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

Resort to computingResort to computingon a problem of Euler and on a problem of Euler and

of Lucasof Lucas

Christian BoyerChristian Boyer

Lecture of March 31st 2005, Amiens (Picardy, France)Lecture of March 31st 2005, Amiens (Picardy, France)

Jules Verne, le Centenaire (1905 – 2005)Jules Verne, le Centenaire (1905 – 2005)

ASSOCIATION pour leDEVELOPPEMENT de laCULTURE SCIENTIFIQUE

Union Régionaledes Ingénieurset Scientifiquesde Picardie

March 31, 2005March 31, 2005 © Christian Boyer© Christian Boyer 22

Jules Verne - Edouard LucasJules Verne - Edouard Lucas

(Nantes 1828 – Amiens 1905) (Amiens 1842 – Paris 1891)(Nantes 1828 – Amiens 1905) (Amiens 1842 – Paris 1891)


E. Lu

cas

E. Lu

cas

Am

iens P

aris

Am

iens P

aris

Edouard LucasEdouard Lucas

1842 : Born in Amiens, studied at the imperial school1842 : Born in Amiens, studied at the imperial schoolof Amiens (current Louis Thuillier school)of Amiens (current Louis Thuillier school)

1859 : Scientific diploma (« Bac ») at Amiens,1859 : Scientific diploma (« Bac ») at Amiens,then math studies at Douaithen math studies at Douai

1861 : Passed at Polytechnique and at Normale Sup,1861 : Passed at Polytechnique and at Normale Sup,Chose Normale Sup, and left AmiensChose Normale Sup, and left Amiens

1864 : Paris Observatory1864 : Paris Observatory 1870 : Artillery lieutenant, took part in the battles1870 : Artillery lieutenant, took part in the battles

of the Loire (Orléans, Blois, Le Mans, …)of the Loire (Orléans, Blois, Le Mans, …) 1872 : Math teacher at Mougins1872 : Math teacher at Mougins 1876 : Math teacher at the Charlemagne school1876 : Math teacher at the Charlemagne school

and at the Saint-Louis school, in Parisand at the Saint-Louis school, in Paris 1891 : Accidental death, following a banquet1891 : Accidental death, following a banquet

J. Vern

e

J. Vern

e N

an

tes P

aris Le

Cro

toy / A

mie

ns

Nan

tes P

aris Le

Cro

toy / A

mie

ns


Unsolved problemUnsolved problemof Martin Gardnerof Martin Gardner

Is it possible to construct a 3x3 magic square usingIs it possible to construct a 3x3 magic square using9 distinct square integers?9 distinct square integers?

Gardner asks this question inGardner asks this question in Quantum Quantum, 1996, 1996then again in then again in Scientific AmericanScientific American, 1998, 1998

He attributes the initial problem to Martin LaBar,He attributes the initial problem to Martin LaBar,Southern Wesleyan University, USASouthern Wesleyan University, USA• problem 270 published on 2 linesproblem 270 published on 2 lines

in in The College Mathematics JournalThe College Mathematics Journal, 1984, 1984 He has been offering a $100 prize since 1996He has been offering a $100 prize since 1996

4 9 2

3 5 7

8 1 6

A² B² C²

D² E² F²

G² H² I²


Two ways to see the problemTwo ways to see the problem

1)1) Impose Impose 8 magic sums8 magic sums(3 rows, 3 columns, and(3 rows, 3 columns, and 2 diagonals 2 diagonals ►► sums = 3∙a) sums = 3∙a)

• And try to use the maximum of And try to use the maximum of square integers among thesquare integers among the9 distinct integers used9 distinct integers used

2)2) Impose Impose 9 distinct square integers9 distinct square integers• And try to get the maximum of And try to get the maximum of

magic sums among the 8 sumsmagic sums among the 8 sums

a – b a + b + c a – c

a + b – c a a – b + c

a + c a – b – c a + b

A² B² C²

D² E² F²

G² H² I²


Near solution with 9 squaresNear solution with 9 squares

OK for the 9 square integers,OK for the 9 square integers,but… only but… only 7 correct sums7 correct sums out of out of 88• 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 independentlyBy 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 typeA 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

127² 46² 58²2² 113² 94²74² 82² 97²


Near solution with 8 sumsNear solution with 8 sums

OK for the 8 sums (3 rows, 3 columns, 2 diagonals),OK for the 8 sums (3 rows, 3 columns, 2 diagonals),but… only but… only 7 square integers7 square integers out of out of 99• S2 = 3 ∙ Center = 3 ∙ 425² = 541 875S2 = 3 ∙ Center = 3 ∙ 425² = 541 875

By computing, and independentlyBy 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 typeOnly known solution of this type

373² 289² 565²360721 425² 23²

205² 527² 222121


Link with 3 otherLink with 3 othermathematical problemsmathematical problems

John Robertson, USA, John Robertson, USA, Mathematics MagazineMathematics Magazine::• Arithmetic progressionsArithmetic progressions• Right triangles with the same areaRight triangles with the same area• Congruent numbersCongruent numbers

and elliptic curves yand elliptic curves y22 = x = x33 – n – n22xx


No possible solutionNo possible solutionfor powers for powers ≥≥ 3 3

In each 3x3 magic square, we must have xIn each 3x3 magic square, we must have xnn + y + ynn = 2z = 2znn

• Because xBecause xnn + z + znn + y + ynn = Magic sum = Magic sum = 3 ∙ Center = 3z = 3 ∙ Center = 3znn

xx22 + y + y22 = 2z = 2z22 possible (i.e.: 1 possible (i.e.: 122 + 7 + 722 = 2 ∙ 5 = 2 ∙ 522))

• No usable conclusionNo usable conclusion xx33 + y + y33 = 2z = 2z33 impossible with x≠y≠z (Euler) impossible with x≠y≠z (Euler)

• Implies no 3x3 magic square of cubesImplies no 3x3 magic square of cubes xx44 + y + y44 = 2z = 2z22 impossible (Legendre) impossible (Legendre) ► ► of course 2zof course 2z44 impossible impossible

• Implies no 3x3 magic square of 4th powersImplies no 3x3 magic square of 4th powers xxnn + y + ynn = 2z = 2znn impossible for n ≥ 3 (Noam Elkies, using the Andrew impossible for n ≥ 3 (Noam Elkies, using the Andrew

Wiles proof of the last theorem of Fermat)Wiles proof of the last theorem of Fermat)• Implies no 3x3 magic square of powers ≥3Implies no 3x3 magic square of powers ≥3

x n

z n

y n


Computing researchComputing research

Duncan BuellDuncan Buell• Department of Computer Science and Engineering,Department of Computer Science and Engineering,

University of South Carolina, USAUniversity of South Carolina, USA Background code through most of calendar year 1998Background 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 integersNo « magic hourglass » having 7 square integersfor all a < 2.5 ∙ 10for all a < 2.5 ∙ 102525

a – b a + b + c a – c

a

a + c a – b – c a + b


Computing researchComputing research Andrew Bremner, Andrew Bremner, Acta ArithmeticaActa Arithmetica, 2001, 2001 Simple remarkSimple remark

• In order that a 3x3 magic square could have In order that a 3x3 magic square could have 99 square integers, it is necessary that all the possible square integers, it is necessary that all the possible combinations of combinations of 66 square integers, among the 9, have solutions 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


4x4 solution4x4 solution

Andrew Bremner, 2001Andrew Bremner, 2001 S2 = 2823 for the 4 rows, 4 columns and 2 diagonalsS2 = 2823 for the 4 rows, 4 columns and 2 diagonals

37² 23² 21² 22²1² 18² 47² 17²38² 11² 13² 33²3² 43² 2² 31²


Much research and many Much research and many publications from 1996 to 2004publications from 1996 to 2004

Andrew BremnerAndrew Bremner• Acta ArithmeticaActa Arithmetica (2 articles) (2 articles)

Duncan BuellDuncan Buell Martin GardnerMartin Gardner

• QuantumQuantum then then Scientific AmericanScientific American Richard Guy, problem D15Richard Guy, problem D15

• Unsolved Problems in Number TheoryUnsolved Problems in Number Theory, 3, 3rdrd edition edition• American Math MonthlyAmerican Math Monthly

Landon RabernLandon Rabern• Rose-Hulman Institute Math JournalRose-Hulman Institute Math Journal

John RobertsonJohn Robertson• Mathematics MagazineMathematics Magazine

Lee SallowsLee Sallows• The Mathematical IntelligencerThe Mathematical Intelligencer

For all the authors, Martin LaBar is always introduced as the For all the authors, Martin LaBar is always introduced as the original submitter of the problemoriginal submitter of the problem


My own researchMy own researchof 2004 - 2005of 2004 - 2005

3x3 case: alas no progress3x3 case: alas no progress 4x4 case: first simple parametric solutions4x4 case: first simple parametric solutions 5x5 case: first known solution5x5 case: first known solution First solutions with prime numbers (^²)First solutions with prime numbers (^²) First solutions with cubesFirst solutions with cubes Discovery that Euler, then Lucas, had already workedDiscovery that Euler, then Lucas, had already worked

on the subject (far before LaBar and Gardner)on the subject (far before LaBar and Gardner)

∑ ∑ = Publication in 2005 of an article in= Publication in 2005 of an article inThe Mathematical IntelligencerThe Mathematical Intelligencer


3x3 case: alas no progress3x3 case: alas no progress Among the 8 possible configurations of Among the 8 possible configurations of 77 square integers, only one square integers, only one

config. allowed the construction of an example (but already known)config. allowed the construction of an example (but already known)

Numerous attempts with centers reaching 10Numerous attempts with centers reaching 102020 or 10 or 103030

(however non exhaustive) for nothing…(however non exhaustive) for nothing…

So, still very far from a supposed example of So, still very far from a supposed example of 88 square integers square integers

(Magichourglass)


S2 = 85(k² + 29)S2 = 85(k² + 29)

With k = 3 S2 = 85(3² + 29) = 3230With k = 3 S2 = 85(3² + 29) = 3230

4x4 case: first simple4x4 case: first simpleparametric solutionsparametric solutions

(2k + 42)² (4k + 11)² (8k – 18)² (k + 16)²(k – 24)² (8k + 2)² (4k + 21)² (2k – 38)²

(4k – 11)² (2k – 42)² (k – 16)² (8k + 18)²(8k – 2)² (k + 24)² (2k + 38)² (4k – 21)²

48² 23² 6² 19²21² 26² 33² 32²1² 36² 13² 42²22² 27² 44² 9²


5x5 case: first known solution5x5 case: first known solution

S2 = 1375S2 = 1375 Distinct integers from 1 to 31Distinct integers from 1 to 31

(only 4, 18, 26, 28, 29, 30 are missing)(only 4, 18, 26, 28, 29, 30 are missing)

1² 2² 31² 3² 20²22² 16² 13² 5² 21²11² 23² 10² 24² 7²12² 15² 9² 27² 14²25² 19² 8² 6² 17²


First solutions withFirst solutions withprime numbers only (^²)prime numbers only (^²)

Why? Only to strengthen the difficulty…Why? Only to strengthen the difficulty… 3x3: impossible3x3: impossible

4x4: OK !4x4: OK !S2 = 509 020S2 = 509 020

5x5: OK !5x5: OK !S2 = 34 229S2 = 34 229

See See www.primepuzzles.netwww.primepuzzles.net

29² 191² 673² 137²71² 647² 139² 257²

277² 211² 163² 601²653² 97² 101² 251²

11² 23² 53² 139² 107²13² 103² 149² 31² 17²71² 137² 47² 67² 61²

113² 59² 41² 97² 83²127² 29² 73² 7² 109²


First solutions with cubesFirst solutions with cubes

3x3 proved impossible 3x3 proved impossible (Reminder: x(Reminder: x33 + y + y33 = 2z = 2z33 impossible with x≠y≠z) 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…

19 3 (-3) 3 (-10) 3 (-18) 3

(-42) 3 21 3 28 3 35 3

42 3 (-21) 3 (-28) 3 (-35) 3

(-19) 3 3 3 10 3 18 3

11 3 (-20) 3 12 3 13 3 14 3

(-15) 3 21 3 3 3 (-10) 3 (-17) 3

(-5) 3 (-4) 3 0 3 4 3 5 3

17 3 10 3 (-3) 3 (-21) 3 15 3

(-14) 3 (-13) 3 (-12) 3 20 3 (-11) 3

x is « distinct » from –x,and x3 + (-x)3 = 0


Edouard Lucas was the first to Edouard Lucas was the first to propose the 3x3 problem!propose the 3x3 problem!

Edouard Lucas in 1876, in Edouard Lucas in 1876, in Nouvelle Correspondance Nouvelle Correspondance MathématiqueMathématique (magazine of the Belgian mathematician (magazine of the Belgian mathematician Eugène Catalan), more than one century before LaBarEugène Catalan), more than one century before LaBar

• Jules Verne in Amiens: 1872 Around the World in 80 Days,Jules Verne in Amiens: 1872 Around the World in 80 Days,1874 The Mysterious Island, 1876 Michael Strogoff1874 The Mysterious Island, 1876 Michael Strogoff

Parametric solution of a semi-magic squareParametric solution of a semi-magic square• 66 correct sums S2 = (p²+q²+r²+s²)² correct sums S2 = (p²+q²+r²+s²)²

Example with (p, q, r, s) = (6, 5, 4, 2), so S2 = 81Example with (p, q, r, s) = (6, 5, 4, 2), so S2 = 812 2 = 3= 388

• then moving rows and columnsthen moving rows and columns

(p² + q² – r² – s²)² [2(qr + ps)]² [2(qs – pr)]²

[2(qr – ps)]² (p² – q² + r² – s²)² [2(rs + pq)]²

[2(qs + pr)]² [2(rs – pq)]² (p² – q² – r² + s²)²

1² 68² 44²76² 16² 23²28² 41² 64²


Smallest possible squareSmallest possible squarewith the Lucas methodwith the Lucas method

With With 66 sums 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 With 88 sums, Lucas mathematically proves that his sums, Lucas mathematically proves that his method does not allow themmethod does not allow them

But with But with 77 sums, Lucas had not seen that his method did sums, Lucas had not seen that his method did allow themallow them• (p, q, r, s) = (1, 3, 4, 11), we retrieve exactly the Sallows and (p, q, r, s) = (1, 3, 4, 11), we retrieve exactly the Sallows and

Schweitzer square!Schweitzer square!

• And it explains why S2 was a squareAnd it explains why S2 was a squareS2 = (1²+3²+4²+11²)² = 147²S2 = (1²+3²+4²+11²)² = 147²

47² 28² 16²4² 23² 52²32² 44² 17²

127² 46² 58²2² 113² 94²74² 82² 97²


Leonhard Euler was the first to Leonhard Euler was the first to construct a square of squares!construct a square of squares!

Letter sent to Lagrange in Letter sent to Lagrange in 1770, without the method1770, without the method

« « Permettez-moi, Monsieur, Permettez-moi, Monsieur, que je vous parle d’un que je vous parle d’un problème fort curieux et problème fort curieux et digne de toute attentiondigne de toute attention » »

S2 = 8515S2 = 8515

68² 29² 41² 37²17² 31² 79² 32²59² 28² 23² 61²11² 77² 8² 49²


Euler’s letter to Lagrange, of 1770Euler’s letter to Lagrange, of 1770

Original found in the Bibliothèque de l’Institut de France,Original found in the Bibliothèque de l’Institut de France,within the correspondence of Lagrangewithin the correspondence of Lagrange

Bibliothèque de l’Institut de FrancePhoto C. Boyer


Euler’s 4x4 methodEuler’s 4x4 method Published in 1770Published 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 Lagrangeconjecture which will be completely demonstrated by Lagrange

from the partial results of Eulerfrom the partial results of Euler Precursor to Hamilton’s quaternions theoryPrecursor 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 LagrangeSquare 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

(+ap+bq+cr+ds)² (+ar–bs–cp+dq)² (–as-br+cq+dp)² (+aq–bp+cs–dr)²

(–aq+bp+cs–dr)² (+as+br+cq+dp)² (+ar–bs+cp-dq)² (+ap+bq–cr–ds)²

(+ar+bs–cp–dq)² (–ap+bq–cr+ds)² (+aq+bp+cs+dr)² (+as–br–cq+dp)²

(–as+br–cq+dp)² (–aq–bp+cs+dr)² (–ap+bq+cr–ds)² (+ar+bs+cp+dq)²


Euler’s 3x3 semi-magic squaresEuler’s 3x3 semi-magic squares Also in 1770, studies the 3x3 semi-magic squares of squaresAlso in 1770, studies the 3x3 semi-magic squares of squares Method linked to his works in physics and mechanicsMethod 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 sumsConsiders only the case of 6 magic sums

• 3 rows, 3 columns3 rows, 3 columns

Does not speak of the 8 magic sums problemDoes 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 problemLucas will be the first to completely submit the problem

+47/57 +28/57 -16/57+4/57 +23/57 +52/57

+32/57 -44/57 +17/57


Euler’s publication of 1770Euler’s publication of 1770 Academy of Sciences of Saint PetersburgAcademy of Sciences of Saint Petersburg

Bibliothèque de l’École PolytechniquePhotos C. Boyer


Smallest possible squareSmallest possible squarewith the Euler methodwith the Euler method

In 1770In 1770: S2 = 8515: S2 = 8515• Minimum found by EulerMinimum found by Euler

In 1942In 1942: S2 = 7150: S2 = 7150• Minimum found by Gaston Benneton, Acad. Sciences Paris and SMFMinimum found by Gaston Benneton, Acad. Sciences Paris and SMF

In 2004In 2004: S2 = 3230: 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

48² 23² 6² 19²21² 26² 33² 32²1² 36² 13² 42²22² 27² 44² 9²


Some unsolved problemsSome unsolved problems Magic squares of squaresMagic squares of squares

Magic squares of cubesMagic squares of cubes(of positive integers)(of positive integers)

3x33x3 Who?Who?

4x44x4 Euler, 1770Euler, 1770

5x55x5 Boyer, 2004Boyer, 2004

6x66x6 Who?Who?

7x77x7 Who?Who?

8x8 et +8x8 et + Bimagic knownBimagic known

3x33x3 ImpossibleImpossible

4x4 … 11x114x4 … 11x11 Who?Who?

12x12 et +12x12 et + Trimagic knownTrimagic known


Unsolved problems (more)Unsolved problems (more) 3x3 magic square with 3x3 magic square with 99 distinct square integers distinct square integers

or proof ot its impossibilityor 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 ∙ 10If solution, then center is > 2.5 ∙ 102525

Or an « easier » problem:Or an « easier » problem:

Another 3x3 magic square with Another 3x3 magic square with 77 distinct square integers distinct square integers ( (different from the already known square of Sallows and Bremner,different from the already known square of Sallows and Bremner, and of its rotations, symmetries, and k² multiples and of its rotations, symmetries, and k² multiples))or a 3x3 magic square with or a 3x3 magic square with 88 distinct square integers distinct square integers• €€100 prize offered by Christian Boyer100 prize offered by Christian Boyer• + a bottle of Champagne+ a bottle of Champagne! !


InIn

(Springer, New York) (Springer, New York)

In In www.multimagie.comwww.multimagie.com

END To follow…END To follow…

Documents

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