“Hilbert’s problems today” (April 6, 2001)

Open Diophantine Problems

by

Michel WALDSCHMIDT

Content §1. Diophantine Equations §2. Diophantine Approximation §3. Transcendence §4. Heights

  http://www.math.jussieu.fr/∼miw/articles/ps/odp.ps

1. Diophantine Equations

Hilbert’s tenth problem:

Given a Diophantine equation with any number of unknown quantities and with integral numerical coefficients: To devise a process according to which it can be determined by a finite number of operations whether the equation is solvable in rational integers.

Open problem:

To answer Hilbert’s tenth problem for the special case of plane curve, which means to give an algorithm to decide whether a given Diophantine equation

f(x, y) = 0

has a solution (in ℤ, and the same problem for ℚ).

Problem. Let f ∈ ℤ[X, Y] be a polynomial such that the equation f(x, y) = 0 has only finitely many solutions (x, y) ∈ ℤ × ℤ. Give an upper bound for max{|x|, |y|} when (x, y) is such a solution, in terms of the degree of f and of the maximal absolute value of the coefficients of f.

Lettre adressée à l’Éditeur par Monsieur E. Catalan, Répétiteur à l’école polytechnique de Paris^{⟩⟩}, published in Crelle Journal (1844):

⟨⟨Je vous prie, Monsieur, de bien vouloir énoncer, dans votre recueil, le théorème suivant, que je crois vrai, bien que je n’aie pas encore réussi à le démontrer complètement: d’autres seront peut-être plus heureux:

Deux nombres entiers consécutifs, autres que 8 et 9, ne peuvent être des puissances exactes; autrement dit: l’équation x^{m} − y^{n} = 1, dans laquelle les inconnues sont entières et positives, n’admet qu’une seule solution.^{⟩⟩}

Perfect powers: 1, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 81, 100, 121, 125, 128, 144, 169, …

Conjecture (Catalan). The equation

x^{p} − y^{q} = 1,

where the unknowns x, y, p and q are integers all ≥ 2, has only one solution (x, y, p, q) = (3, 2, 2, 3).

R. Tijdeman (1976): only finitely many solutions.

Conjecture (Pillai). Let k be a positive integer. The equation

x^{p} − y^{q} = k,

where the unknowns x, y, p and q are integers all ≥ 2, has only finitely many solutions (x, y, p, q).

The Diophantine equation

x^{p} + y^{q} = z^{r}

has 10 known solutions (x, y, z, p, q, r) in positive integers for which

\frac{1}{p} + \frac{1}{q} + \frac{1}{r} < 1

and such that x, y, z are relatively prime:

1 + 2^{3} = 3^{2}  2^{5} + 7^{2} = 3^{4}  7^{3} + 13^{2} = 2^{9}

2^{7} + 17^{3} = 71^{2} 3^{5} + 11^{4} = 122^{2}

17^{7} + 76271^{3} = 21063928^{2}
1414^{3} + 2213459^{2} = 65^{7}
9262^{3} + 15312283^{2} = 113^{7}
43^{8} + 96222^{3} = 30042907^{2}
33^{8} + 1549034^{2} = 15613^{3}.

R. Tijdeman and D. Zagier Conjecture : there is no solution with the further restriction that each of p, q and r is ≥ 3.

Markoff Spectrum x^{2} + y^{2} + z^{2} = 3xyz Sequence:

1, 2, 5, 13, 29, 34, 89, 169, 194,

233, 433, 610, 985, 1325, 1597, …

Conjecture. Fix a positive integer m for which the equation

m^{2} + m_{1}^{2} + m_{2}^{2} = 3mm_{1}m_{2}

has a solution in positive integers (m_{1}, m_{2}) with

0 < m_{1} ≤ m_{2} ≤ m.

Then such a pair (m_{1}, m_{2}) is unique. True for m ≤ 10^{105}. Connection with Diophantine Approximation:

μ_{m} = \sqrt{9m^{2} − 4}/m :

Sequence:

\sqrt{5}, \sqrt{8}, \sqrt{221}/5, \sqrt{1517}/13, …

lim sup_{q→∞}|q(qα_{m} − p)| = \frac{m}{\sqrt{9m^{2} − 4}}·

2. Diophantine Approximation

abc Conjecture

(D.W. Masser and J. Œsterlé, 1987).

For a positive integer n, denote by

R(n) = ∏_{p|n} p

the radical or squarefree part of n.

Conjecture (abc Conjecture). For each ϵ > 0 there exists a positive number κ(ϵ) which has the following property: if a, b and c are three positive rational integers which are relatively prime and satisfy a + b = c, then

c < κ(ϵ)R(abc)^{1+ϵ}.

Triples (a, b, c) with 0 < a < b < c, a + b = c and gcd(a, b) = 1.

− ∘ −

λ(a, b, c) = \frac{log c}{log R(abc)}·

(?) Finitely many (a, b, c) with λ(a, b, c) > 1 + ϵ. Largest known value for λ (É. Reyssat):

2 + 3^{10} · 109 = 23^{5}, λ = 1.629912 … 140 known values of λ(a, b, c) which are ≥ 1.4.

− ∘ −

ϱ(a, b, c) = \frac{log abc}{log R(abc)}·

(?) Finitely many (a, b, c) with ϱ(a, b, c) > 3 + ϵ. Largest known value for ϱ (A. Nitaj): 13 · 19^{6} + 2^{30} · 5 = 3^{13} · 11^{2} · 31, ϱ = 4.41901 … 46 known triples (a, b, c) with 0 < a < b < c and gcd(a, b) = 1 satisfying ϱ(a, b, c) > 4.

Conjecture (Erdős-Woods). There exists a positive integer k such that, for m and n positive integers, the conditions

R(m + i) = R(n + i) (i = 0, …, k − 1)

imply m = n.

Remark: k ≥ 3:

R(75) = 15 = R(1215), R(76) = 2 · 19 = R(1216).

Also m = 2^{h} − 2, n = 2^{h}m, n + 1 = (m + 1)^{2}

R(m) = R(n) and R(m + 1) = R(n + 1)

Arithmetic progressions (T.N. Shorey): (?) Does there exist a positive integer k such that,

for any m, n, d and d′ positive integers satisfying gcd(m, d) = gcd(n, d′) = 1, the conditions

R(m + id) = R(n + id′) (i = 0, …, k − 1)

imply m = n and d = d′? Remark: k ≥ 4:

R(2) = R(4) = R(8),
R(2 + 79) = R(4 + 23) = R(9),
R(2 + 2 · 79) = R(4 + 2 · 23) = R(10).

Theorem (Thue-Siegel-Roth). For any ϵ > 0 and any irrational algebraic number α, there is a positive constant C(α, ϵ) > 0 such that, for any rational number p/q,

|α − \frac{p}{q}| > \frac{C(ϵ)}{q^{2+ϵ}}·

Consequence of abc (E. Bombieri, M. Langevin):

|α − \frac{p}{q}| > \frac{C(ϵ)}{R(pq)q^{ϵ}} ·

Main Open Problem: Effectivity

(?) Does there exist an algebraic number of degree

≥ 3 with bounded partial quotients? (?) Does there exist one with unbounded partial quo-

tients?

Let ψ(q) be a continuous positive real valued function. Assume that the function qψ(q) is nonincreasing.

Conjecture. Let θ be real algebraic number of degree at least 3. Then inequality

|θ − \frac{p}{q}| > \frac{ψ(q)}{q}·

has infinitely many solutions in integers p and q with q > 0 if and only if the integral

∫_{1}^{∞} ψ(x)dx

diverges.

Schmidt’s Subspace Theorem Consequence: finiteness of solutions of the equation

x_{1} + ⋯ + x_{n} = 1

where the unknowns are integers (or S-integers) in a number field and no proper subsum vanishes.

Open problem: effective version for n ≥ 3?

Waring’s problem (1770):

“Every integer is a cube or the sum of two, three, … nine cubes; every integer is also the square of a square, or the sum of up to nineteen such; and so forth. Similar laws may be affirmed for the correspondingly defined numbers of quantities of any like degree.”

n = x_{1}^{k} + ⋯ + x_{g(k)}^{k}.

I(k) = 2^{k} + [(3/2)^{k}] − 2.

Easy to check g(k) ≥ I(k):

3^{k} = 2^{k}q + r with 0 < r < 2^{k}, q = [(3/2)^{k}]

N = 2^{k}q − 1 = (q − 1)2^{k} + (2^{k} − 1)1^{k}

I(k) = (q − 1) + (2^{k} − 1)

Known: g(k) = I(k) for 2 ≤ k ≤ 471 600 000.

(?)  ∥(\frac{3}{2})^{k}∥ ≥ 2 · (\frac{3}{4})^{k}

3. Transcendence Irrationality Problems:

Euler’s constant

\begin{aligned}γ & = lim_{n→∞} (1 + \frac{1}{2} + \frac{1}{3} + ⋯ + \frac{1}{n} − log n) \\ & = 0.5772157 …, \end{aligned}

Catalan’s constant

G = ∑_{n≥0} \frac{(−1)^{n}}{(2n + 1)^{2}}

= \frac{π}{4} ∫_{0}^{1} _{2}F_{1} ( \begin{matrix}1/2 , 1/2 \\ 1 \end{matrix} |t) \frac{dt}{\sqrt{4t}}

= 0.915965594 …,

Γ(1/5) = ∫_{0}^{∞} e^{−t}t^{−4/5}dt = 4.59084371 … e + π = 5.8598744 …, e^{γ} = 1.781072 …

∑_{n≥1} \frac{σ_{k}(n)}{n!} (k = 1, 2) where σ_{k}(n) = ∑_{d|n} d^{k}

Conjecture (Schanuel). Let x_{1}, …, x_{n} be ℚ-linearly independent complex numbers. Then the transcendence degree over ℚ of the field

ℚ(x_{1}, …, x_{n}, e^{x_{1}}, …, e^{x_{n}})

is at least n.

Conjecture (Algebraic Independence of Logarithms of Algebraic Numbers). Let λ_{1}, …, λ_{n} be ℚ-linearly independent complex numbers. Assume that the numbers e^{λ_{1}}, …, e^{λ_{n}} are algebraic. Then the numbers λ_{1}, …, λ_{n} are algebraically independent.

D. Roy:  L = \{log α ; α ∈ \overline{ℚ}^{×}\}

(?) For any algebraic subvariety V of ℂ^{n} defined over the field \overline{ℚ} of algebraic numbers, the set V ∩ L^{n} is the union of the sets E ∩ L^{n}, where E ranges over the set of vector subspaces of ℂ^{n} which are contained in V.

Conjecture (Gel’fond-Schneider). Let β be an irrational algebraic number of degree d and α a nonzero algebraic number. Let log α be a nonzero logarithm of α. Then the d numbers

log α, α^{β}, α^{β^{2}}, …, α^{β^{d−1}}

are algebraically independent.

Elimination Theory

Hilbert Nulstellensatz.

Conjecture (Blum, Cucker, Shub and Smale). Given an absolute constant c and polynomials P_{1}, …, P_{m} with a total of N coefficients and no common complex zeros, there is no program to find, in at most N^{c} step, the coefficients of polynomials A_{i} satisfying Bézout’s relation

A_{1}P_{1} + ⋯ + A_{m}P_{m} = 1.

Complexity in theoretical computer science and Diophantine approximation (W.D. Brownawell)

|x − \frac{p}{q}| > ψ(p/q).

Polyzeta (or Multiple Zeta) Values

L. Euler, K. Nielsen, D. Zagier, A.B. Goncharov, M. Kontsevich, M. Petitot, Minh Hoang Ngoc, P. Cartier,…

ζ(\underline{s}) = ∑_{n_{1}>···>n_{k}≥1} n_{1}^{−s_{1}} ⋯ n_{k}^{−s_{k}},

\underline{s} = (s_{1}, …, s_{k}) ∈ ℤ^{k} with

s_{1} ≥ 2, s_{2} ≥ 1, …, s_{k} ≥ 1.

Let ℨ_{p} denote the ℚ-vector subspace spanned by the real numbers ζ(\underline{s}) with s_{1} + ⋯ + s_{k} = p. Set ℨ_{0} = ℚ and ℨ_{1} = {0}. The ℚ-subspace ℨ spanned by all ℨ_{p}, p ≥ 0, is a subalgebra of ℝ

Conjecture (A.B. Goncharov). As a ℚ-algebra, ℨ is the direct sum of ℨ_{p} for p ≥ 0.

Conjecture (D. Zagier). For p ≥ 3 the dimension d_{p} of the ℚ-vector space ℨ_{p} is given by

d_{p} = d_{p−2} + d_{p−3}

with d_{0} = 1, d_{1} = 0, d_{2} = 1.

Conjecture. The numbers

π, ζ(3), ζ(5), …, ζ(2n + 1), …

are algebraically independent.

Known:

• (Euler– Lindemann) ζ(2n) is transcendental for n ≥ 1.

• (Apéry, 1978) ζ(3) is irrational.

• (T. Rivoal, CRAS 2000; K. Ball & T. Rivoal, to appear). The ℚ-vector space spanned by the n+1 numbers 1, ζ(3), ζ(5), …, ζ(2n+1) has dimension

≥ \frac{1 − ϵ}{1 + log 2} log n

for n ≥ n_{0}(ϵ).

For instance infinitely many of these numbers ζ(2n+1) (n ≥ 1) are irrational.

Gamma Function

Set

G(z) = \frac{1}{\sqrt{2π}} Γ(z).

For N > 0 and x ∈ ℂ such that Nx ≢ 0 (mod ℤ),

∏_{i=0}^{N−1} G (x + \frac{i}{N} ) = N^{(1/2)−Nx}G(Nx).

Then

\overline{G} : (ℚ/ℤ) \ \{0\} → ℂ^{×}/\overline{ℚ}^{×}

is an odd distribution on (ℚ/ℤ) \ {0}:

∏_{i=0}^{N−1} \overline{G} (x + \frac{i}{N} ) = \overline{G}(Nx) for x ∈ (ℚ/ℤ) \ \{0\}

and

\overline{G}(−x) = \overline{G}(x)^{−1}.

Conjecture (Rohrlich). \overline{G} is the universal odd distribution with values in groups where multiplication by 2 is invertible.

Conjecture. Three at least of the four numbers

π, Γ(1/5), Γ(2/5), e^{π\sqrt{5}}

are algebraically independent.

Algebraic Independence and Modular Forms

\begin{aligned}P (q) & = 1 − 24 ∑_{n=1}^{∞} \frac{nq^{n}}{1 − q^{n}}, \\ Q(q) & = 1 + 240 ∑_{n=1}^{∞} \frac{n^{3}q^{n}}{1 − q^{n}}, \\ R(q) & = 1 − 504 ∑_{n=1}^{∞} \frac{n^{5}q^{n}}{1 − q^{n}} · \end{aligned}

Conjecture (Nesterenko). Let τ ∈ ℂ have positive imaginary part. Assume that τ is not quadratic. Set q = e^{2iπτ}. Then 4 at least of the 5 numbers

ℚ(τ, q, P (q), Q(q), R(q))

are algebraically independent.

Fibonacci Numbers F_{n+2} = F_{n+1} + F_{n}, F_{0} = 0, F_{1} = 1.

Special values

∑_{n=1}^{∞} \frac{1}{F_{n}F_{n+2}} = 1

∑_{n=0}^{∞} \frac{1}{F_{2^{n}}} = \frac{7 − \sqrt{5}}{2},  ∑_{n=1}^{∞} \frac{(−1)^{n}}{F_{n}F_{n+1}} = \frac{1 − \sqrt{5}}{2},

∑_{n=1}^{∞} \frac{1}{F_{2n−1} + 1} = \frac{\sqrt{5}}{2}·

Each of the numbers

∑_{n=0}^{∞} \frac{1}{F_{n}}, ∑_{n=1}^{∞} \frac{1}{F_{n} + F_{n+2}} and ∑_{n≥1} \frac{1}{F_{1}F_{2} ⋯ F_{n}}

is irrational (transcendental ?). The numbers

∑_{n=0}^{∞} \frac{1}{F_{2n−1}} , ∑_{n=0}^{∞} \frac{1}{F_{n}^{2}} , ∑_{n=0}^{∞} \frac{(−1)^{n}}{F_{n}^{2}} , ∑_{n=0}^{∞} \frac{n}{F_{2n}},
∑_{n=0}^{∞} \frac{1}{F_{2^{n}−1} + F_{2^{n}+1}} and ∑_{n=0}^{∞} \frac{1}{F_{2^{n}+1}}

are all transcendental Series of Rational Fractions

∑_{n=1}^{∞} \frac{1}{n(n + 1)} = 1

while

∑_{n=0}^{∞} \frac{1}{(2n + 1)(2n + 2)} = log 2,

∑_{n=0}^{∞} \frac{1}{(n + 1)(2n + 1)(4n + 1)} = \frac{π}{3}

∑_{n=0}^{∞} \frac{1}{n^{2}} = \frac{π^{2}}{6} ,  ∑_{n=0}^{∞} \frac{1}{n^{2} + 1} = \frac{1}{2} + \frac{π}{2} · \frac{e^{π} + e^{−π}}{e^{π} − e^{−π}},
∑_{n=0}^{∞} \frac{1}{(6n + 1)(6n + 2)(6n + 3)(6n + 4)(6n + 5)(6n + 6)}

= \frac{1}{4320}(192 log 2 − 81 log 3 − 7π\sqrt{3})

are transcendental.

4. Heights Mahler’s measure of

\begin{aligned}f(X) & = a_{0}X^{d} + a_{1}X^{d−1} + ⋯ + a_{d−1}X + a_{d} \\ & = a_{0} ∏_{i=1}^{d}(X − α_{i}) \end{aligned}

is

M(f) = |a_{0}| ∏_{i=1}^{d} max{1, |α_{i}|}
= exp (∫_{0}^{1} log |f(e^{2iπt})|dt) .

Lehmer (1933): Is-it true that for every positive ϵ there exists an algebraic integer α for which

1 < M(α) < 1 + ϵ?

Smallest known value > 1 for M(α) is

α_{0} = 1.1762808 …,

root of X^{10} + X^{9} − X^{7} − X^{6} − X^{5} − X^{4} − X^{3} + X + 1.

Logarithmic height

h(α) = \frac{1}{d} log M(α).

Conjecture (Lehmer’s Problem). There exists a positive absolute constant c such that, for any nonzero algebraic number α of degree at most d which is not a root of unity,

h(α) ≥ \frac{c}{d}·

Conjecture (Amoroso-David). For each positive integer n ≥ 1 there exists a positive number c(n) having the following property. Let α_{1}, …, α_{n} be multiplicatively independent algebraic numbers. Define

D = [ℚ(α_{1}, …, α_{n}) : ℚ].

Then

∏_{i=1}^{n} h(α_{i}) ≥ \frac{c(n)}{D}·

Heights of Subvarieties

Bogomolov, height of small points, Philippon’s alternative heights. Group varieties, height of translate of algebraic subgroups.

Conjecture (Amoroso-David). For each integer n ≥ 1 there exists a positive constant c(n) such that, for any algebraic subvariety V of 𝔾_{m}^{n} which is defined over ℚ, which is ℚ-irreducible, and which is not a union of translates of algebraic subgroups by torsion points,

V̂ ≥ c(n) deg(V)^{(s−dim V −1)/(s−dim V)},

where s is the dimension of the smallest algebraic subgroup of 𝔾_{m}^{n} containing V.

Mazur Density Problem

Topology of rational points. Connection with the rational version of Hilbert’s tenth problem.

Let K be a number field with a given real embedding. Let V be a smooth variety over K. Denote by Z the closure, for the real topology, of V (K) in V (ℝ).

Question (Mazur). Assume that K = ℚ and that V (ℚ) is Zariski dense; is Z a union of connected components of V (ℝ)?

Colliot-Thélène, Skorobogatov and Swinnerton-Dyer

(?) Let A be a simple abelian variety over ℚ. Assume the Mordell-Weil group A(ℚ) has rank ≥ 1. Then A(ℚ) ∩ A(ℝ)^{0} is dense in the neutral component A(ℝ)^{0} of A(ℝ).

Conjecture. Let A be a simple abelian variety over ℚ, exp_{A} : ℝ^{g} → A(ℝ)^{0} the exponential map of the Lie group A(ℝ)^{0} and Ω = ℤω_{1} + ⋯ + ℤω_{g} its kernel. Let u = u_{1}ω_{1} + ⋯ + u_{g}ω_{g} ∈ ℝ^{g} satisfy exp_{A}(u) ∈ A(ℚ). Then 1, u_{1}, …, u_{g} are linearly independent over ℚ.