 3x3 semi-magic squares:  Euler’s 3x3 semi-magic squares Also in 1770, studies the 3x3 semi-magic squares of squares Method linked to his works in physics and mechanics Rotation of a solid body around a fixed point Considers only the case of 6 magic sums 3 rows, 3 columns Does not speak of the 8 magic sums problem 3 rows, 3 columns, AND 2 diagonals Lucas will be the first to completely submit the problem Euler’s publication of 1770:  Euler’s publication of 1770 Academy of Sciences of Saint Petersburg Bibliothèque de l’École Polytechnique Photos C. Boyer Smallest possible squarewith the Euler method:  Smallest possible square with the Euler method In 1770: S2 = 8515 Minimum found by Euler In 1942: S2 = 7150 Minimum found by Gaston Benneton, Acad. Sciences Paris and SMF In 2004: S2 = 3230 Absolute minimum (a, b, c, d, p, q, r, s) = (2, 3, 5, 0, 1, 2, 8, -4) Generates the same square already seen with parameter k = 3 Some unsolved problems:  Some unsolved problems Magic squares of squares Magic squares of cubes (of positive integers) Unsolved problems (more):  Unsolved problems (more) 3x3 magic square with 9 distinct square integers or proof ot its impossibility Proposed by Lucas in 1876 Gardner has been offering a $100 prize since 1996! If solution, then center is andgt; 2.5 ∙ 1025 Or an « easier » problem: Another 3x3 magic square with 7 distinct square integers (different from the already known square of Sallows and Bremner, and of its rotations, symmetries, and k² multiples) or a 3x3 magic square with 8 distinct square integers €100 prize offered by Christian Boyer + a bottle of Champagne! Slide30:  In (Springer, New York) In www.multimagie.com END To follow…

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