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Abstract Updated 2016 On the Marko↵ Equation x 2 + y 2 + z 2 = 3xyz It is easy to check that the equation x 2 + y 2 + z 2 = 3xyz, where the three unknowns x, y , z are positive integers, has infinitely many solutions. There is a simple algorithm which produces all of them. However, this does not answer to all questions on this equation : in particular, Frobenius asked whether it is true that for each integer z > 0, there is at most one pair (x, y ) such that x < y < z and (x, y , z) is a solution. This question is an active research topic nowadays. Michel Waldschmidt Institut de Mathématiques de Jussieu (Paris VI) http://www.math.jussieu.fr/⇠miw/ 1 / 87 Abstract (continued) 2 / 87 The sequence of Marko↵ numbers A Marko↵ number is a positive integer z such that there exist two positive integers x and y satisfying Marko↵’s equation occurred initially in the study of minima of quadratic forms at the end of the XIX–th century and the beginning of the XX–th century. It was investigated by many a mathematician, including Lagrange, Hermite, Korkine, Zolotarev, Marko↵, Frobenius, Hurwitz, Cassels. The solutions are related with the Lagrange-Marko↵ spectrum, which consists of those quadratic numbers which are badly approximable by rational numbers. It occurs also in other parts of mathematics, in particular free groups, Fuchsian groups and hyperbolic Riemann surfaces (Ford, Lehner, Cohn, Rankin, Conway, Coxeter, Hirzebruch and Zagier. . .). We discuss some aspects of this topic without trying to cover all of them. Andrei Andreyevich Marko↵ (1856–1922) x 2 + y 2 + z 2 = 3xyz. For instance 1 is a Marko↵ number, since (x, y , z) = (1, 1, 1) is a solution. Photos : http://www-history.mcs.st-andrews.ac.uk/history/ 3 / 87 4 / 87 The On-Line Encyclopedia of Integer Sequences Integer points on a surface 1, 2, 5, 13, 29, 34, 89, 169, 194, 233, 433, 610, 985, 1325, 1597, 2897, 4181, 5741, 6466, 7561, 9077, 10946, 14701, 28657, 33461, 37666, 43261, 51641, 62210, 75025, 96557, 135137, 195025, 196418, 294685, . . . The sequence of Marko↵ numbers is available on the web The On-Line Encyclopedia of Integer Sequences Neil J. A. Sloane Given a Marko↵ number z, there exist infinitely many pairs of positive integers x and y satisfying x 2 + y 2 + z 2 = 3xyz. This is a cubic equation in the 3 variables (x, y , z), of which we know a solution (1, 1, 1). There is an algorithm producing all integer solutions. http://oeis.org/A002559 5 / 87 Marko↵’s cubic variety 6 / 87 Algorithm producing all solutions Let (m, m1 , m2 ) be a solution of Marko↵’s equation : The surface defined by Marko↵’s equation A.A. Marko↵ (1856–1922) m2 + m12 + m22 = 3mm1 m2 . Fix two coordinates of this solution, say m1 and m2 . We get a quadratic equation in the third coordinate m, of which we know a solution, hence, the equation x 2 + y 2 + z 2 = 3xyz. is an algebraic variety with many automorphisms : permutations of the variables, changes of signs and (x, y , z) 7! (3yz x 2 + m12 + m22 = 3xm1 m2 . has two solutions, x = m and, say, x = m0 , with m + m0 = 3m1 m2 and mm0 = m12 + m22 . This is the cord and tangente process. x, y , z). Hence, another solution is (m0 , m1 , m2 ) with m0 = 3m1 m2 7 / 87 m. 8 / 87 Three solutions derived from one New solutions We shall see that any solution di↵erent from (1, 1, 1) and from (2, 1, 1) yields three new di↵erent solutions – and we shall see also that, in each other solution, the three numbers m, m1 and m2 are pairwise distinct. Starting with one solution (m, m1 , m2 ), we derive three new solutions : (m0 , m1 , m2 ), (m, m10 , m2 ), (m, m1 , m20 ). Two solutions are called neighbors if they share two components. If the solution we start with is (1, 1, 1), we produce only one new solution, (2, 1, 1) (up to permutation). For instance • (1, 1, 1) has a single neighbor, namely (2, 1, 1), • (2, 1, 1) has two neighbors : (1, 1, 1) et (5, 2, 1), • any other solution has exactly three neighbors. If we start from (2, 1, 1), we produce only two new solutions, (1, 1, 1) and (5, 2, 1) (up to permutation). A new solution means distinct from the one we start with. 9 / 87 Marko↵’s tree 10 / 87 This algorithm yields all the solutions Assume we start with (m, m1 , m2 ) satisfying m > m1 > m2 . We shall check m20 > m10 > m > m0 . Conversely, starting from any solution other than (1, 1, 1), the algorithm produces a smaller solution. Hence, by induction, we get a sequence of smaller and smaller solutions, until we reach (1, 1, 1). We order the solution according to the largest coordinate. Then two of the neighbors of (m, m1 , m2 ) are larger than the initial solution, the third one is smaller. Therefore the solution we started from was in Marko↵’s tree. Hence, if we start from (1, 1, 1), we produce infinitely many solutions, which we organize in a tree : this is Marko↵’s tree. 11 / 87 12 / 87 First branches of Marko↵’s tree Marko↵’s tree starting from (2, 5, 29) FIGURE 2 Markoff triples ( p , q, r ) with max( p , q ) 13 / 87 14 / 87 Conversely, given fractions a Markoff and triplethe ( p , Marko↵ q, r ) with tree r > 1, one checks easily th Continued 3pq - r < r; and from t h s it follows by induction that all Markoff triples occu and occur only once, on this tree (for a fuller discussion of t h s and other propertie of the Markoff tree, see [2]). E. Bombieri, To prove the theorem we must analyze the asymptotic behavior of the Marko Continued fractions and the(1) we find that 3r2 2 3pqr or r 2 pq; if p is larg tree. From the Markoff equation Marko↵ tree, for all but a small portion of the tree, contributing O(log x ) (which will happen Expo.this Math. 25 (2007), M(x)), then implies that r is much larger than q and hence (1) gives r 2 < 3pqr r 2 o(r2) r 3pq. Multiplying both sides of this equation by 3 and takin no. 3,or187–213. logarithms gives Marko↵’s tree up to 100 000 DON ZAGIER Don Zagier, On the number of Marko↵ numbers below a given bound. Mathematics of Computation, 39 160 (1982), 709–723. + - log(3p) FIGURE 2 Markoff triples ( p , q, r ) with max( p , q ) 100000 + log(3q) = log(3r) + o(1) ( p large) 100000 Conversely, given a Markoff triple ( p , q, r ) with r > 1, one checks easily that 3pq - r < r; and from t h s it follows by induction that all Markoff triples occur, and occur only once, on this tree (for a fuller discussion of t h s and other properties of the Markoff tree, see [2]). To prove the theorem we must analyze the asymptotic behavior of the Markoff tree. From the Markoff equation (1) we find that 3r2 2 3pqr or r 2 pq; if p is large 15 / 87 16 / 87 Marko↵’s tree Letter from Masanobu Kaneko to Yuri Manin 26/06/2008 17 / 87 a2 + b 2 + c 2 = 3abc X2 18 / 87 The Fibonacci sequence and the Marko↵ equation The smallest Marko↵ number is 1. When we impose z = 1 in the Marko↵ equation x 2 + y 2 + z 2 = 3xyz, we obtain the equation x 2 + y 2 + 1 = 3xy . 3abX + a2 + b 2 = (X c)(X 3ab + c) Going along the Marko↵’s tree starting from (1, 1, 1), we obtain the subsequence of Marko↵ numbers 1, 2, 5, 13, 34, 89, 233, 610, 1597, 4181, 10946, 28657, . . . which is the sequence of Fibonacci numbers with odd indices F1 = 1, F3 = 2, F5 = 5, F7 = 13, F9 = 34, F11 = 89, . . . 19 / 87 20 / 87 Leonardo Pisano (Fibonacci) The Fibonacci sequence (Fn )n 0 : Encyclopedia of integer sequences (again) 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, . . . Leonardo Pisano (Fibonacci) (1170–1250) The Fibonacci sequence is available online The On-Line Encyclopedia of Integer Sequences 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233 . . . is defined by Neil J. A. Sloane F0 = 0, F1 = 1, Fn = F n 1 + Fn (n 2 2). http://oeis.org/A000045 21 / 87 Fibonacci numbers with odd indices Order of the new solutions Let (m, m1 , m2 ) be a solution of Marko↵’s equation Fibonacci numbers with odd indices are Marko↵’s numbers : Fm+3 Fm 1 2 Fm+1 = ( 1)m for m m2 + m12 + m22 = 3mm1 m2 . 1 and Fm+3 + Fm 1 = 3Fm+1 for m 22 / 87 Denote by m0 the other root of the quadratic polynomial 1. X2 3m1 m2 X + m12 + m22 . Set y = Fm+1 , x = Fm 1 , x 0 = Fm+3 , so that, for even m, x + x 0 = 3y , Hence, xx 0 = y 2 + 1 X2 and X2 3yX + y 2 + 1 = (X x)(X x 0 ). 3m1 m2 X + m12 + m22 = (X m)(X m0 ) and m + m0 = 3m1 m2 , 23 / 87 mm0 = m12 + m22 . 24 / 87 m1 6= m2 Two larger, one smaller Assume m1 > m2 . Question : Do we have m0 > m1 or else m0 < m1 ? Let us check that if m1 = m2 , then m1 = m2 = 1 : this holds only for the two exceptional solutions (1, 1, 1), (2, 1, 1). Consider the number a = (m1 Assume m1 = m2 . We have m2 + 2m12 = 3mm12 m)(m1 m0 ). Since m + m0 = 3m1 m2 , and mm0 = m12 + m22 , we have hence m2 = (3m 2)m12 . a= m12 m1 (m + m0 ) + mm0 = 2m12 + m22 Therefore m1 divides m. Let m = km1 . We have k 2 = 3km1 2, hence k divise 2. For k = 1 we get m = m1 = 1. For k = 2 we get m1 = 1, m = 2. = (2m12 3m12 m2 2m12 m2 ) + (m22 m12 m2 ). However 2m12 < 2m12 m2 and m22 < m12 m2 , hence a < 0. Consider now a solution distinct from (1, 1, 1) or (2, 1, 1) : hence m1 6= m2 . This means that m1 is in the interval defined by m and m0 . 25 / 87 Order of the solutions 26 / 87 Prime factors Remark. Let m be a Marko↵ number with m2 + m12 + m22 = 3mm1 m2 . If m > m1 , we have m1 > m0 and the new solution (m0 , m1 , m2 ) is smaller than the initial solution (m, m1 , m2 ). If m < m1 , we have m1 < m0 and the new solution (m0 , m1 , m2 ) is larger than the initial solution (m, m1 , m2 ). The same proof shows that the gcd of m, m1 and m2 is 1 : indeed, if p divides m1 , m2 and m, then p divides the new solutions which are produced by the preceding process – going down in the tree shows that p would divide 1. The odd prime factors of m are all congruent to 1 modulo 4 (since they divide a sum of two relatively prime squares). If m is even, then the numbers m , 2 3m 2 , 4 3m + 2 , 8 are odd integers. 27 / 87 28 / 87 Marko↵’s Conjecture Frobenius’s work The previous algorithm produces the sequence of Marko↵ numbers. Each Marko↵ number occurs infinitely often in the tree as one of the components of the solution. Marko↵’s Conjecture does not occur in Marko↵’s 1879 and 1880 papers but in Frobenius’s one in 1913. Ferdinand Georg Frobenius (1849–1917) According to the definition, for a Marko↵ number m > 2, there exist a pair (m1 , m2 ) of positive integers with m > m1 > m2 such that m2 + m12 + m22 = 3mm1 m2 . Question : Given m, is such a pair (m1 , m2 ) unique ? The answer is yes, as long as m 10105 . 29 / 87 Special cases 30 / 87 Powers of a prime number The Conjecture has been proved for certain classes of Marko↵ numbers m like pn ± 2 pn , 3 for p prime. A. Baragar (1996), P. Schmutz (1996), J.O. Button (1998), M.L. Lang, S.P. Tan (2005), Ying Zhang (2007). Arthur Baragar Anitha Srinivasan, 2007 A really simple proof of the Marko↵ conjecture for prime powers Number Theory Web Created and maintained by Keith Matthews, Brisbane, Australia www.numbertheory.org/pdfs/simpleproof.pdf http://www.nevada.edu/ baragar/ 31 / 87 32 / 87 Marko↵ Equation and Nilpotent Matrices arXiv :0709.1499 Marko↵ Equation and Nilpotent Matrices arXiv :0709.1499 [math.NT] From : Norbert Riedel Submission history [v1] Mon, 10 Sep 2007 22 :11 :39 GMT (11kb) [v2] Thu, 13 Sep 2007 18 :45 :29 GMT (11kb) [v3] Tue, 4 Dec 2007 17 :43 :40 GMT (15kb) [v4] Tue, 5 Aug 2008 21 :24 :23 GMT (15kb) [v5] Thu, 12 Mar 2009 14 :08 :48 GMT (15kb) [v6] Tue, 28 Jul 2009 18 :49 :17 GMT (15kb) [v7] Fri, 29 Mar 2013 12 :36 :56 GMT (0kb,I) Comments : Most of the (correct) portion of this paper has been incorporated into the paper ”On the Marko↵ equation” (arXiv :1208.4032) Norbert Riedel http://fr.arxiv.org/abs/0709.1499 A triple (a, b, c) of positive integers is called a Marko↵ triple i↵ it satisfies the diophantine equation a2 + b 2 + c 2 = abc. Recasting the Marko↵ tree, whose vertices are Marko↵ triples, in the framework of integral upper triangular 3 ⇥ 3 matrices, it will be shown that the largest member of such a triple determines the other two uniquely. This answers a question which has been open for almost 100 years. 33 / 87 On the Marko↵ Equation arXiv :1208.4032 34 / 87 Serge Perrine http://www.tessier-ashpool.fr/html/marko↵.html [v1] Mon, 20 Aug 2012 15 :05 :47 GMT (51kb) [v2] Mon, 1 Apr 2013 10 :21 :39 GMT (52kb) [v3] Wed, 15 May 2013 17 :39 :35 GMT (52kb) [v4] Mon, 8 Jul 2013 17 :35 :21 GMT (53kb) [v5] Sun, 13 Oct 2013 21 :23 :28 GMT (50kb) [v6] Sun, 20 Oct 2013 13 :37 :31 GMT (50kb) [v7] Mon, 25 Nov 2013 19 :11 :53 GMT (53kb) [v8] Sun, 12 Jan 2014 14 :59 :52 GMT (53kb) [v9] Mon, 15 Dec 2014 18 :50 :40 GMT (58kb) [v10] Fri, 6 Feb 2015 20 :16 :28 GMT (59kb) [v11] Sat, 5 Sep 2015 12 :30 :49 GMT (59kb) La théorie de Marko↵ et ses développements Tessier et Ashpool, 2002. 35 / 87 36 / 87 http://hal.archives-ouvertes.fr/hal-00406601 Why the coefficient 3 ? Let n be a positive integer. If the equation x 2 + y 2 + z 2 = nxyz has a solution in positive integers, then either n = 3 and x, y , z are relatively prime, or n = 1 and the gcd of the numbers x, y , z is 3. Norbert Riedel [v1] Mon, 10 Sep 2007 22 :11 :39 GMT (11kb) [v2] Thu, 13 Sep 2007 18 :45 :29 GMT (11kb) [v3] Tue, 4 Dec 2007 17 :43 :40 GMT (15kb) [v4] Tue, 5 Aug 2008 21 :24 :23 GMT (15kb) [v5] Thu, 12 Mar 2009 14 :08 :48 GMT (15kb) [v6] Tue, 28 Jul 2009 18 :49 :17 GMT (15kb) Comments by Serge Perrine (23/07/2009) On the version 2 on Novembre 7, 2007, 32 p. On the version 3 on April 25, 2008, 57 p. On the version 4 on March 10, 2009, 103 p. On the versions 5 and 6 on July 29, 2009, 145 p. Friedrich Hirzebruch & Don Zagier, The Atiyah–Singer Theorem and elementary number theory, Publish or Perish (1974) 37 / 87 38 / 87 Equations x 2 + ay 2 + bz 2 = (1 + a + b)xyz Marko↵ type equations If we insist that (1, 1, 1) is a solution, then up to permutations there are only two more Diophantine equations of the type Bijection between the solutions for n = 1 and those for n = 3 : • if x 2 + y 2 + z 2 = 3xyz, then (3x, 3y , 3z) is solution of X 2 + Y 2 + Z 2 = XYZ , since (3x)2 + (3y )2 + (3z)2 = (3x)(3y )(3z). x 2 + ay 2 + bz 2 = (1 + a + b)xyz having infinitely many integer solutions, namely those with (a, b) = (1, 2) and (2, 3) : • if X 2 + Y 2 + Z 2 = XYZ , then X , Y , Z are multiples of 3 and (X /3)2 + (Y /3)2 + (Z /3)2 = 3(X /3)(Y /3)(Z /3). x 2 + y 2 + 2z 2 = 4xyz The squares modulo 3 are 0 and 1. If X , Y and Z are not multiples of 3, then X 2 + Y 2 + Z 2 is a multiple of 3. and x 2 + 2y 2 + 3z 2 = 6xyz • x 2 + y 2 + z 2 : tessalation of the plane by equilateral triangles • x 2 + y 2 + 2z 2 = 4xyz : tessalation of the plane by isoceles rectangle triangles • x 2 + 2y 2 + 3z 2 = 6xyz : tessalation ? If one or two (not three) integers among X , Y , Z are multiples of 3, then X 2 + Y 2 + Z 2 is not a multiple of 3. 39 / 87 40 / 87 Hurwitz’s equation x12 + x22 + · · · + xn2 = kx1 · · · xn Hurwitz’s equation (1907) For each n equation 2 the set Kn of positive integers k for which the x12 + x22 + · · · + xn2 = kx1 · · · xn When there is a solution in positive integers, there are infinitely many solutions, which can be organized in finitely many trees. has a solution in positive integers is finite. The largest value of k in Kn is n — with the solution (1, 1, . . . , 1). A. Baragar proved that there exists such equations which require an arbitrarily large number of trees : J. Number Theory (1994), 49 No 1, 27-44. Examples : K3 = {1, 3}, K4 = {1, 4}, K7 = {1, 2, 3, 5, 7}. The analog for the rank of elliptic curves over the rational number field is yet a conjecture. 41 / 87 42 / 87 Growth of Marko↵’s sequence 1978 : order of magnitude of m, m1 and m2 for m2 + m12 + m22 = 3mm1 m2 with m1 < m2 < m, Euclidean tree Start with (0, 1, 1). From a triple (a, b, c) satisfying a + b = c and a b c, one produces two larger such triples (a, c, a + c) and (b, c, b + c) and a smaller one (a, b a, b) or (b a, a, b). Harvey Cohn (0, 1, 1) | (1, 1, 2) | (1, 2, 3) | log(3m1 ) + log(3m2 ) = log(3m) + o(1). To identify primitive words in a free group with two generators, H. Cohn used Marko↵ forms. | | (1, 3, 4) | x 7! log(3x) : (m1 , m2 , m) 7! (a, b, c) with a + b ⇠ c. .. . 43 / 87 | (2, 3, 5) | | (1, 4, 5) (3, 4, 7) (2, 5, 7) | | | .. .. . . .. .. . . | | (3, 5, 8) .. .. . . | .. . 44 / 87 Marko↵ and Euclidean trees Growth of Marko↵’s sequence Don Zagier (1982) : estimating theOFnumber MATHEMATICS COMPUTATIONof the VOLUME 39, NUMBER 160 OCTOBER 1982, PAGES 709-723 Marko↵ triples bounded by x : c(log x)2 + O log x(log log x)2 ,of Markoff Numbers On the Number Below a Given Bound c = 0.180 717 0 . . . By Don Zagier Abstract.:According to a famous theoremnumber of Markoff, the Conjecture the n-th Marko↵ mnindefinite is quadratic forms with exceptionally large minima (greater than f of the square root of the discriminant) are in 1 : 1 correspondence withpthe solutions of the Diophantine equation p2 q 2 r 2 = 3pqr. By n relating Markoff's algorithm for finding solutions of this equation to a problem of countn points in triangles, it is shown that the number of solutions less than x ing lattice equals C l o g 2 3 x O(log x loglog2 x ) with an explicitly computable constant C = 0.1807170471 1507. . . . Numerical data up to is presented which suggests that the true error term is considerably 2 smaller. + + m ⇠A Tom Cusik & Mary Flahive, The Marko↵ and Lagrange spectra, Math. Surveys and Monographs 30, AMS (1989). + with A = 10.510 150 4 · · · Remark : c = 1/(log A) . 1. By a Markoff triple we mean a solution ( p , q, r ) of the Markoff equation 45 / 87 Marko↵ and Diophantine approximation 46 / 87 (1) P2+q2+r2=3pqr ( p , q , r ~ Z , l ~ p ~ q ~ r ) ; a Markoff number is a member of such a triple. These numbers, of which the first few are 1, 2 , 5 , 13,29, 34, 89, 169, 194,233,433,610,985,. . . , Connection with Hurwitz’s Theorem play a role in a famous theorem of Markoff [lo] (see also Frobenius [6], Cassels 121): the GL2(Z)-equivalence classes of real indefinite binary quadratic forms Q of J.W.S. Cassels, An introduction to Diophantine approximation, Cambridge Univ. Press (1957) discriminant 1 for whlch the invariant Don Zagier, On the number of Marko↵ numbers below a given bound. Mathematics of Computation, 39 160 (1982), 709–723. is greater than 3 are in one-to-one correspondence with the Markoff triples, John William Scott Cassels the invariant p(Q) for the form corresponding to ( p , q, r ) being (9 - 4r-2)-'/2. Thus the part of the Markoff spectrum (the set of all p(Q)) lying above 3 is described exactly by the Markoff numbers. An equivalent theorem is that, under the action of SL2(Z) on R U {a)given by x * (ax b)/(cx d), the SL2(Z)-equivalence classes of real numbers x for which the approximation measure + + is > are in 1 : 1 correspondence with the Markoff triples, the spectrum being the same as above (e.g. p(x) = 5-'I2 for x equivalent to the golden ratio and p(x) G 8-'12 for all other x). Thus the Markoff numbers are important both in the theory of quadratic forms and in the theory of Diophantine approximation. They have also arisen in connection with problems in several other branches of mathematics, e.g. the M.W., Open Diophantine Problems, Moscow Mathematical 4198N1. 1, 2004, 245–305. Received July 15, 198 1; revisedJournal December 16, 1980 Muthemutics Suhject Classification. Primary 10BIO; Secondary 10A21, 10C25, 10525, 10F99. 01982 American Mathematical Society 48 / 87 0025-571 8/82/ooOO-0259/$03.75 47 / 87 709 Historical origin : rational approximation Hurwitz’s Theorem (1891) : For any real irrational number x, there exist infinitely many rational numbers p/q such that x Lagrange’s constant p For x 2 R \ Q denote by (x) 2 [ 5, +1] the least upper bound of the numbers > 0 such that there exist infinitely many p/q 2 Q satisfying Adolf Hurwitz (1859–1919) x p 1 p · q 5q 2 p 1 2· q q This means 1 = lim inf q min |qx q!1 p2Z (x) Golden ratio p = (1 + 5)/2 = 1.618 033 9 . . . Hurwitz’s result is optimal. Hurwitz p : (x) ( ) = 5. p p| . 5 = 2.236 067 9 · · · for any x and 49 / 87 Badly approximable numbers Badly approximable numbers An irrational real number x is badly approximable by rational numbers if its Lagrange’s constant is finite. This means that there exists > 0 such that, for any p/q 2 Q, x p q 50 / 87 Any quadratic irrational real number has a finite Lagrange’s constant (= is badly approximable). It is not known whether there exist real algebraic numbers of degree 3 which are badly approximable. 1 · q2 It is not known whether there exist real algebraic numbers of degree 3 which are not badly approximable . . . For instance Liouville’s numbers have an infinite Lagrange’s constant. A real number is badly approximable if and only if the sequence (an )n 0 of partial quotients in its continued fraction expansion One conjectures that any irrational real number which is not quadratic and which is badly approximable is transcendental. This means that one conjectures that no real algebraic number of degree 3 is badly approximable. x = [a0 , a1 , a2 , . . . , an , . . . ] is bounded. 51 / 87 52 / 87 Lebesgue measure Properties of the Lagrange’s constant We have The set of badly approximable real numbers has zero measure for Lebesgue’s measure. Henri Léon Lebesgue (1875–1941) (x + 1) = (x) : x +1 and ( x) = (x) : x p = x q p+q q p p = x+ , q q Also (1/x) = (x) : p2 1 x q p = q2 · x p qx p . q 53 / 87 54 / 87 ✓ The modular group The multiplicative group generated by the three matrices ✓ ◆ ✓ ◆ ✓ ◆ 1 1 1 0 0 1 , , 0 1 0 1 1 0 ◆ ax + b a b x= c d cx + d ✓ ◆ 1 1 x =x +1 0 1 (x + 1) = (x) ✓ 1 0 ◆ 0 x= 1 x ( x) = (x) ✓ ◆ 1 0 1 x= 1 0 x (1/x) = (x) is the group GL2 (Z) of 2 ⇥ 2 matrices ✓ ◆ a b with coefficients in Z c d and determinant ±1. Consequence : Let x 2 R \ Q and let a, b, c, d be rational integers satisfying ad bc = ±1. Set J-P. Serre – Cours d’arithmétique, Coll. SUP, Presses Universitaires de France, Paris, 1970. Then (x) = (y ). y= 55 / 87 ax + b · cx + d 56 / 87 Hurwitz’s work (continued) p The inequality (x) 5 for all real irrational x is optimal for the Golden ratio and for all the noble irrational numbers whose continued fraction expansion ends with an infinite sequence of 1’s – these numbers are roots of the quadratic polynomials having discriminant 5 : Noble numbers Adolf Hurwitz, 1891 A noble number, whose continued fraction expansion ends with an infinite sequence of 1’s, is a number related to the Golden ratio by a homography of determinant ±1 : ✓ ◆ a +b a b with 2 GL2 (Z). c d c +d = [1, 1, 1, . . . ] = [1]. 57 / 87 The first elements of the spectrum Lagrange Spectrum For all the real numbers which are not noble numbers, a stronger inequality than Hurwitz’s inequality p (x) 5 = 2, 236 067 977 . . . The constant p 5 is best possible in Hurwitz’s Theorem. If ↵ is notpGL(2, Z) equivalent to the Golden ratio, the p constant 5 improves to 8. is valid, namely (x) 58 / 87 p 2 2 = 2, 828 427 125 . . . If we also p exclude the numberspwhich are GL(2, Z) equivalent to 1 + 2, then we can take 221/5 as the constant. This is optimal for p 2 = 1.414 213 562 373 095 048 801 688 724 209 698 078 . . . The sequence p p p 5 = 2.236 . . . , 8 = 2.828 . . . , 221/5 = 2.973 . . . whose continued fraction expansion is continues as follows. [1; 2, 2, . . . , 2, . . . ] = [1; 2]. 59 / 87 60 / 87 Lagrange Spectrum Lagrange Spectrum There is an infinite increasing sequence of real numbers (Li )i with limit 3, and an associated sequence of quadratic irrationalities (✓i )i 1 , such that for any i 1, if ↵ is not GL(2, Z) equivalent to any of ✓1 , . . . , ✓i 1 , then there is infinitely many rational numbers that satisfy ↵ Denoting by m1 , m2 , . . . the sequence of Marko↵ numbers, one has q Li = 9 (4/mi2 ) 1 and ✓i = 3mi + 2ki + 2mi p 9mi2 4 , where ki is an integer satisfying ai ki ⌘ bi (mod mi ) and (ai , bi , mi ) is a solution of Marko↵’s equation with mi max{ai , bi }. p 1 , < q Li q 2 and this is best possible when ↵ is GL(2, Z) equivalent to ✓i . Here one assumes Marko↵’s Conjecture to get unicity of (ai , bi ). 61 / 87 First values q Li = 9 Minima of quadratic forms (4/mi2 ), ✓i = Fi (X , Y ) = mi X 2 + (2ki i 1 2 3 mi 1 2 5 ki 1 1 2 62 / 87 Li p p5 p 8 221/5 3mi + 2ki + 2mi 3mi )XY + ki2 ✓pi (1 + p5)/2 1 +p 2 ( 11 + 221)/10 p 9mi2 Let f (X , Y ) = aX 2 + bXY + cY 2 be a quadratic form with real coefficients. Denote by (f ) its discriminant b 2 4ac. 4 Consider the minimum m(f ) of |f (x, y )| on Z2 \ {(0, 0)}. Assume (f ) 6= 0 and set p C (f ) = m(f )/ | (f )|. 3ki mi + 1 2 Y . mi Fi X XY Y 2 2(X 2 2XY Y 2 ) 5X 2 + 11XY 5Y 2 2 63 / 87 Let ↵ and ↵0 be the roots of f (X , 1) : f (X , Y ) = a(X ⇢ 1 0 {↵, ↵ } = 2a ↵Y )(X b± p ↵0 Y ), (f ) . 64 / 87 Example with <0 Definite quadratic forms ( The form If the discriminant is negative, J.L. Lagrange and Ch.pHermite (letter to Jacobi, August 6, 1845) proved C (f ) 1/ 3 with 2 2 equality for p f (X , Y ) = X + XY + Y . For each % 2 (0, 1/ 3], there exists such a form f with C (f ) = %. f (X , Y ) = X 2 + XY + Y 2 has discriminant (f ) = For m(f ) 1 C (f ) = p =p · 3 | (f )| 3 and minimum m(f ) = 1, hence Joseph-Louis Lagrange (1736–1813) < 0, the form r Charles Hermite (1822–1901) Carl Gustav Jacob Jacobi (1804–1851) | | 2 (X + XY + Y 2 ) 3 p and minimum | |/3. Again f (X , Y ) = has discriminant < 0) 1 C (f ) = p · 3 65 / 87 Example with >0 Indefinite quadratic forms ( The form f (X , Y ) = X 2 Assume Y2 XY has discriminant (f ) = 5 and minimum m(f ) = 1, hence For m(f ) 1 C (f ) = p =p · 5 (f ) f (X , Y ) = r 5 (X 2 and minimum p XY Y 2) /5. Again 1 C (f ) = p · 5 67 / 87 > 0) >0 A. Korkine and E.I.. Zolotarev p proved in 1873 C (f ) 1/ 5 with equality for f0 (X , Y ) = X 2 XY Y 2 . For all forms which are not equivalent to f0 under GL(2, Z), p they prove C (f ) 1/ 8. p 1/ 5 = 0.447 213 595 . . . p 1/ 8 = 0.353 553 391 . . . > 0, the form has discriminant 66 / 87 Egor Ivanovich Zolotarev (1847–1878) 68 / 87 Indefinite quadratic forms ( The works by Korkine and Zolotarev inspired Marko↵ who pursued the study of this question. He produced infinitely many values C (fi ), pi = 0, 1, . . ., between 1/ 5 and 1/3, with the same property as f0 . These values form a sequence which converges to 1/3. He constructed them by means of the tree of solutions of the Marko↵ equation. > 0). Indefinite quadratic forms ( > 0) Assume f ((X , Y ) = aX 2 + bXY + cY 2 2 R[X , Y ] with a > 0 has discriminant > 0. A. Marko↵, 1879 and 1880. If |f (x, y )| is small with y 6= 0, then x/y is close to a root of f (X , 1), say ↵. Then and ↵ ↵0 = p |x ↵0 y | ⇠ |y | · |↵ ↵0 | /a. Hence |f (x, y )| = |a(x ↵y )(x ↵0 y )| ⇠ y 2 69 / 87 Lagrange spectrum and Marko↵ spectrum p ↵ x . y 70 / 87 Chronologie Marko↵ spectrum = set of values taken by J. Liouville, 1844 J-L. Lagrange et Ch. Hermite, 1845 A. Korkine et E.I. Zolotarev, 1873 A. Marko↵, 1879 F. Frobenius, 1915 L. Ford, 1917, 1938 R. Remak, 1924 J.W.S. Cassels, 1949 J. Lehner, 1952, 1964 H. Cohn, 1954, 1980 R.A. Rankin, 1957 A. Schmidt, 1976 S. Perrine, 2002 p 1 = (f )/m(f ) C (f ) when f runs over the set of quadratic forms ax 2 + bxy + cy 2 with real coefficients of discriminant (f ) = b 2 4ac > 0 and m(f ) = inf (x,y )2Z2 \{0} |f (x, y )|. Lagrange spectrum = set of values taken by Lagrange’s constant. (x) = 1/ lim inf q(min |qx q!1 p2Z p|) when x runs over the set of real numbers. The Marko↵ spectrum contains the p Lagrange spectrum. The intersection with the intervall [ 5, 3[ is the same for both of them, and is a discrete sequence. 71 / 87 72 / 87 Fraction continue et géométrie hyperbolique The Geometry of Marko↵ Numbers Caroline Series, The Geometry of Marko↵ Numbers, The Mathematical Intelligencer 7 N.3 (1985), 20–29. Référence : Caroline Series 73 / 87 Fuchsian groups and hyperbolic Riemann surfaces Marko↵’s tree can be seen as the dual of the triangulation of the hyperbolic upper half plane by the images of the fundamental domain of the modular invariant under the action of the modular group. Lazarus Immanuel Fuchs (1833–1902) 74 / 87 Triangulation of polygons, metric properties of polytopes Harold Scott MacDonald Coxeter (1907–2003) 75 / 87 Robert Alexander Rankin (1915–2001) John Horton Conway 76 / 87 Fricke groups The subgroup Free groups. Fricke proved that if A and B are two generators of , then their traces satisfy of SL2 (Z) generated by the two matrices ✓ ◆ ✓ ◆ 1 1 2 1 and 1 2 1 1 (trA)2 + (trB)2 + (trAB)2 = (trA)(trB)(trAB) Harvey Cohn showed that p quadratic forms with a Marko↵ constant C (f ) 2]1/3, 1/ 5] are equivalent to is the free group with two generators. cx 2 + (d The Riemann surface quotient of the Poincaré upper half plane by is a punctured torus. The minimal lengths of the closed geodesics are related to the C (f ), for f indefinite quadratic form. where is a generator of . ✓ a)xy a b c d by 2 ◆ 77 / 87 Fundamental domain of a punctured disc 78 / 87 A simple curve on a punctured disc 79 / 87 80 / 87 Ford circles Farey sequence of order 5 The Ford circle associated to the irreducible fraction p/q is tangent to the real axis at the point p/q and has radius 1/2q 2 . Ford circles associated to two consecutive elements in a Farey sequence are tangent. Lester Randolph Ford (1886–1967) Amer. Math. Monthly (1938). 0 1 1 1 2 1 3 2 3 4 1 , , , , , , , , , , 1 5 4 3 5 2 5 3 4 5 1 81 / 87 Complex continued fraction 82 / 87 Laurent’s phenomenon Connection with Laurent polynomials. James Propp, The combinatorics of frieze patterns and Marko↵ numbers, http://fr.arxiv.org/abs/math/0511633 The third generation of Asmus Schmidt’s complex continued fraction method. If f , g , h are Laurent polynomials in two variables x and y , i.e., polynomials in x, x 1 , y , y 1 , in general h f (x, y ), g (x, y ) is not a Laurent polynomial : x2 + 1 1 =x+ , x x ✓ ◆2 1 x+ +1 x 4 + 3x 2 + 1 x f f (x) = = · 1 x(x 2 + 1) x+ x f (x) = http://www.maa.org/editorial/mathgames/mathgames 03 15 04.html 83 / 87 84 / 87 Simultaneous rational approximation and Marko↵ spectrum Relation between Marko↵ numbers and extremal numbers : simultaneous approximation of x and x 2 by rational numbers with the same denominator. Damien Roy Greatest prime factor of Marko↵ pairs Pietro Corvaja and Umberto Zannier, 2006 : The greatest prime factor of the product xy when x, y , z is a solution of Marko↵’s equation tends to infinity with max{x, y , z}. Equivalent statement : If S denotes a finite set of prime numbers, the equation x 2 + y 2 + z 2 = 3xyz has only finitely many solutions in positive integers x, y , z, such that xy has no prime divisor outside S. (The integers x and y are called S–units. ) Marko↵–Lagrange spectrum and extremal numbers, arXiv.0906.0611 [math.NT] 2 June 2009 85 / 87 Updated 2016 On the Marko↵ Equation x 2 + y 2 + z 2 = 3xyz Michel Waldschmidt Institut de Mathématiques de Jussieu (Paris VI) http://www.math.jussieu.fr/⇠miw/ 87 / 87 86 / 87