Multidimensional generalisation of the Schmidt conjecture is true

You need to know: Euclidean space ${\mathbb R}^n$ , notation ${\mathbb N}$ for the set of positive integers. In addition, you need to know the concept of Hausdorff dimension of set $A\subset {\mathbb R}^n$ to understand the “moreover” part of the Theorem.

Background: For $x\in{\mathbb R}$ , let $||x||$ denote the distance from x to the nearest integer. Let ${\cal R}_n \subset {\mathbb R}^n$ be the set of $r=(r_1, \dots, r_n)$ such that $r_i\geq 0, \, i=1,\dots,n$ and $\sum\limits_{i=1}^n r_i=1$ . Given $\textbf{r}\in {\cal R}_n$ , we say that point $\textbf{y}=(y_1,\dots, y_n) \in {\mathbb R}^n$ is $\textbf{r}$ -badly approximable if there exists $c=c(\textbf{y})>0$ such that $\max\limits_{1 \leq i \leq n}||qy_i||^{1/r_i} \geq c/q$ for all $q \in {\mathbb N}$ , with convention that $||qy_i||^{1/0}=0$ . Let Bad(r) be be set of all $\textbf{r}$ -badly approximable points in ${\mathbb R}^n$ .

The Theorem: On 2nd April 2013, Victor Beresnevich submitted to arxiv a paper in which he proved that for any finite subset $W\subset {\cal R}_n$ , the set $\bigcap\limits_{\textbf{r}\in W} \textbf{Bad(r)}$ is non-empty. Moreover, this set has Hausdorff dimension n.

Short context: A real number x is said to be badly approximable if there exists a constant $c(x)>0$ such that $||qx|| > c(x)/q$ for all $q \in {\mathbb N}$ . Sets Bad(r) are natural generalisations containing n-tuples $(y_1, \dots, y_n)$ of simultaneously badly approximable numbers. In 1983, Schmidt conjectured that $\textbf{Bad}\left(\frac{1}{3}, \frac{2}{3}\right) \cap \textbf{Bad}\left(\frac{2}{3}, \frac{1}{3}\right) \neq \emptyset$ . In 2011, Badziahin, Pollington, and Velani proved the $n=2$ case of the Theorem, which implies the Schmidt’s conjecture as a special case. The Theorem is a generalisation of this result to all dimensions.

Links: Free arxiv version of the original paper is here, journal version is here.

Go to the list of all theorems

Most odd degree hyperelliptic curves have no finite rational points

You need to know: Polynomial, degree of a polynomial.

Background: Let ${\cal P}$ be the set of odd-degree polynomials $P(x)=\sum\limits_{i=0}^{2g+1} a_i x^{2g+1-i}$ with rational coefficients $a_i$ and leading coefficient $a_0=1$ . With change of variables $x'=u^2x$ , $y'=u^{2g+1}y$ , we can have new coefficients $a'_i=u^{2i}a_i$ , $i=1,2,\dots,2g+1$ , and, by selecting $u$ to be the common denominator of $a_1, a_2, \dots, a_{2g+1}$ , we can make all coefficients integers. After this, define height of $P\in{\cal P}$ by $H(P) = \max\{|a_1|, |a_2|^{1/2}, \dots, |a_{2g+1}|^{1/(2g+1)}\}$ . For fixed integer $g>0$ and real $X>0$ , let $\mu(X,g)$ be a fraction of the polynomials $P\in {\cal P}$ of degree $2g+1$ and height less than $X$ for which the equation $y^2=P(x)$ has no rational solutions.

The Theorem: On 1st February 2013, Bjorn Poonen and Michael Stoll submitted to arxiv a paper in which they proved that $\lim\limits_{g\to\infty}(\lim\limits_{X\to\infty}\mu(X,g)) = 1$ .

Short context: Set of real solutions to $y^2=P(x)$ for $P \in {\cal P}$ is known as odd degree hyperelliptic curve, and rational solutions are called finite rational points on this curve. In this terminology, the Theorem states that most odd degree hyperelliptic curves have no finite rational points. Moreover, Poonen and Stoll also proved for “almost all” polynomials $P\in{\cal P}$ in the same sense as in the Theorem, there is an explicit algorithm, with polynomial P as an input, and output certifying that there are indeed no rational solutions to $y^2=P(x)$ . In other words, there exists a universal method able to solve almost all equations in the form $y^2=P(x), \, P \in {\cal P}$ at once!

Links: Free arxiv version of the original paper is here, journal version is here.

Go to the list of all theorems

For every e>0, there is a set of n integers with no sum-free subset of size (1/3+e)n

You need to know: Integers, addition of integers.

Background: A set B of integers is called sum-free, if there are no distinct $x,y,z \in B$ such that $x+y=z$ .

The Theorem: On 19th January 2013, Sean Eberhard, Ben Green, and Freddie Manners submitted to arxiv a paper in which they proved that for every $\epsilon > 0$ , there is a set A of n integers such that every set $B \subset A$ with at least $(\frac{1}{3}+\epsilon)n$ elements is not sum-free (that is, B contains three distinct elements $x, y, z$ with $x + y = z$ ).

Short context: In 1965, Erdős proved that every set A of n integers has a sum-free subset B of size at least $n/3$ , and asked if the constant $1/3$ in this theorem can be improved. In 2011, Lewko found a 28-element set with maximal sum-free subset of size 11, showing that the constant cannot be greater than $\frac{11}{28} \approx 0.39$ . The theorem states that no constant greater than $\frac{1}{3}$ can work, so Erdős theorem is essentially the best possible.

Links: Free arxiv version of the original paper is here, journal version is here.

Go to the list of all theorems

A set of density one satisfies the local-global conjecture for integral Apollonian packings

You need to know: Greatest common divisor (gcd) of a possibly infinite set of integers, circle in the plane, radius of a circle, curvature of a circle ( $1/r$ where r is the radius). Also, see this previous theorem description for the definition of bounded Apollonian circle packing (ACP).

Background: Let $b(C)$ denote the curvature of a circle C. The ACP is called integer if all circles in it has integer curvatures. For an integer ACP ${\cal P}$ , let ${\cal B}=\{b(C) : C \in {\cal P}\}$ be the set of all curvatures in ${\cal P}$ . The integer ACP ${\cal P}$ is called primitive if $\text{gcd}({\cal B})=1$ . A positive integer m is called admissible for ${\cal P}$ if for any integer $q\geq 1$ , there exists $k \in {\cal B}$ such that $k-m$ is divisible by q. Let $f_{\cal P}(N)$ be the number of admissible integers at most N which does not belong to ${\cal B}$ .

The Theorem: On 20th May 2012, Jean Bourgain and Alex Kontorovich submitted to arxiv a paper in which they proved that for any primitive integer ACP ${\cal P}$ there exist constants $\eta>0$ and $C<\infty$ such that $f_{\cal P}(N) \leq C N^{1-\eta}$ for all $N>0$ .

Short context: Apollonian circle packing is named after Apollonius of Perga, who lived more than 2000 years ago, and is studied by many researchers since that. Important research directions are counting circles with the curvature at most X (see here), as well as studying the set ${\cal B}$ of distinct integers occurring as curvatures. A central conjecture in the area is the the local-global conjecture stating that every sufficiently large admissible integer belongs to ${\cal B}$ . Previously, it was known that a positive percentage of integers satisfy this conjecture. The Theorem implies that the percentage of integers up to N that satisfy it approaches $100\%$ as N grows.

Links: Free arxiv version of the original paper is here, journal version is here.

Go to the list of all theorems

Erdős conjecture on square-free values of f(p), f cubic, p prime, is true

You need to know: Set ${\mathbb Z}$ of integers, greatest common divisor $\text{gcd}(a,b)$ of integers $a,b$ , notation $|S|$ for the number of elements in finite set S, primes, infinite product, cubic polynomial and its roots, big O notation, small o notation.

Background: For integer $m>0$ , let $P(m)$ be the set of integers $0<k<m$ such that $\text{gcd}(k,m)=1$ . For function $f:{\mathbb Z} \to {\mathbb Z}$ , let $P_f(m)$ be the set of $k\in P(m)$ such that $f(k)$ is divisible by m. Let $c_f = \prod\limits_{p\in {\cal P}} \left(1-\frac{|P_f(p^2)|}{|P(p^2)|}\right)$ , where ${\cal P}=\{2,3,5,\dots\}$ is the set of all primes. An integer is called square-free if it is not divisible by the square $d^2$ of any integer $d>1$ .

The Theorem: On 16th December 2011, Harald Helfgott submitted to arxiv a paper in which he proved the following result. Let $f$ be a cubic polynomial with integer coefficients without repeated roots. Then the number of prime numbers $p\leq N$ such that $f(p)$ is square-free is $(1+o(1))c_f \frac{N}{\log N} + O(1)$ .

Short context: For linear polynomial $f(x)=mx+a$ , $f(n)$ is square-free for infinitely many integers n provided that $\text{gcd}(a,m)$ is square-free. In 1931, Estermann proved that $f(n)$ is square-free infinitely often (subject to easy-to-check necessary conditions on f) for quadratic polynomials f. In 1953, Erdős did the same for cubic polynomials. Erdős also conjectured that if cubic polynomial f has no repeated roots and for every prime q there exists $k\in P(q^2)$ such that $f(k)$ is not divisible by $q^2$ , then $f(p)$ is square-free for infinitely many primes p. The Theorem confirms this conjecture.

Links: Free arxiv version of the original paper is here, journal version is here.

Go to the list of all theorems

Zaremba’s conjecture is true for a set of density one

You need to know: Notation ${\mathbb N}$ for the set of positive integers, greatest common divisor $\text{gcd}(a,b)$ for $a,b\in{\mathbb N}$ , notation $\left \lfloor{x}\right \rfloor$ for the largest integer not exceeding real number x.

Background: For a real number x, define a sequence $\{x_n\}$ by the rules: (i) $x_0=x$ , and (ii) for $n\geq 0$ , if $x_n$ is an integer then stop, otherwise let $x_{n+1}=\frac{1}{x_n-\left\lfloor{x_n}\right \rfloor}$ . This sequence may be finite or infinite depending on x. Let $a_n=\left\lfloor{x_n}\right \rfloor$ for all n. Notation $x=[a_0; a_1, a_2, \dots, a_n,\dots]$ in called continued fraction expansion of x. For $A>0$ , let $D(A)$ be the set of $d \in {\mathbb N}$ , for which there exist an integer $b$ , such that $0<b<d$ , $\text{gcd}(b,d)=1$ , and $\frac{b}{d}$ has a finite continuous fraction expansion $\frac{b}{d}=[0; a_1, \dots a_n]$ such that $a_k \leq A$ , $k=1,\dots,n$ .

The Theorem: On 19th July 2011, Jean Bourgain and Alex Kontorovich submitted to arxiv a paper in which they proved the following result. Let $K(N)$ be the number of integers less than $N$ belonging to $D(50)$ . Then $\lim\limits_{N\to\infty} \frac{K(N)}{N} = 1$ .

Short context: If we fix bound A and study rational numbers in the form $[a_0; a_1, a_2, \dots, a_n]$ such that $a_i\leq A$ , $\forall i$ , what positive integers can occur as a denominator of such a number in reduced form? Zaremba conjectured in 1971 that every number can! More formally, Zaremba’s conjecture states that there exists $A>0$ such that $D(A)={\mathbb N}$ . In fact, Zaremba suggested that $A=5$ may work. While this conjecture remains open, the Theorem states that, for $A=50$ , “almost every” positive integer belongs to $D(A)$ .

Links: Free arxiv version of the original paper is here, journal version is here.

Go to the list of all theorems

zeta(2,…,2,3,2,…,2) is a rational linear combination of products zeta(m)pi^(2n) with m odd

You need to know: Sum of infinite series, factorial $n!=1\cdot 2 \cdot \dots \cdot n$ with convention that $0!=1$ , notation ${n}\choose{k}$ for $\frac{n!}{k!(n-k)!}$ , with convention that ${{n}\choose{k}}=0$ if $n<k$ .

Background: Zeta function is the sum of infinite series $\zeta(s) = \sum\limits_{n=1}^\infty \frac{1}{n^s}$ for $s>1$ . Multiple zeta function is $\zeta(k_1, k_2, \dots k_n) = \sum\limits_{0<m_1<\dots<m_n} \frac{1}{m_1^{k_1} m_2^{k_2} \dots m_n^{k_n}}$ . Let $H(n)=\zeta(\underbrace{2,\dots, 2}_{n})=\frac{\pi^{2n}}{(2n+1)!}$ . Let $H(a,b)=\zeta(\underbrace{2,\dots, 2,}_{a},3,\underbrace{2,\dots, 2}_{b})$ .

The Theorem: On 3rd March 2011, Don Zagier submitted to the Annals of Mathematics a paper in which he proved that for all integers $a,b \geq 0$ , we have $H(a,b)=\sum\limits_{r=1}^{a+b+1}(-1)^r \left[{{2r}\choose{2a+2}}-(1-2^{-2r}){{2r}\choose{2b+1}} \right] H(a+b-r+1)\zeta(2r+1)$ .

Short context: Zeta function in one variable $\zeta(s)$ is one of the central functions in the whole mathematics which is studied for centuries. Multiple zeta functions are natural generalisations which are also well-studied, see, for example, here. It is a rare achievement to derive an exact analytical expression for $\zeta(k_1, k_2, \dots k_n)$ . The Theorem achieves this for $\zeta(2,\dots,2,3,2,\dots,2)$ .

Links: The original paper is available here.

Go to the list of all theorems

There exist e>0 and C such that if A is a cap set in F_3^N, then |A|/3^N < C/N^(1+e)

You need to know: Addition modulo 3, vectors, notation $|A|$ for the number of elements in finite set A.

Background: Let ${\mathbb F}_3$ be the set $\{0,1,2\}$ with addition defined modulo 3. Let ${\mathbb F}_3^N$ be the set of N-component vectors $x=(x_1, \dots, x_N)$ with each $x_i \in {\mathbb F}_3$ , and with addition defined component-wise. We say that three different points $x,y,z\in {\mathbb F}_3^N$ form a line if $x+y=z+z$ . A set $A \subseteq {\mathbb F}_3^N$ is called a cap set if it contains no lines.

The Theorem: On 31st January 2011, Michael Bateman and Nets Katz submitted to arxiv a paper in which they proved the existence of $\epsilon>0$ and $C<\infty$ such that if $A \subseteq {\mathbb F}_3^N$ is a cap set, then $\frac{|A|}{3^N} \leq \frac{C}{N^{1+\epsilon}}$ .

Short context: What can the maximal size of a cap set in ${\mathbb F}_3^N$ ? This problem is interesting in its own, but also studied because of hope that methods to solve it may be useful for the similar problem of finding dense sets of integers without 3-term arithmetic progressions, see here. A cap set conjecture predicts the existence of constant $c<3$ such that $|A|<c^N$ for every cap set $A \subseteq {\mathbb F}_3^N$ . However, before 2011, the best upper bound was $\frac{|A|}{3^N} \leq \frac{C}{N}$ , proved by Meshulam in 1995. The Theorem improves this bound.

Links: Free arxiv version of the original paper is here, journal version is here.

Go to the list of all theorems

The main conjecture in Vinogradov’s Mean Value Theorem is true if s>=k(k+1)

You need to know: Just basic arithmetic for the Theorem. You may like to learn what is Vinogradov’s Mean Value Theorem and Waring’s problem to better understand the context.

Background: For positive integers s,k, and X, let $J_{s,k}(X)$ be the number of integral solutions of the system of equations $x_1^j+\dots+x_s^j = y_1^j+\dots+y_s^j$ , $1 \leq j \leq k$ , such that $1\leq x_i,y_i \leq X$ for $1\leq i \leq s$ .

The Theorem: On 3rd December 2010, Trevor Wooley submitted to the Annals of Mathematics a paper in which he proved that for any natural numbers $k\geq 2$ and $s \geq k(k + 1)$ , and any real $\epsilon>0$ , there exist a constant C such that $J_{s,k}(X) \leq C X^{2s-\frac{1}{2}k(k+1)+\epsilon}$ .

Short context: A famous conjecture, known as the main conjecture in Vinogradov’s Mean Value Theorem, predicts that $J_{s,k}(X) \leq C X^\epsilon(X^s+X^{2s-\frac{1}{2}k(k+1)})$ for any $s\geq 1$ and $k\geq 2$ . This estimate, if true, would be optimal up to constant and $X^\epsilon$ factors. The Theorem implies that it holds if $s \geq k(k + 1)$ . In the same paper, Wooley demonstrated several applications of the Theorem, for example, to Waring’s problem. In a later work, Bourgain, Demeter, and Guth proved the conjecture in general.

Links: Free arxiv version of the original paper is here, journal version is here.

Go to the list of all theorems

The size of subset of {1,…,N} without 3-term arithmetic progressions is O(N/(log n)^(1-o(1)))

You need to know: A (nontrivial) 3-term arithmetic progression, big O notation, small o notation.

Background: For integer $N>0$ , let $r_3(N)$ be the cardinality of the largest subset of $\{1, 2, \dots , N\}$ which contains no nontrivial 3-term arithmetic progressions.

The Theorem: On 30th October 2010, Tom Sanders submitted to arxiv a paper in which he proved that $r_3(N) = O\left(\frac{N(\log\log N)^6}{\log N}\right)$ .

Short context: In 1936, Erdős and Turán conjectured that any set containing a positive proportion of integers must contain a 3-term arithmetic progression. This is equivalent to $\lim\limits_{N\to\infty}\frac{r_3(N)}{N}=0$ . In 1953, Roth confirmed this conjecture by proving that $r_3(N) = O\left(\frac{N}{\log\log N}\right)$ . In 1999, Bourgain improved this to $r_3(N) = O\left(\frac{N\sqrt{\log\log N}}{\sqrt{\log N}}\right)$ . The Theorem proves a much stronger upper bound, the first one in the form $r_3(N) = O\left(\frac{N}{\log^{1-o(1)} N}\right)$ . In a later work, Bloom improved the bound slightly to $r_3(N) = O\left(\frac{N(\log\log N)^4}{\log N}\right)$ .

Links: Free arxiv version of the original paper is here, journal version is here.

Go to the list of all theorems