Solutions of the Markov equation over polynomial rings
Ricardo Conceição, R. Kelly, S. VanFossen
49th John H. Barrett Memorial Lectures University of Tennessee, Knoxville, TN
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Markov equation Introduction
In his work on Diophantine approximation, Andrei Markov studied the equation
x2 + y2 + z2 = 3xyz.
The structure of the integral solutions of the Markov equation is very interesting.
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Markov equation Automorphisms
x2 + y2 + z2 = 3xyz
(x1, x2, x3) −→ (xπ(1), xπ(2), xπ(3))
(x, y, z) −→ (−x, −y, z)
A Markov triple (x, y, z) is a solution of the Markov equation that satisfies 0 < x ≤ y ≤ z.
(x, y, 3xy − z)
(x, y, z) (x, 3xz − y, z)
(3zy − x, y, z)
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Markov Tree
x2 + y2 + z2 = 3xyz (3zy − x, y, z) (x, y, 3xy − z) (x, y, z) (x, 3xz − y, z)
··· (2,29,169) ··· (2,5,29) ··· (5,29,433) ··· (1,1,1) (1,1,2) (1,2,5) ··· (5,13,194) ··· (1,5,13) ··· (1,13,34) ···
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Markov Tree
x2 + y2 + z2 = 3xyz (3zy − x, y, z) (x, y, 3xy − z) (x, y, z) (x, 3xz − y, z)
··· (2,29,169) ··· (2,5,29) ··· (5,29,433) ··· (1,1,1) (1,1,2) (1,2,5) ··· (5,13,194) ··· (1,5,13) ··· (1,13,34) ···
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Markov Tree
x2 + y2 + z2 = 3xyz (3zy − x, y, z) (x, y, 3xy − z) (x, y, z) (x, 3xz − y, z)
··· (2,29,169) ··· (2,5,29) ··· (5,29,433) ··· (1,1,1) (1,1,2) (1,2,5) ··· (5,13,194) ··· (1,5,13) ··· (1,13,34) ···
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Markov Tree
x2 + y2 + z2 = 3xyz (3zy − x, y, z) (x, y, 3xy − z) (x, y, z) (x, 3xz − y, z)
··· (2,29,169) ··· (2,5,29) ··· (5,29,433) ··· (1,1,1) (1,1,2) (1,2,5) ··· (5,13,194) ··· (1,5,13) ··· (1,13,34) ···
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Markov Tree
x2 + y2 + z2 = 3xyz (3zy − x, y, z) (x, y, 3xy − z) (x, y, z) (x, 3xz − y, z)
··· (2,29,169) ··· (2,5,29) ··· (5,29,433) ··· (1,1,1) (1,1,2) (1,2,5) ··· (5,13,194) ··· (1,5,13) ··· (1,13,34) ···
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Markov Tree
x2 + y2 + z2 = 3xyz (3zy − x, y, z) (x, y, 3xy − z) (x, y, z) (x, 3xz − y, z)
··· (2,29,169) ··· (2,5,29) ··· (5,29,433) ··· (1,1,1) (1,1,2) (1,2,5) ··· (5,13,194) ··· (1,5,13) ··· (1,13,34) ···
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Markov numbers Definition and properties
Theorem (Markov) All Markov triples lie on the Markov tree.
The entries of all Markov triples form the sequence of Markov numbers
1, 2, 5, 13, 29, 34, 89, 169,... (A002559 in the OEIS)
Every Markov number mn is the maximum of some Markov triple (x, y, mn). Conjecture (Frobenius) Given a Markov number m, there is exactly one Markov triple of the form (x, y, m).
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Markov numbers Subsequence
All odd-indexed Fibonacci number is a Markov number:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233... (A000045)
··· (2,29,169) ··· (2,5,29) ··· (5,29,433) ··· (1,1,1) (1,1,2) (1,2,5) ··· (5,13,194) ··· (1,5,13) ··· (1,13,34) ···
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Markov numbers Subsequence
All odd-indexed Pell number is a Markov number:
0, 1, 2, 5, 12, 29, 70, 169, 408, 985, ... (A000129)
··· (2,29,169) ··· (2,5,29) ··· (5,29,433) ··· (1,1,1) (1,1,2) (1,2,5) ··· (5,13,194) ··· (1,5,13) ··· (1,13,34) ···
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Related work
The equation x2 + y2 + z2 = axyz has positive integral solutions if and only if a = 1, 3. The positive integral solutions of
2 2 x1 + ··· + xn = ax1 ··· xn
are also “finitely generated”. Over orders in a number field, the set of solutions
2 2 x1 + ··· + xn = ax1 ··· xn
may or may not be “finitely generated”. See work by Silverman and Baragar. Several authors have investigated the solutions of Markov equation and its variations over finite fields.
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Markov equation over polynomial rings
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Rational numbers vs Function Fields
Close your eyes and think of the few slides from Matilde Lalín’s and Allysa Lumley’s talks.
Goal: Find all “positive” solutions to the Markov equation equation over the polynomial ring K[t], where K is a field. Every non-zero element of K[t] can be written as
αf(t)
where f(t) is monic and α ∈ K∗. ∗ (Every element n ∈ Z satisfies n = (±1)˜n, with n˜ > 0.)
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Markov equation over polynomials Monic solutions
Let K be a field of characteristic 6= 2. For A monic, x2 + y2 + z2 = Axyz has a monic solution if and only if A = 1.
x2 + y2 + z2 = xyz √ has a monic solution if and only if i = −1 ∈ K.
If β = 2i then (t, t + β, t2 + βt − 2) is a monic solution.
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Markov equation New solutions from old solutions
x2 + y2 + z2 = xyz
(x1, x2, x3) −→ (xπ(1), xπ(2), xπ(3))
(zy − x, y, z)
(x, y, z) (x, xz − y, z)
(x, y, xy − z)
(x, y, z) −→ (x(f), y(f), z(f)), where f is any monic polynomial.
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Markov tree of monic solutions
(zy − x, y, z) (x, y, xy − z) (x, y, z) (x, xz − y, z)
··· (t, t3 + βt2 − 3t − β, t4 + βt3 − 4t2 − 2βt + 2) ··· (t, t2 + βt − 2, t3 + βt2 − 3t − β) ··· (t2 + βt − 2, t3 + βt2 − 3t − β, t5 + 2βt4 − 9t3 − 6βt2 + 9t + 2β) ··· (t, t + β, t2 + βt − 2) ··· (t2 + βt − 2, t3 + 2βt2 − 7t − 2β, t5 + 3βt4 − 17t3 − 13βt2 + 21t + 3β) ··· (t + β, t2 + βt − 2, t3 + 2βt2 − 7t − 2β) ··· (t + β, t3 + 2βt2 − 7t − 2β, t4 + 3βt3 − 16t2 − 10βt + 10) ···
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Markov tree of monic solutions
(zy − x, y, z) (x, y, xy − z) (x, y, z) (x, xz − y, z)
··· (t, t3 + βt2 − 3t − β, t4 + βt3 − 4t2 − 2βt + 2) ··· (t, t2 + βt − 2, t3 + βt2 − 3t − β) ··· (t2 + βt − 2, t3 + βt2 − 3t − β, t5 + 2βt4 − 9t3 − 6βt2 + 9t + 2β) ··· (t, t + β, t2 + βt − 2) ··· (t2 + βt − 2, t3 + 2βt2 − 7t − 2β, t5 + 3βt4 − 17t3 − 13βt2 + 21t + 3β) ··· (t + β, t2 + βt − 2, t3 + 2βt2 − 7t − 2β) ··· (t + β, t3 + 2βt2 − 7t − 2β, t4 + 3βt3 − 16t2 − 10βt + 10) ···
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Markov tree of monic solutions
(zy − x, y, z) (x, y, xy − z) (x, y, z) (x, xz − y, z)
··· (t, t3 + βt2 − 3t − β, t4 + βt3 − 4t2 − 2βt + 2) ··· (t, t2 + βt − 2, t3 + βt2 − 3t − β) ··· (t2 + βt − 2, t3 + βt2 − 3t − β, t5 + 2βt4 − 9t3 − 6βt2 + 9t + 2β) ··· (t, t + β, t2 + βt − 2) ··· (t2 + βt − 2, t3 + 2βt2 − 7t − 2β, t5 + 3βt4 − 17t3 − 13βt2 + 21t + 3β) ··· (t + β, t2 + βt − 2, t3 + 2βt2 − 7t − 2β) ··· (t + β, t3 + 2βt2 − 7t − 2β, t4 + 3βt3 − 16t2 − 10βt + 10) ···
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Markov tree of monic solutions
(zy − x, y, z) (x, y, xy − z) (x, y, z) (x, xz − y, z)
··· (t, t3 + βt2 − 3t − β, t4 + βt3 − 4t2 − 2βt + 2) ··· (t, t2 + βt − 2, t3 + βt2 − 3t − β) ··· (t2 + βt − 2, t3 + βt2 − 3t − β, t5 + 2βt4 − 9t3 − 6βt2 + 9t + 2β) ··· (t, t + β, t2 + βt − 2) ··· (t2 + βt − 2, t3 + 2βt2 − 7t − 2β, t5 + 3βt4 − 17t3 − 13βt2 + 21t + 3β) ··· (t + β, t2 + βt − 2, t3 + 2βt2 − 7t − 2β) ··· (t + β, t3 + 2βt2 − 7t − 2β, t4 + 3βt3 − 16t2 − 10βt + 10) ···
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings (x, y, z) −→ (x(t2 + 1), y(t2 + 1), z(t2 + 1))
(t2 + 1 + β, t4 + (β + 2) t2 + β − 1, t6 + (β + 3) t4 + 2βt2 − 2) (t2 + 1, t2 + 1, t4 + (β + 2) t2 + β − 1) (t2 + 1, t4 + (β + 2) t2 + β − 1, t6 + (2β + 3) t4 + (4β − 4) t2 − 6)
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Monic solutions of the Markov equation
Theorem (C., Kelly, VanFossen) Up to permutations, all monic solutions of
x2 + y2 + z2 = xyz
over K lie on a tree with root
(f, f + β, f 2 + βf − 2),
where f is a monic polynomial and β = 2i.
Remark Up to permutation and composition, the monic polynomial solutions of the Markov equation “branch out” of the solution
(t, t + β, t2 + βt − 2)
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Monic solutions of the Markov equation
Theorem (C., Kelly, VanFossen) Up to permutations, all monic polynomial solutions of
x2 + y2 + z2 = xyz
lie on a tree with root
(f, f + β, f 2 + βf − 2),
where f is a monic polynomial and β = 2i. “The sketchiest of all proofs” (zy − x, y, z) (x, y, xy − z) (x, y, z) (x, xz − y, z)
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Markov Polynomials
Definition A Markov polynomial is a monic polynomial on the “main Markov tree”. They are polynomials defined over Z[i]. ··· (t, t3 + βt2 − 3t − β, t4 + βt3 − 4t2 − 2βt + 2) ··· (t, t2 + βt − 2, t3 + βt2 − 3t − β) ··· (t2 + βt − 2, t3 + βt2 − 3t − β, t5 + 2βt4 − 9t3 − 6βt2 + 9t + 2β) ··· (t, t + β, t2 + βt − 2) ··· (t2 + βt − 2, t3 + 2βt2 − 7t − 2β, t5 + 3βt4 − 17t3 − 13βt2 + 21t + 3β) ··· (t + β, t2 + βt − 2, t3 + 2βt2 − 7t − 2β) ··· (t + β, t3 + 2βt2 − 7t − 2β, t4 + 3βt3 − 16t2 − 10βt + 10) ···
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Markov Polynomials
M0 = t
M1 = t + 2i 2 M2 = t + 2it − 2 3 2 M3 = t + 2it − 3t − 2i 3 2 M4 = t + 4it − 7t − 4i 4 3 2 M5 = t + 2it − 4t − 4it + 2 4 3 2 M6 = t + 6it − 16t − 20it + 10 5 4 3 2 M7 = t + 2it − 5t − 6it + 5t + 2i 5 4 3 2 M8 = t + 4it − 9t − 12it + 9t + 4i 5 4 3 2 M9 = t + 6it − 17t − 26it + 21t + 6i 5 4 3 2 M10 = t + 8it − 29t − 56it + 57t + 24i
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Unanswered questions – Episode I
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Unanswered questions
Does the uniqueness conjecture hold for Markov polynomials? That is, does a Markov polynomial z determine the triple (x, y, z)? We can prove that if z or z ± 2 are a power of an irreducible polynomial then (x, y, z) is unique.
How many irreducible or power of irreducible Markov polynomials are there? Except for the first two, the first 500 Markov polynomials are reducible over Z[i].
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Markov-Fibonacci polynomials
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Markov polynomial vs Fibonacci numbers
Do (t, fn−1, fn) behave like “Fibonacci numbers”? ··· (t, t3 + βt2 − 3t − β, t4 + βt3 − 4t2 − 2βt + 2) ··· (t, t2 + βt − 2, t3 + βt2 − 3t − β) ··· (t2 + βt − 2, t3 + βt2 − 3t − β, t5 + 2βt4 − 9t3 − 6βt2 + 9t + 2β) ··· (t, t + β, t2 + βt − 2) ··· (t2 + βt − 2, t3 + 2βt2 − 7t − 2β, t5 + 3βt4 − 17t3 − 13βt2 + 21t + 3β) ··· (t + β, t2 + βt − 2, t3 + 2βt2 − 7t − 2β) ··· (t + β, t3 + 2βt2 − 7t − 2β, t4 + 3βt3 − 16t2 − 10βt + 10) ··· Definition They are called the Markov-Fibonacci polynomials.
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Markov-Fibonacci polynomials
The polynomials fn satisfy the recurrence relation
fn = tfn−1 − fn−2
with f0 = 2 and f1 = t + 2i.
They satisfy −n fn(i) · i = Fn+2 −n fn(2i) · i = 2Pn+1
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Markov-Fibonacci polynomials and Chebyshev polynomials
We have
fn+1(t) = (−t + β)Un(t/2) + 2Un+1(t/2),
where Un(t) be the n-th Chebyshev polynomial of the second kind, that is, Un−1(cos θ) · sin θ = sin nθ.
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Divisibility properties
Recall: The Fibonacci numbers satisfy:
Fn | Fm if and only if m ≡ 0 mod n.
Do Markov-Fibonacci polynomials satisfy any divisibility property? Not for general K. If K has characteristic zero then fn | fm if and only if m = n.
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Divisibility properties over Finite fields
Theorem Suppose√ K is a finite field of odd characteristic containing i = −1. For every n, there exists a positive integer r such that fn | fm if and only if m ≡ n mod r.
Example Over F13, f1 = t + 10 divides
. . . , f−13, f−6,f 1, f8, f15, f22,...
Example
Over F13, f2 = (t + 4)(t + 6) divides
. . . , f−82, f−40,f 2, f44, f86, f128,...
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Divisibility properties over Finite fields
Theorem Suppose√ K is a finite field of odd characteristic containing i = −1. For every n, there exists a positive integer r such that fn | fm if and only if m ≡ n mod r.
Example Over F13, f1 = t + 10 divides
. . . , f−13, f−6,f 1, f8, f15, f22,...
Example
Over F13, f2 = (t + 4)(t + 6) divides
. . . , f−82, f−40,f 2, f44, f86, f128,...
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Divisibility properties
Theorem (C., Kelly, VanFossen)
Let α ∈ Fq. If α is a root of fn, for some n ∈ N, then there exists an integer r = r(α, q) > 1 such that
fn+kr(α) = 0
for all integers k.
Example Over F5, t = 1 is a root of
. . . , f−5, f−2,f 1, f4, f7, f10, ···
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Sketch of a proof
Theorem (C., Kelly, VanFossen)
Let α ∈ Fq. If α is a root of fn, for some n ∈ N, then there exists an integer r = r(α, q) > 1 such that
fn+kr(α) = 0
for all integers k ≥ 0.
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Sketch of a proof
Assume α 6= 0, ±2. We have the generating function
∞ X (−α + β)x + 2 (−α + β)x + 2 f (α)xn = = n x2 − αx + 1 (x − ω )(x − ω ) n=0 α α where √ √ α + α2 − 4 α − α2 − 4 ω = and ω = . α 2 α 2
If α is a root of fn then
n n n+1 n+1 0 = (−α + β)(ωα − ωα) + 2(ωα − ωα )
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Sketch of a proof
n This is equivalent to an equation Wα = Vα, where n √ 2 √ ! 2 α + α2 − 4 α − 8 − 4i α2 − 4 = . 2 −α2 | {z } | {z } Vα Wα
Thus if r = ord(Wα) < ∞ and m ≡ n mod r then
fm(α) = 0.
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Unanswered questions – Episode II
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Unanswered questions
Can we find an asymptotic formula for the number Nq of α ∈ Fq that are a root of fn, for some n? Over Fp, computations show that
Np ∼ cp, p −→ ∞
where 0.5 < c < 0.8. √ 2 The√ group generated by a unit a + b α − 4 in K( α2 − 4) can be embedded in the group of K-rational points of the Pell curve
x2 − (α2 − 4)y2 = 1,
via the map p a + b α2 − 4 7−→ (a, b).
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Unanswered questions
√ Recall: If f (α) = 0 then W = α+ α2−4 and n √ α 2 α2−8−4i α2−4 n Vα = −α2 are units such that Wα = Vα.
n For what values of α ∈ Fq, does Wα = Vα, for some n?
When do two sections of the surface
x2 − (t2 − 4)y2 = 1
have the same specialization?
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings Thank you!
Unanswered questions – Episode III?
Ricardo Conceição ettysburg College Solutions of the Markov equation over polynomial rings