Markoff’s Numbers and Quadratic Forms

G.V.Belyi

April 25, 1997

Introduction

The natural solutions of the Markoff equation x^{2} + y^{2} + z^{2} = 3xyz (x ≥ y ≥ z) are called the “Markoff numbers”. A.A.Markoff investigated this equation in [ 1 ] , while studying rational approximations of real numbers. There is a conjecture asserting that any solution of the Markoff equation is determined by x. This assertion is often refered to as the uniqueness of the Markoff numbers. Recently the interest to this topic has grown in connection with the A.L.Gorodentsev and A.N.Rudakov works [ 6 ], [ 7 ], where they had proved that the dimensions of exceptional bundles on ℙ^{2} are Markoff’s numbers.

The aim of this article is to explain some Markoff’s numbers properties in a very explicit way. We give another proof of some results of Hurwitz and Frobenius [ 2 ], [ 3 ] relating to the Markoff numbers. It seems that the construction in our proof can be modified to study this equation over the ring of Gaussian integers for example. Next, we obtain a certain convex curve like a level line and discuss the Markoff numbers growth. Finally, we add some speculations relating to the uniqueness conjecture.

I am grateful to S. Kuleshov, B.Karpov, and D.Orlov for valuable discussions and the invaluable help with L^{A}TEX . I am also grateful to A.N.Tyurin for his stimulating interest in my work. I would like to thank the MPI for hospitality.

1 Preliminary

1.1 For any natural number S ≥ 3 , we choose the unit

ε = \frac{S + \sqrt{(S − 2)(S + 2)}}{2} = \frac{S + k\sqrt{D_{0}}}{2} ,

where D_{0} is a square-free integer. The index f_{0} of the order 𝔒_{1} = ℤ[ε] in the maximal order 𝔒 of the field ℚ(\sqrt{D_{0}}) is equal to k whenever D_{0} ≡ 1 (mod 4). In other cases (D_{0} ≡ 2, 3 (mod 4)) these index f_{0} is equal to k/2. For each divisor f of the index f_{0} = ff_{1} there exist an unique intermediate order 𝔒_{f} such that

[𝔒_{f} : 𝔒_{1}] = f, [𝔒 : 𝔒_{f}] = f_{1}.

1.2 For all these orders, we consider the corresponding class groups Ĉl of oriented modules M up to strict similarity; this means that the modules M and αM are equivalent if the norm of α ∈ ℚ(\sqrt{D_{0}}) is positive. The orientation is a choice of such a basis ⟨λ, μ⟩ = M of the module that the determinant

det ( \begin{matrix}λ & μ \\ λ̃ & μ̃ \end{matrix} ) > 0 ,

were ∼ denotes the conjugation. Moreover, we can choose the second basis element with positive norm and divide by it. So, without loss of generality, we suppose it is 1. We refer to the textbook [ 4 ] for further details. Let us choose the basis for the main modules 𝔒_{f} as follows: μ = 1, λ = (ε − d)/f, where d is a rational integer solution of the congruense 2d ≡ S (mod f). Then the matrix representing the unit ε on the main module 𝔒_{f} has the folloving form

A_{1} = ( \begin{matrix}fa + d & fbc \\ f & d \end{matrix} ) ,

and λ^{2} = aλ + bc . When we consider the action of the unit ε on an arbitrary module M with the coefficient ring 𝔒_{f} it’s matrix will be of the form

A = ( \begin{matrix}fa + d & fb \\ fc & d \end{matrix} ) ∈ SL(2, ℤ)

with 2d + fa = S and gcd(a, b, c) = 1. The corresponding module M = ⟨μ, 1⟩ = ⟨λ/c, 1⟩ is determined by the matrix A : μ is a root of the quadratic equation cμ^{2} − aμ − b = 0 , or in other words, it is a fixed point of the linear-fractional transform which corresponds to the matrix A. Indeed,

ε · 1 = \frac{fa + 2d + \sqrt{f^{2}a^{2} + (4fad + 4d^{2} − 4)}}{2} = fc\frac{a + \sqrt{a^{2} + 4bc}}{2c} + d = fc · μ + d · 1.

Thus we obtain the correspondence between classes of modules and the conjugation classes of matrices in the group SL(2, ℤ).

1.3 We need to find a matrix Ã corresponding to the module M̃ inverse to the M. It is proved in [ 4 ] that the module M̃ = ⟨ −μ̃, 1 ⟩ is such a module. Obviously, it it is positively oriented, so

Ã = ( \begin{matrix}d & fb \\ fc & fa + d \end{matrix} )

Next we find the composition rule for modules over the same order:

1.4 Lemma. Suppose that the matrices

( \begin{matrix}fa + d & fbc_{2} \\ fc_{1} & d \end{matrix} ) , ( \begin{matrix}fa + d & fbc_{1} \\ fc_{2} & d \end{matrix} ) , ( \begin{matrix}fa + d & fb \\ fc_{1}c_{2} & d \end{matrix} ) ∈ SL(2, ℤ)

correspond to three modules M_{1}, M_{2}, and M_{3} over the same order, then the third module is the composition of the first and the second one.

Proof. We have M_{1} = ⟨ λ/c_{1}, 1 ⟩, M_{2} = ⟨ λ/c_{2}, 1 ⟩, M_{3} = ⟨ λ/c_{1}c_{2}, 1 ⟩.

M_{1} ∗ M_{2} = ⟨ 1, c_{1} \frac{λ}{c_{1}c_{2}} , c_{2} \frac{λ}{c_{1}c_{2}} , \frac{λ^{2}}{c_{1}c_{2}} ⟩ = ⟨ 1, c_{1} \frac{λ}{c_{1}c_{2}} , c_{2} \frac{λ}{c_{1}c_{2}} , a\frac{λ}{c_{1}c_{2}} + b ⟩ = M_{3},

as gcd(a, c_{1}, c_{2}) = 1.

1.5 Definition. A matrix A = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ GL(2, ℤ) with tr A ≥ det A + 2 is called

reduced if a ≥ b , c ≥ d ≥ 0 .

1.6 Lemma.

i) Each conjugation class of matrices with trace tr A ≥ det A + 2 contains a reduced matrix Ā.

ii) For the module M = ⟨ μ, 1 ⟩ associated to the reduced matrix Ā, μ > 1 and −1 < μ̃ < 0.

Proof. If the absolute value |d| > |b| , let d̄be a residue of d modulo |b|. The matrices

A ∼ ( \begin{matrix}a + d − d̄ & b \\ ∗ & d̄ \end{matrix} ) are conjugated in SL(2, ℤ). The rest is evident.

1.7 The Euclidean algorithm provides for each reduced matrix A ∈ SL(2, ℤ) the unique decomposition into the product :

A = ( \begin{matrix}a_{1} & 1 \\ 1 & 0 \end{matrix} ) · ( \begin{matrix}a_{2} & 1 \\ 1 & 0 \end{matrix} ) · . . . · ( \begin{matrix}a_{2m} & 1 \\ 1 & 0 \end{matrix} ) , a_{i} ∈ ℕ ,

and this decomposition is nothing else as the continued fraction for the rational number a/c , or some period of the continued fraction for the irrational number μ. We write, for brevity, A = [a_{1}, a_{2}, …, a_{2m}] . Henceforth we do not distinguish the matrices and the associated modules. Consider the cyclic permutation σ = (1, 2, …, 2m). It acts on the period A in an evident way. And each even power of this action gives us a module equivalent to A. Moreover, each equivalent reduced module is obtained in such a way. Note that 1.3 gives us that the inverse module Ã ∼ [a_{1}, a_{2m}, …, a_{3}, a_{2}].

1.8 Consider now the natural epimorphism ψ : Ĉl → Cl onto the usual ideal class group of the order 𝔒_{f} which forgets the orientation and changes the strict similarity to usual one. If the order 𝔒_{f} contains an unit with the norm −1 , then the epimorphism ψ is an isomorphism. Otherwise, the kernel of this epimorphism ψ is the group {±1}. Denote by A_{−1} the generator of this kernel. So ker(ψ) = {A_{1} , A_{−1}} . If we take the module A as in 1.7 , then A ∗ A_{−1} ∼ [a_{2}, a_{3}, …, a_{2m}, a_{1}]. It follows easily from 1.4 .

1.9 Definition. The module A is said to be ambiguous if A ∗ A ∼ A_{1}, i.e., if it is of the second order. We shall call the module A semi-ambiguous if A ∗ A ∼ A_{−1} or equivalently Ã ∼ A ∗ A_{−1},

1.10 Lemma. Each semi-ambiguous module is equivalent to the symmetric module [a_{1}, a_{2}, …, a_{m}, a_{m}, …, a_{2}, a_{1}] and each ambiguous module is equivalent to the symmetric module [a_{1}, a_{2}, …, a_{m}, a_{m+1}, a_{m} …, a_{2}].

Proof. Making use of 1.7 and 1.8 we obtain the equality

[a_{2m}, … a_{2}, a_{1}]^{τ} = [a_{1}, a_{2}, …, a_{2m}],

where τ is an even power of σ for semi-ambiguous modules and is an odd power of σ for ambiguous modules. Since a product of a reflection and a rotation is a reflection again, this concludes the proof.

Note that if we started with the unit ϵ = \frac{T +\sqrt{T^{2}+4}}{2} with norm −1, then the notions of ambiguous modules and semi-ambiguous modules coincide. In this case, each ambiguous module is equivalent to the symmetric module [a_{1}, …, a_{m}, …, a_{1}].

1.11 Note that for matrices A = ( \begin{matrix}a & b \\ c & d \end{matrix} ) , B = ( \begin{matrix}m & ∗ \\ n & ∗ \end{matrix} ) ∈ SL(2, ℤ) we have

B^{−1} · A · B = ( \begin{matrix}∗ & ∗ \\ t & ∗ \end{matrix} ) , where t = cm^{2} − (a − d)mn − bn^{2}.

2 Markoff’s numbers and semi-ambiguous forms

2.1 Now, let (x, y, z, p) be a natural solution of the following equation

x^{2} + y^{2} + z^{2} = pxyz

and suppose y, z are coprime. Than there exist a matrix B = ( \begin{matrix}y & ∗ \\ z & ∗ \end{matrix} ) ∈ SL(2, ℤ) such

that

B^{−1} · ( \begin{matrix}px & −1 \\ 1 & 0 \end{matrix} ) · B = ( \begin{matrix}a & b \\ −x^{2} & d \end{matrix} ) , see 1.11 . Note that the matrix ( \begin{matrix}px & −1 \\ 1 & 0 \end{matrix} )

is equivalent to the matrix A_{1} from 1.2 and corresponds to the main module 𝔒_{1} = ℤ[ε] with trace S = px = a + d.

2.2 In the cases x = 1 or x = 2 the equation 2.1 turns in the “negative Pell’s equation” and means that the order 𝔒_{2} contains an unit with negative norm. Denote this unit by

ϵ = \frac{T + \sqrt{T^{2} + 4}}{2}.  We have  ϵ^{2} = \frac{T^{2} + 2 + T \sqrt{T^{2} + 4}}{2}.

As this natural number T divides f = 2 , we obtain T = 1, f = 1, S = 3, p = 3 , for x = 1 , and T = 2 , f = 2, S = 6, p = 3 for x = 2 . Henceforth we assume that x > 2 .

2.3 Lemma. The matrix ( \begin{matrix}a & −xb \\ x & d \end{matrix} ) ∈ SL(2, ℤ) is semi-ambiguous and therefore it is

conjugated with a reduced symmetrical matrix A = L^{T} · L by Lemma 1.10 .

Proof. Let f = gcd ( x , a − d) be the corresponding index. If f = 1, then the lemma follows from 2.1 and 1.4 .

Note that the index f can be at most 2, as x divides S = a + d. In this case the numbers y, z, a, d have to be odd, ad ≡ 1 (mod 4), S ≡ 2 (mod 4), and we can write

x = 2x_{0} , with odd x_{0}. Now the matrix A_{2} = ( \begin{matrix}S − 1 & ( S − 2 )/2 \\ 2 & 1 \end{matrix} ) corresponds to the

main module 𝔒_{2}. We set C = ( \begin{matrix}1 & −1 \\ 0 & 2 \end{matrix} ) · B · ( \begin{matrix}1/2 & 0 \\ 0 & 1 \end{matrix} ) . It is easy to check that

C ∈ SL(2, ℤ) and C^{−1} · A_{2} · C = ( \begin{matrix}a & 2b \\ −2x_{0}^{2} & d \end{matrix} ) ∼ A ∗ A ∗ A_{−1} . We denote by α the

symmetry axis.

2.4 Next we show that matrix A from 2.3 has another axis β of the “almost” symmetry. Let d̄ be a residue of d modulo x . We can assume without loss of generality that 0 < d̄ < x/2 , otherwise we change A to Ã , see 1.3 . Further, consider the following product

C = ( \begin{matrix}p & −1 \\ 1 & 0 \end{matrix} )^{−1} · ( \begin{matrix}px − d̄ & ∗ \\ x & d̄ \end{matrix} ) = ( \begin{matrix}x & d̄ \\ d̄ & ∗ \end{matrix} ) ∈ SL(2, ℤ).

We have the reduced symmetrical matrix C. Taking into account 1.7 , we obtain C = [b, b_{1}, …, b_{n}, b_{n}, …, b_{1}, b]. The condition d̄ < x/2 gives us b ≥ 2 . We see that the matrix Ã is equivalent to the usual product C_{1} · C_{2} , where C_{2} is equal to either the identity matrix Id whenever n = 0 or to C_{2} = [b_{1}, …, b_{n}, b_{n}, …, b_{1}] , and

C_{1} = ( \begin{matrix}b & 1 \\ 1 & 0 \end{matrix} ) · ( \begin{matrix}p & −1 \\ 1 & 0 \end{matrix} ) · ( \begin{matrix}b & 1 \\ 1 & 0 \end{matrix} ) . So we have C_{1} = [b + 1, b − 1] for p = 1 and

C_{1} = [b, p − 1, 1, b − 1] for p > 1 . This other axis β gives us the following.

2.5 Theorem. (Markoff, Hurwitz) The equation from 2.1 has solutions with coprime y, z iff p = 3 . In these case the equation turns in “Markoff’s equation” and we have 3x = trA for the matrix A from 2.4 .

Proof. Let the matrix A have two such axes. We arrange its factors ( see 1.7 ) equidistantly along a circle. Consider two transformations II , III of the matrix A , as it is represented on the pictures

, where C_{1} , α , β are as in 2.4 . The other notations are evident. It is clear that these transformations give the new matrixes with the two such axes and have the inverse; denote them by I_{l} , I_{r} . We can write L_{x} = M · N and denote by m , n the quantities of factors in M , N . If we only use the transformations I_{l} , I_{r} the Euclidean algorithm provides us, the axes α and β become orthogonal after finitely many steps. We see on the next picture

that the upper semicircle can not contain the matrix C_{1} as C_{1}^{T} is not symmetrical. This means that b = 2 , p = 3 , and m , n are coprime; and we obtain Markoff’s number 5 corresponding to the matrix C_{1} = [2, 2, 1, 1] .

2.6 For a solution (x, y, z) with x > y > z the transformation II corresponds to the mutation (x, y, z) → (3xz − y, x, z) , and the transformation III corresponds to the mutation (x, y, z) → (3xy − z, x, y) . The transformations I corresponds to the mutation (x, y, z) → (y, z, 3yz − x) ; in more detail, we have

L_{z} = M^{T} , L_{y} = M N , L_{x} = M^{T} M N , Lx̃ = N for I_{l} and
L_{y} = M N , L_{z} = N^{T} , L_{x} = M N N^{T} , Lx̃ = M for I_{r} .

2.7 Having applied the transformations I to the matrix C_{1} , we obtain the main modules [1, 1] and [2, 2] corresponding to the singular Markoff numbers 1 and 2 . We denote them

X = [1, 1] = ( \begin{matrix}2 & 1 \\ 1 & 1 \end{matrix} ) and Y = [2, 2] = ( \begin{matrix}5 & 2 \\ 2 & 1 \end{matrix} ) .

It is not clear how we could define the “almost” symmetry axis β for them. So the matrix C_{1} can be restored explicitly neither by X , nor by Y .

2.8 Next we find the relations between quantities of the twos, denote them by k, k_{m}, k_{n} , and the ones, denote them by l, l_{m}, l_{n} , in the decompositions of L, M, N . Denote by m, n the quantities of factors in M, N . It is clear that k + l = m + n . Show that

det ( \begin{matrix}l_{m} & l_{n} \\ k_{m} & k_{n} \end{matrix} ) = 1 ,

For L = [1, 2] we have det Id = 1 . The transformation II corresponds to addition of the first column to the second and the transformation III corresponds to addition of the second column to the first one. We obtain

km − 1 = (k_{m} + k_{n})(k_{m} + l_{m}) − l_{m}k_{n} + l_{n}k_{m} = k_{m}(k_{m} + k_{n} + l_{n} + l_{m}) = k_{m}(m + n)

or km ≡ ln ≡ 1 (mod (m + n) ) .

Now we can consider the pair (k, l) as the coordinates for the Markoff number x.

2.9 Examples:

tr[1, 1] = 3 · 1(0, 1) ,  tr[2, 2] = 3 · 2(1, 0) ,  tr[2, 1, 1, 2] = 3 · 5(1, 1) ,

tr[2, 1, 1, 1, 1, 2] = 3 · 13(1, 2) , tr[1, 2, 2, 2, 2, 1] = 3 · 29(2, 1) , tr[2, 1, 1, 1, 1, 1, 1, 2] = 3 · 34(1, 3) , tr[2, 1, 1, 1, 1, 1, 1, 1, 1, 2] = 3 · 89(1, 4) , tr[1, 2, 2, 2, 2, 2, 2, 1] = 3 · 169(3, 1) , tr[1, 2, 2, 1, 1, 1, 1, 2, 2, 1] = 3 · 194(2, 3) , tr[2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] = 3 · 233(1, 5) , tr[2, 1, 1, 2, 2, 2, 2, 1, 1, 2] = 3 · 433(3, 2) .

Compare [ 3 ].

2.10 It is obvious that either A or Ã is equivalent to the product of k pieces of Y and l pieces of X in spite of our axes can cut some factors. Let us imagine each factor to be composed of two equal pieces like an oxygen molecule O_{2}. Taking into account that after each transformation the separated “oxygen atoms” unite again into “oxygen molecules”, we see this.

The proof of theorem 2.5 gives an answer to the following question: there are kr pieces of the matrix Y and lr pieces of the matrix X with coprime k, l; how to intermix them in maximal uniform way to the product having minimal trace? The answer is: A^{r} is such a product.

Besides, we obtain a possibility to compute the Markoff number x (k , l) without computing the intermediate numbers. We divide a circle into 2(k + l) equal pieces. Next we solve the congruenses

km ≡ ln ≡ 1 (mod (k + l) )

by using the Euclidean algorithm. Of course, we obtain the coordinates for the intermediate numbers, but not the numbers itself. Further, we draw the axes α , β , and arrange the twos and the ones along the circle. Example: Let k = 7 , l = 10 ; then m = 5 , n = 12 , and A = Y XY XY XXY XY XXY XY XX , 3x (k , l) = trA .

2.11 Let the matrices M, N be as in 2.5 . The “almost” symmetry axis β provides us NN^{T} M^{T} M = K^{T} C_{1}K for some K ∈ GL(2, ℤ) . We have

C_{1} = [2, 2, 1, 1] = ( \begin{matrix}12 & 7 \\ 5 & 3 \end{matrix} ) = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) + 3( 2 1 )^{T} ( 2 1 ) .

Write (r s) = (2 1)K . As all the elements of the group GL(2, ℤ) are conformally symplectic, we obtain

K^{T} C_{1}K = ( \begin{matrix}3r^{2} & 3rs ± 1 \\ 3rs ∓ 1 & 3s^{2} \end{matrix} ) ,

and x = r^{2}+s^{2} . Besides, NN^{T} M^{T} M = ±W + S , where W = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) and S

is a symmetrical matrix, compare [ 3 ]. It is easily shown that gcd ( x , y + zi) = r ± si , and

Y · K^{T} · K · X = ( \begin{matrix}3x − d̄ & ∗ \\ x & d̄ \end{matrix} ) .

3 Polynomial solutions of a pseudo-Markoff equation

3.1 The straight lines on the Markoff cubic are defined over the integer Gaussian numbers ring ℤ[ i] except three lines at infinity. To stay over ℤ we change the equation 2.1 to

x^{2} − y^{2} − z^{2} = ±pxyz .

For a natural solution (x, y, z, p) with coprime x, y, z we have

( \begin{matrix}a & b \\ x^{2} & d \end{matrix} ) ∼ ( \begin{matrix}px & −1 \\ 1 & 0 \end{matrix} )  and  ( \begin{matrix}a′ & b′ \\ y^{2} & d′ \end{matrix} ) ∼ ( \begin{matrix}py & 1 \\ 1 & 0 \end{matrix} ) ,

see 2.1 . Repeating the arguments in 2.3 and 2.4 , we obtain a symmetry axis α and a “almost” symmetry axis β for the matrices

A ∼ ( \begin{matrix}a & xb \\ x & d \end{matrix} )  and B ∼ ( \begin{matrix}a′ & yb′ \\ y & d′ \end{matrix} )  again .

3.2 There is a slight difference with 2.5 : both matrixes A and B are ambiguous, so these axes can cut some factors. We can repeat the arguments in theorem 2.5 taking into account 2.10 . Using the first transformation we obtain 3 simplest forms for the matrix A with ortogonal axes α, β

i) A = [1, b − 1, 1, 1, b, 1] , p = 2, ii) A = [b − 1, b + 1] , p = 1, iii) A = [1, p − 1] . In case i) we can take

2x = trA = 2(2b^{2} + 2b + 1), 2y = tr[1, b, 1] = 2(b + 1), 2z = tr[1, b − 1, 1] = 2b

We see that this case reduces to the second one, where

x = trA = b^{2} + 1,  y = tr[b + 1] = b + 1,  z = tr[b − 1] = b − 1,

by a linear change of variables. Both these cases correspond to the straight line

y = t + 1, z = t − 1, x = 2 on the cubic y^{2} + z^{2} − x^{2} = xyz .

In case iii) we have singular solution like in 2.7 . The matrix [1, p − 1] can be obtained by the transformation I from the two different matrixes [p, p − 1, 1, p − 1] or [1, p − 1, 1, 0] ∼ [2, p − 1] with the axes angle 45^{o}. As above, the second case reduces to the first one by a linear change of variables. In the first case we have

px = tr[p, p − 1, 1, p − 1] = p(p^{2} + 1), py = tr[p − 1, 1, p − 1] = p^{2}, pz = tr[p] = p .

We obtain the solution of 3.1 corresponding to the straight line

x = t, y = 0, z = t on the same cubic y^{2} + z^{2} − x^{2} = xyz .

It is not hard to check that the other straight lines on the cubic are symmetrical to these two or lie at infinity. Compare [ 8 ].

4 Markoff’s numbers growth

4.1 Let (x, y, z) be a real solution of the equation

x^{2} + y^{2} + z^{2} = xyz ,

where x ≥ y ≥ z > 2 . There exist positive α > β > γ such that

x = 2 cosh α , y = 2 cosh β , z = 2 cosh γ , and
cosh^{2} α + cosh^{2} β + cosh^{2} γ = 2 cosh α cosh β cosh γ .

For the mutation x̃ = yz − x we denote by α̃ the corresponding logarithm, so

cosh α + cosh α̃ = 2 cosh β cosh γ .

4.2 Let 𝔏 be a lattice with two generators a, b , and a ∧ b = 1 . Denote the origin of the lattice 𝔏 by O . We arrange the logarithms α, β, γ and all their mutations in the lattice 𝔏 in the following way. We put γ at the vertices ±a ; we put β at the vertices ±b ; and we put α at the vertices ±(a + b) . Further, suppose we have a fundamental parallelogram of the lattice 𝔏 with three vertices being filled, for example, a, b , and a + b . Take the diagonal of the parallelogram that contains the origin O . It divides the parallelogram into two triangles. Reflect these triangles to obtain two new parallelograms with empty vertices. Put the mutations at these empty places to get triples of logarithms correspondig to solutions of 4.1 at the parallelogram’s vertices. In our example we put the mutation β̃ at the vertex 2a + b and we put the mutation γ̃ at the vertex 2b + a. We do the same for the remaining old parallelograms. Next we do the same for all new parallelograms and so on. Each primitive point of the lattice 𝔏 will be filled by some logarithm. Further, we fill nonprimitive points in the following way. We put nη at the vertex nc if n is natural, c is primitive, and η corresponds to c.

Now we move all the points of the lattice 𝔏 towards the origin O or away from the origin, except the point O itself as follows. We move c to m(c) = c/η , if η corresponds to c .

4.3 Theorem. There exists a convex curve Γ that contains all the points m(c).

Proof. We obtain the curve Γ as a limit sequence of convex (6 · 2^{n}) -gons. First we have to check the hexagon we start with is convex. If we had α = β + γ, the point m(a + b) would be located on the segment m(a) m(b) . But we have

cosh α = cosh β · cosh γ + \sqrt{sinh^{2} β · sinh^{2} γ − 1} < cosh ( β + γ) ,

see 4.1 . This inequality means that the point m(a + b) is situated behind the straight line m(a) m(b) with respect to the origin O .

Next we consider the parallelogram O , (n + 1)a + nb , (2n + 1)(a + b) , (n + 1)b + na with area 2n + 1 . We have

t = 2 cosh (2n + 1)α = tr C^{2n+1} , where C = ( \begin{matrix}x & −1 \\ 1 & 0 \end{matrix} ) .

Let ˜ β̃ , ˜ γ̃ correspond to the other vertices of the parallelogram. We have to check that ˜β̃ + ˜ γ̃ > (2n + 1)α. It can be easily shown that

t_{1} = 2 cosh ˜ β̃ = tr C^{n} · ( \begin{matrix}y & 0 \\ z & 0 \end{matrix} ) and t_{2} = 2 cosh ˜ γ̃ = tr C^{n} · ( \begin{matrix}z & 0 \\ y & 0 \end{matrix} ) .

So we have to obtain the inequality

4 + t t_{1} t_{2} > t^{2} + t_{1}^{2} + t_{2}^{2} ,

which gives us

\frac{t_{1}}{2} \frac{t_{1}}{2} + \sqrt{\frac{t_{1}^{2}}{4} − 1}\sqrt{\frac{t_{2}^{2}}{4} − 1} > \frac{t}{2} .

It is clear that C^{n} = μ · Id − λ · C^{−1} with positive μ, λ. We have

C^{2n+1} = μ^{2}C − 2μλId + λ^{2}C^{−1}, t = x(μ^{2} + λ^{2}) − 4μλ, t_{1} = μy − λz, t_{2} = μz − λy.

As y^{2} + z^{2} = xx̃, yz = x + x̃, μ^{2} + λ^{2} − xμλ = 1, we obtain

t_{1}t_{2} = (μ^{2} + λ^{2})yz − μλ(y^{2} + z^{2}) = (μ^{2} + λ^{2})(x + x̃) − μλxx̃ = x̃ + (μ^{2} + λ^{2})x ,

t_{1}^{2} + t_{2}^{2} = (μ^{2} + λ^{2})(y^{2} + z^{2}) − 4μλyz = (μ^{2} + λ^{2})xx̃ − 4μλ(x + x̃) = x̃t + 4μλx .

Substituting these expressions in our inequality, we see that it means that

4 > t[t − (μ^{2} + λ^{2})x] + 4μλx and 1 > −μλ(t − x).

4.4 Let us put the Markoff numbers in the their growth order. The theorem means that we can regard the convex curve αΓ as a level line. It bounds the domain containing the points of the lattice 𝔏 with the logarithms smaller than α , the points with the logarithms bigger than α lie outside, and the points with the logarithms equal to α lie on the curve. The number of the primitive points belonging to the domain bounded by the curve αΓ is proportional to area of the domain ,i.e., it is proportional to α^{2} , up to o(α^{1+ε}) . It means that the number of the Markoff numbers x with 3x ≤ 2 cosh α is equal to

Cα^{2} + o(α^{1+ε}) (α → ∞) with some constant C , 0.18061 < C < 0.18076 .

To obtain the interval for C we find the area of 24-gon inscribed into Γ and we find the area of 24-gon described around Γ ; then we multiply them by Chebyshev’s 6/π^{2} and divide by 12. Note that Zagier, [ 5 ], used similar arguments and found the constant C without explicit use of this curve Γ , let alone its convexity. In the same article an estimate

∼ e^{c_{1}\sqrt{N}+c_{2}} ,

with some constants c_{1}, c_{2} for the Markoff number enumerated by N was given.

5 Discussion of the uniqueness conjecture

5.1 It seems to be very strange, but no correct proof of this conjecture has been advanced until now, in spite of the fact that the Markoff numbers grow so fast.

Clearly, the triple ( x > y > z) is restored by the couple ( r , s) with x = r^{2}+s^{2} , see 2.11 . We have the similar situation for 3x − 2 and 3x + 2 . For each element L ∈ GL(2, ℤ) we can write

2L = ( L + L^{W} ) + ( L − L^{W} ) = ( \begin{matrix}a & b \\ −b & a \end{matrix} ) + ( \begin{matrix}c & d \\ d & −c \end{matrix} ) ,

so  4 detL = a^{2} − c^{2} − d^{2} + b^{2} ,  2 trL L^{T} = a^{2} + c^{2} + d^{2} + b^{2} ,  and
a^{2} + b^{2} = trL L^{T} + 2 detL , c^{2} + d^{2} = trL L^{T} − 2 detL .

Let us return to the notations in the section 2 . Taking L_{x} as in 2.5 , we obtain

a^{2} + b^{2} = 3 x ± 2 , c^{2} + d^{2} = 3 x ∓ 2 .

Denote by ã , b̃ , c̃, d̃ the corresponding integers for x̃ . Combining 2.6 and 2.11 , we see

det ( \begin{matrix}a & b \\ ã & b̃ \end{matrix} ) = det ( \begin{matrix}c & d \\ c̃ & d̃ \end{matrix} ) = ±2

in the both cases I_{l} and I_{r} .

5.2 We see that each couple

( a , b) with a^{2} + b^{2} = 3 x ± 2 ,  ( c , d) with c^{2} + d^{2} = 3 x ∓ 2 ,  or
( r , s) with r^{2} + s^{2} = x

restores the triple ( x > y > z) . If there exists another triple ( x′ > y′ > z′) with

x′ = x ,

then the corresponding semi-ambiguous forms A and A′ differ by an ambiguous form B . On the other hand, we have two different decompositions into sum of two squares for x , 3x − 2 , 3x + 2 . They provide decompositions into coprime factors

x = x_{1} x_{2} , 3 x − 2 = p_{1} p_{2} , 3 x + 2 = q_{1} q_{2} for odd x and
x = 2 x_{1} x_{2} , 3 x − 2 = 4 p_{1} p_{2} , 3 x + 2 = 8 q_{1} q_{2} for even x ,

where x_{1} , x_{2} , p_{1} , p_{2} , q_{1} , q_{2} ≥ 5 . It is not hard to check that the ambiguous form B corresponds to the decompositions of 3x − 2 , 3x + 2 and represents ±x_{1}^{2} and ∓x_{2}^{2} , so existence of a counterexample to the conjecture is equivalent to the solvability of the system

\{ \begin{matrix}r k^{2} − s l^{2} = 4 x_{1}^{2} \\ r m^{2} − s n^{2} = −4 x_{2}^{2} \\ r s = 9 x_{1}^{2} x_{2}^{2} − 4 \end{matrix}

for odd x , or the system

\{ \begin{matrix}r k^{2} − 2 s l^{2} = x_{1}^{2} \\ r m^{2} − 2 s n^{2} = − x_{2}^{2} \\ 8 r s = 9 x_{1}^{2} x_{2}^{2} − 1 \end{matrix}

for even x . Furthermore, for x_{2} > x_{1} it is easy to prove that x_{2} < x_{1}^{3}/9 .

Finally we note that the corresponding conjecture is true for the polynomial solutions in section 3 .

References

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381-409, 379-399. [ 2 ] A.Hurwitz.Uber eine Aufgabe der unbestimmten Analisis. Archiv der Math. u. Phys.

3/11 (1907), 185-196. [ 3 ] G.Frobenius.Uber die Markoffschen Zahlen. Preuss. Akad. Wiss. Sitzungsberichte.

(1913), 458-487. [ 4 ] Z.I.Borevich, I.R.Shafarevich. Number Theory. Academic Press Inc. (1966) . [ 5 ] D.Zagier. On the Number of Markoff Numbers Below a Given Bound. Math. Comp.

39 (1982), 709-723. [ 6 ] A.L.Gorodentsev, A.N.Rudakov. Exceptional vector bundles on projective

spaces. Duke Math.J. 54 (1987), 115-130. [ 7 ] A.N.Rudakov. The Markov numbers and exceptional bundles on ℙ^{2}. English

transl. in Math. SSSR Izv. 32 (1989), 99-112. [ 8 ] J.H.Silverman. The Markov Equation X^{2} + Y^{2} + Z^{2} = aXY Z over Quadratic

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