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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.

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Presentation on theme: "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."— Presentation transcript:

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2 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

3 March 31, 2005 © Christian Boyer 2 Jules Verne - Edouard Lucas Jules Verne - Edouard Lucas (Nantes 1828 – Amiens 1905) (Amiens 1842 – Paris 1891)

4 March 31, 2005 © Christian Boyer 3 E. Lucas Amiens Paris Edouard Lucas 1842 : Born in Amiens, studied at the imperial school of Amiens (current Louis Thuillier school) 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 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 1861 : Passed at Polytechnique and at Normale Sup, Chose Normale Sup, and left Amiens 1864 : Paris Observatory 1864 : Paris Observatory 1870 : Artillery lieutenant, took part in the battles of the Loire (Orléans, Blois, Le Mans, …) 1870 : Artillery lieutenant, took part in the battles of the Loire (Orléans, Blois, Le Mans, …) 1872 : Math teacher at Mougins 1872 : Math teacher at Mougins 1876 : Math teacher at the Charlemagne school and at the Saint-Louis school, in Paris 1876 : Math teacher at the Charlemagne school and at the Saint-Louis school, in Paris 1891 : Accidental death, following a banquet 1891 : Accidental death, following a banquet J. Verne Nantes Paris Le Crotoy / Amiens

5 March 31, 2005 © Christian Boyer 4 Unsolved problem of Martin Gardner Is it possible to construct a 3x3 magic square using 9 distinct square integers? 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 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 He attributes the initial problem to Martin LaBar, Southern Wesleyan University, USA problem 270 published on 2 lines in The College Mathematics Journal, 1984problem 270 published on 2 lines in The College Mathematics Journal, 1984 He has been offering a $100 prize since 1996 He has been offering a $100 prize since 1996

6 March 31, 2005 © Christian Boyer 5 Two ways to see the problem 1)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 usedAnd try to use the maximum of square integers among the 9 distinct integers used 2)Impose 9 distinct square integers And try to get the maximum of magic sums among the 8 sumsAnd try to get the maximum of magic sums among the 8 sums

7 March 31, 2005 © Christian Boyer 6 Near solution with 9 squares OK for the 9 square integers, but… only 7 correct sums out of 8 OK for the 9 square integers, but… only 7 correct sums out of 8 S2 = 21 609 = 147² for 3 rows, 3 columns, 1 diagonalS2 = 21 609 = 147² for 3 rows, 3 columns, 1 diagonal Unfortunately S2 = 38 307 for the other diagonalUnfortunately S2 = 38 307 for the other diagonal By computing, and independently By computing, and independently 1996: Lee Sallows, University of Nijmegen, Netherlands1996: Lee Sallows, University of Nijmegen, Netherlands 1996: Michaël Schweitzer, Göttingen, Germany1996: Michaël Schweitzer, Göttingen, Germany A lot of other known solutions of this type A lot of other known solutions of this type S2 is often a square, as it is in this solutionS2 is often a square, as it is in this solution

8 March 31, 2005 © Christian Boyer 7 Near solution with 8 sums OK for the 8 sums (3 rows, 3 columns, 2 diagonals), but… only 7 square integers out of 9 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 875S2 = 3 ∙ Center = 3 ∙ 425² = 541 875 By computing, and independently By computing, and independently 1997: Lee Sallows, University of Nijmegen, Netherlands1997: Lee Sallows, University of Nijmegen, Netherlands 1997: Andrew Bremner, Arizona State University, USA1997: Andrew Bremner, Arizona State University, USA Only known solution of this type Only known solution of this type

9 March 31, 2005 © Christian Boyer 8 Link with 3 other mathematical problems John Robertson, USA, Mathematics Magazine: John Robertson, USA, Mathematics Magazine: Arithmetic progressionsArithmetic progressions Right triangles with the same areaRight triangles with the same area Congruent numbers and elliptic curves y 2 = x 3 – n 2 xCongruent numbers and elliptic curves y 2 = x 3 – n 2 x

10 March 31, 2005 © Christian Boyer 9 No possible solution for powers ≥ 3 In each 3x3 magic square, we must have x n + y n = 2z n In each 3x3 magic square, we must have x n + y n = 2z n Because x n + z n + y n = Magic sum = 3 ∙ Center = 3z nBecause x n + z n + y n = Magic sum = 3 ∙ Center = 3z n x 2 + y 2 = 2z 2 possible (i.e.: 1 2 + 7 2 = 2 ∙ 5 2 ) x 2 + y 2 = 2z 2 possible (i.e.: 1 2 + 7 2 = 2 ∙ 5 2 ) No usable conclusionNo usable conclusion x 3 + y 3 = 2z 3 impossible with x≠y≠z (Euler) x 3 + y 3 = 2z 3 impossible with x≠y≠z (Euler) Implies no 3x3 magic square of cubesImplies no 3x3 magic square of cubes x 4 + y 4 = 2z 2 impossible (Legendre) ► of course 2z 4 impossible x 4 + y 4 = 2z 2 impossible (Legendre) ► of course 2z 4 impossible Implies no 3x3 magic square of 4th powersImplies no 3x3 magic square of 4th powers x n + y n = 2z n impossible for n ≥ 3 (Noam Elkies, using the Andrew Wiles proof of the last theorem of Fermat) x n + y n = 2z n impossible for n ≥ 3 (Noam Elkies, using the Andrew Wiles proof of the last theorem of Fermat) Implies no 3x3 magic square of powers ≥3Implies no 3x3 magic square of powers ≥3

11 March 31, 2005 © Christian Boyer 10 Computing research Duncan Buell Duncan Buell Department of Computer Science and Engineering, University of South Carolina, USADepartment of Computer Science and Engineering, University of South Carolina, USA Background code through most of calendar year 1998 Background code through most of calendar year 1998 Multiple processors, SGI Challenge computer of his universityMultiple processors, SGI Challenge computer of his university Result… Result… No « magic hourglass » having 7 square integers for all a < 2.5 ∙ 10 25No « magic hourglass » having 7 square integers for all a < 2.5 ∙ 10 25

12 March 31, 2005 © Christian Boyer 11 Computing research Andrew Bremner, Acta Arithmetica, 2001 Andrew Bremner, Acta Arithmetica, 2001 Simple remark 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 solutionsIn 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… Result… Numerous solutions for each of these 16 possible combinationsNumerous solutions for each of these 16 possible combinations So there is no « locking » at this levelSo there is no « locking » at this level

13 March 31, 2005 © Christian Boyer 12 4x4 solution Andrew Bremner, 2001 Andrew Bremner, 2001 S2 = 2823 for the 4 rows, 4 columns and 2 diagonals S2 = 2823 for the 4 rows, 4 columns and 2 diagonals

14 March 31, 2005 © Christian Boyer 13 Much research and many publications from 1996 to 2004 Andrew Bremner Andrew Bremner Acta Arithmetica (2 articles)Acta Arithmetica (2 articles) Duncan Buell Duncan Buell Martin Gardner Martin Gardner Quantum then Scientific AmericanQuantum then Scientific American Richard Guy, problem D15 Richard Guy, problem D15 Unsolved Problems in Number Theory, 3 rd editionUnsolved Problems in Number Theory, 3 rd edition American Math MonthlyAmerican Math Monthly Landon Rabern Landon Rabern Rose-Hulman Institute Math JournalRose-Hulman Institute Math Journal John Robertson John Robertson Mathematics MagazineMathematics Magazine Lee Sallows Lee Sallows The Mathematical IntelligencerThe Mathematical Intelligencer For all the authors, Martin LaBar is always introduced as the original submitter of the problem For all the authors, Martin LaBar is always introduced as the original submitter of the problem

15 March 31, 2005 © Christian Boyer 14 My own research of 2004 - 2005 3x3 case: alas no progress 3x3 case: alas no progress 4x4 case: first simple parametric solutions 4x4 case: first simple parametric solutions 5x5 case: first known solution 5x5 case: first known solution First solutions with prime numbers (^²) First solutions with prime numbers (^²) First solutions with cubes First solutions with cubes Discovery that Euler, then Lucas, had already worked on the subject (far before LaBar and Gardner) 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 ∑ = Publication in 2005 of an article in The Mathematical Intelligencer

16 March 31, 2005 © Christian Boyer 15 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) 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 10 20 or 10 30 (however non exhaustive) for nothing… Numerous attempts with centers reaching 10 20 or 10 30 (however non exhaustive) for nothing… So, still very far from a supposed example of 8 square integers So, still very far from a supposed example of 8 square integers (Magic hourglass)

17 March 31, 2005 © Christian Boyer 16 S2 = 85(k² + 29) S2 = 85(k² + 29) With k = 3 S2 = 85(3² + 29) = 3230 With k = 3 S2 = 85(3² + 29) = 3230 4x4 case: first simple parametric solutions

18 March 31, 2005 © Christian Boyer 17 5x5 case: first known solution S2 = 1375 S2 = 1375 Distinct integers from 1 to 31 (only 4, 18, 26, 28, 29, 30 are missing) Distinct integers from 1 to 31 (only 4, 18, 26, 28, 29, 30 are missing)

19 March 31, 2005 © Christian Boyer 18 First solutions with prime numbers only (^²) Why? Only to strengthen the difficulty… Why? Only to strengthen the difficulty… 3x3: impossible 3x3: impossible 4x4: OK ! S2 = 509 020 4x4: OK ! S2 = 509 020 5x5: OK ! S2 = 34 229 5x5: OK ! S2 = 34 229 See www.primepuzzles.net See www.primepuzzles.netwww.primepuzzles.net

20 March 31, 2005 © Christian Boyer 19 First solutions with cubes 3x3 proved impossible (Reminder: x 3 + y 3 = 2z 3 impossible with x≠y≠z) 3x3 proved impossible (Reminder: x 3 + y 3 = 2z 3 impossible with x≠y≠z) 4x4 first solution… hmmm… S3 = 0… 4x4 first solution… hmmm… S3 = 0… 5x5 first solution… hmmm… S3 = 0… 5x5 first solution… hmmm… S3 = 0… x is « distinct » from –x, and x 3 + (-x) 3 = 0

21 March 31, 2005 © Christian Boyer 20 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 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 StrogoffJules Verne in Amiens: 1872 Around the World in 80 Days, 1874 The Mysterious Island, 1876 Michael Strogoff Parametric solution of a semi-magic square Parametric solution of a semi-magic square 6 correct sums S2 = (p²+q²+r²+s²)²6 correct sums S2 = (p²+q²+r²+s²)² Example with (p, q, r, s) = (6, 5, 4, 2), so S2 = 81 2 = 3 8 Example with (p, q, r, s) = (6, 5, 4, 2), so S2 = 81 2 = 3 8 then moving rows and columnsthen moving rows and columns

22 March 31, 2005 © Christian Boyer 21 Smallest possible square with the Lucas method With 6 sums With 6 sums (p, q, r, s) = (1, 2, 4, 6)(p, q, r, s) = (1, 2, 4, 6) S2 = (1²+2²+4²+6²)² = 57²S2 = (1²+2²+4²+6²)² = 57² With 8 sums, Lucas mathematically proves that his method does not allow them 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 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!(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 = 147²And it explains why S2 was a square S2 = (1²+3²+4²+11²)² = 147²

23 March 31, 2005 © Christian Boyer 22 Leonhard Euler was the first to construct a square of squares! Letter sent to Lagrange in 1770, without the method 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 » « Permettez-moi, Monsieur, que je vous parle d’un problème fort curieux et digne de toute attention » S2 = 8515 S2 = 8515

24 March 31, 2005 © Christian Boyer 23 Euler’s letter to Lagrange, of 1770 Original found in the Bibliothèque de l’Institut de France, within the correspondence of Lagrange Original found in the Bibliothèque de l’Institut de France, within the correspondence of Lagrange Bibliothèque de l’Institut de France Photo C. Boyer

25 March 31, 2005 © Christian Boyer 24 Euler’s 4x4 method Published in 1770 Published in 1770 Method linked to his works attempting to demonstrate that Method linked to his works attempting to demonstrate that each positive integer is the sum of at most 4 square integerseach positive integer is the sum of at most 4 square integers old conjecture of Diophantus, then Bachet and Fermatold conjecture of Diophantus, then Bachet and Fermat conjecture which will be completely demonstrated by Lagrange from the partial results of Eulerconjecture which will be completely demonstrated by Lagrange from the partial results of Euler Precursor to Hamilton’s quaternions theory Precursor to Hamilton’s quaternions theory S2 = (a²+b²+c²+d²)(p²+q²+r²+s²) S2 = (a²+b²+c²+d²)(p²+q²+r²+s²) Square sent to Lagrange Square sent to Lagrange (a, b, c, d, p, q, r, s) = (5, 5, 9, 0, 6, 4, 2, -3)(a, b, c, d, p, q, r, s) = (5, 5, 9, 0, 6, 4, 2, -3) S2 = 131∙65 = 8515S2 = 131∙65 = 8515

26 March 31, 2005 © Christian Boyer 25 Euler’s 3x3 semi-magic squares Also in 1770, studies the 3x3 semi-magic squares of squares Also in 1770, studies the 3x3 semi-magic squares of squares Method linked to his works in physics and mechanics Method linked to his works in physics and mechanics Rotation of a solid body around a fixed pointRotation of a solid body around a fixed point Considers only the case of 6 magic sums Considers only the case of 6 magic sums 3 rows, 3 columns3 rows, 3 columns Does not speak of the 8 magic sums problem Does not speak of the 8 magic sums problem 3 rows, 3 columns, AND 2 diagonals3 rows, 3 columns, AND 2 diagonals Lucas will be the first to completely submit the problem Lucas will be the first to completely submit the problem

27 March 31, 2005 © Christian Boyer 26 Euler’s publication of 1770 Academy of Sciences of Saint Petersburg Academy of Sciences of Saint Petersburg Bibliothèque de l’École Polytechnique Photos C. Boyer

28 March 31, 2005 © Christian Boyer 27 Smallest possible square with the Euler method In 1770: S2 = 8515 In 1770: S2 = 8515 Minimum found by EulerMinimum found by Euler In 1942: S2 = 7150 In 1942: S2 = 7150 Minimum found by Gaston Benneton, Acad. Sciences Paris and SMFMinimum found by Gaston Benneton, Acad. Sciences Paris and SMF In 2004: S2 = 3230 In 2004: S2 = 3230 Absolute minimum (a, b, c, d, p, q, r, s) = (2, 3, 5, 0, 1, 2, 8, -4)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 = 3Generates the same square already seen with parameter k = 3

29 March 31, 2005 © Christian Boyer 28 Some unsolved problems Magic squares of squares Magic squares of squares Magic squares of cubes (of positive integers) Magic squares of cubes (of positive integers) 3x3Who? 4x4 Euler, 1770 5x5 Boyer, 2004 6x6Who? 7x7Who? 8x8 et + Bimagic known 3x3Impossible 4x4 … 11x11 Who? 12x12 et + Trimagic known

30 March 31, 2005 © Christian Boyer 29 Unsolved problems (more) 3x3 magic square with 9 distinct square integers or proof ot its impossibility 3x3 magic square with 9 distinct square integers or proof ot its impossibility Proposed by Lucas in 1876Proposed by Lucas in 1876 Gardner has been offering a $100 prize since 1996!Gardner has been offering a $100 prize since 1996! If solution, then center is > 2.5 ∙ 10 25If solution, then center is > 2.5 ∙ 10 25 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 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€100 prize offered by Christian Boyer + a bottle of Champagne !+ a bottle of Champagne !

31 March 31, 2005 © Christian Boyer 30 In (Springer, New York) In (Springer, New York) In www.multimagie.com In www.multimagie.comwww.multimagie.com END To follow… END To follow…


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