There is no Diophantine quintuple

You need to know: Perfect square

Background: A set of $m$ distinct positive integers $\{a_1, a_2, \dots, a_m\}$ is called a Diophantine $m$ -tuple if $a_ia_j+1$ is a perfect square for all $1\leq i<j\leq m$ . In particular, it is called Diophantine quadruple, quintuple, and sextuple for $m=4$ , $m=5$ , and $m=6$ , respectively.

The Theorem: On 13th October 2016, Bo He, Alain Togbè, and Volker Ziegler submitted to arxiv a paper in which they proved that there is no Diophantine quintuple.

Short context: More than two thousands years ago, Diophantus noticed that the set of rational numbers $\{\frac{1}{16}, \frac{33}{16}, \frac{17}{4}, \frac{105}{16}\}$ has the property that the product of any two of them plus one is a square of a rational number. Later, Fermat found positive integers $\{1,3,8,120\}$ with this property, and Euler proved that there are infinitely many such quadruples. A long-standing folklore conjecture predicts that no five positive integers with this property exist. As a partial progress, Dujella proved in 2004 that there is no Diophantine sextuple and that there can be at most finitely many Diophantine quintuples. The Theorem confirms the conjecture in full.

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

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The Furstenberg’s conjecture on the intersections of invariant sets is true

You need to know: Set ${\mathbb Q}$ of rational numbers, closed subset of $[0,1]$ , multiplication of a set by a constant $s\cdot A = \{s\cdot a: a \in A\}$ , sum of two sets $A + B = \{a + b : a \in A; b \in B\}$ , Hausdorff dimension $\text{dim}(A)$ of set $A \subset [0,1]$ .

Background: For real number $x$ , let $\left \lfloor{x}\right \rfloor$ be the largest integer not exceeding x, and let $\text{frac}(x)=x-\left \lfloor{x}\right \rfloor$ . For integer m, let $T_m:[0,1)\to[0,1)$ be a function given by $T_m(x)=\text{frac}(mx)$ . We say that set $S \subset [0,1]$ is invariant under $T_m$ if $T_m(x) \in S$ for every $x \in S$ .

The Theorem: On 25th September 2016, Pablo Shmerkin submitted to arxiv a paper in which he proved the following result. Let $p,q \geq 2$ be positive integers such that $\frac{\log p}{\log q}\not\in{\mathbb Q}$ . Let $A; B \subseteq [0,1]$ be closed sets which are invariant under $T_p$ and $T_q$ , respectively. Then for all real numbers $u$ and $v$ , $\text{dim}((u\cdot A + v) \cap B) \leq \max\{0, \text{dim}(A) + \text{dim}(B) - 1\}$ .

Short context: The Theorem confirms a long-standing conjecture of Furstenberg made in 1969. In fact, the Furstenberg’s conjecture corresponds to the case $u=1$ and $v=0$ of the Theorem. On 26th September 2016, Meng Wu submitted to arxiv a paper with independent and different proof of the same result. Also, see here for an earlier theorem resolving another related conjecture of Furstenberg.

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

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The Pomerance’s conjecture is true: the threshold for making squares is sharp

You need to know: Perfect square, limits, notation P for the probability, independent events, selection uniformly at random.

Background: We say that a finite sequence S of positive integers has a square dependence if it has a subset $A\subset S$ such that the product $\prod\limits_{n\in A}n$ of all integers in A is a perfect square. For $x>1$ , let us select integers $a_1,a_2,\dots$ independently uniformly at random from $[1,x]$ , and let T be the smallest integer such that the sequence $a_1,a_2,\dots, a_T$ has a square dependence.

The Euler-Mascheroni constant is the limit $\gamma=\lim\limits_{n\to\infty}\left(-\log n+\sum\limits_{k=1}^n\frac{1}{k}\right) = 0.577...$ . Integer n is called y-smooth if all of its prime factors are at most y. Let $\Psi(x,y)$ be the number of y-smooth integers up to x, $\pi(y)$ be the number of primes up to y, and let $J(x) = \min\limits_{2\leq y \leq x} \frac{\pi(y)x}{\Psi(x, y)}$ .

The Theorem: On 12th August 2016, Paul Balister, Béla Bollobás, and Robert Morris submitted to arxiv a paper in which they proved that for any $\epsilon>0$ , we have $\lim\limits_{x\to\infty} P\left((e^{-\gamma}-\epsilon)J(x) \leq T \leq (e^{-\gamma}+\epsilon)J(x)\right) = 1$ , where $\gamma$ is the Euler-Mascheroni constant.

Short context: How many random integers between 1 and x we should select such that the product of some selected integers is a perfect square? This question is important for understanding the performance of fastest known factorisation algorithms. In 1994, Pomerance conjectured that the number T of integers needed for this exhibits a sharp threshold, that is, $\lim\limits_{x\to\infty} P\left((1-\epsilon)f(x) \leq T \leq (1+\epsilon)f(x)\right) = 1$ for some function $f(x)$ and any $\epsilon>0$ . See here for the best partial result towards this conjecture before 2016. The Theorem confirms the conjecture in full.

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

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The cap set conjecture is true: the size of any cap set in F_3^n is o(2.756^n)

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

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 30th May 2016, Jordan Ellenberg and Dion Gijswijt submitted to arxiv a paper in which they proved that if $A \subseteq {\mathbb F}_3^N$ is a cap set, then $|A|=o(2.756^n)$ .

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 predicted the existence of constant $c<3$ such that $|A|<c^n$ for every cap set $A \subseteq {\mathbb F}_3^n$ . The Theorem confirms this conjecture, building on an earlier similar result for subsets of ${\mathbb Z}_4^n$ . Before 2016, the best upper bound was $|A|\leq C\frac{3^n}{n^{1+\epsilon}}$ , see here.

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

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Progression-free sets in Z_4^n are exponentially small

You need to know: Addition modulo 4, vectors, notation $|A|$ for the number of elements in finite set A, notation $\log_2 x$ for the base-2 logarithm of x.

Background: Let ${\mathbb Z}_4$ be the set $\{0,1,2,3\}$ with addition defined modulo 4. Let ${\mathbb Z}_4^N$ be the set of N-component vectors $x=(x_1, \dots, x_N)$ with each $x_i \in {\mathbb Z}_4$ , and with addition defined component-wise. We say that three different points $x,y,z\in {\mathbb Z}_4^N$ form a 3-term arithmetic progression if $x+y=z+z$ . A set $A \subseteq {\mathbb Z}_4^N$ is called progression-free if it contains no 3-term arithmetic progressions.

For $x\in(0,1)$ , let $H(x)=-x \log_2 x - (1-x) \log_2(1-x)$ . Let $\gamma = \max\limits_{0<\epsilon<0.25}\left(\frac{1}{2}(H(0.5-\epsilon)+H(2\epsilon))\right)$ . Note that $\gamma \approx 0.926$ . In particular, $\gamma<1$ .

The Theorem: On 5th May 2016, Ernie Croot, Vsevolod Lev, and Peter Pach submitted to arxiv and the Annals of Mathematics a paper in which they proved that if $n\geq 1$ and $A\subseteq {\mathbb Z}_4^n$ is progression-free, then $|A| \leq 4^{\gamma n}$ .

Short context: What can be the maximal size of a progression-free set in ${\mathbb Z}_4^N$ ? This problem is interesting in its own, and also closely related to the problems of bounding progression-free sets in ${\mathbb Z}_3^N$ (known as the cap set conjecture, see here) and in the sets of integers (see here). The Theorem proves, for the first time, that progression-free sets in ${\mathbb Z}_4^n$ are exponentially small comparing to $|{\mathbb Z}_4^n|=4^n$ . In just few weeks after the Theorem was published online, its proof was extended to solve the cap set conjecture.

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

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For any digit d, there are infinitely many primes which do not have d in their decimal expansion

You need to know: Prime numbers

Background: Any positive integer $n$ can be written in a unique way as $n=\sum\limits_{k=1}^{m}a_k10^{m-k}$ , where $m\geq 0$ is an integer, and $a_1, \dots, a_m$ are integers between $0$ and $9$ . This is called decimal expansion of n with digits $a_1, \dots, a_m$ , and this is the usual way we write integers.

The Theorem: On 4th April 2016, James Maynard submitted to arxiv a paper in which he proved that given any digit $d\in\{0,1,\dots,9\}$ , there are exist infinitely many primes p which do not have the digit d in their decimal expansion.

Short context: For $d\in\{0,1,\dots,9\}$ , let ${\cal A}_d$ be the set of positive integers which do not have the digit d in their decimal expansion. For $x \approx 10^k$ , there are about $9^k \approx x^{0.954}$ integers less than x in ${\cal A}_d$ . Sets of integers with at most $x^c$ integers up to x for some $c<1$ are called sparse. Usually, it is hard to prove that there are infinitely many primes in sparse sets. The Theorem achieves this for sets ${\cal A}_d$ , $d=0,1,\dots,9$ . See here for a result about infinitely many primes in (somewhat similar but not sparse) sets of integers with even/odd sum of digits.

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

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The primes contain arbitrarily long multivariate polynomial progressions

You need to know: Prime number, polynomial in r variables, degree of a polynomial, notation ${\mathbb Z}^r$ for the set of vectors $\textbf{m}=(m_1,\dots,m_r)$ with r integer components $m_i$ .

Background: Let ${\mathbb Z}[m_1, \dots, m_r]$ denote the set of polynomials in r variables with integer coefficients.

The Theorem: On 25th March 2016, Terence Tao and Tamar Ziegler submitted to arxiv a paper in which they proved the following result. Let $d, r$ be natural numbers,
$P_1,\dots ,P_k \in {\mathbb Z}[m_1, \dots, m_r]$ be polynomials of degree at most d, such that $P_i-P_j$ has degree exactly d for all $1\leq i<j\leq k$ . Suppose that for each prime p there exist $n\in{\mathbb Z}$ and $\textbf{m}\in {\mathbb Z}^r$ such that $n+P_1(\textbf{m}),\dots, n + P_1(\textbf{m})$ are all not divisible by p. Then there exist infinitely many natural numbers $n, m_1,\dots, m_r$ such that $n+P_1(m_1,\dots, m_r),\dots, n+P_k(m_1,\dots, m_r)$ are simultaneously prime.

Short context: For $n,m \in {\mathbb Z}$ , the sequence $n, n+m,\dots , n+(k-1)m$ is called an arithmetic progression. By analogy, we can call sequence $n+P_1(m),\dots , n+P_k(m)$ for polynomials $P_1,\dots ,P_k$ in one variable polynomial progression, and the sequence $n+P_1(\textbf{m}),\dots, n+P_k(\textbf{m})$ as in the Theorem – multivariate polynomial progression. In an earlier work, Green and Tao proved that primes contains arbitrary long arithmetic progressions (with $m\neq 0$ ). Later, Tao and Ziegler generalised this to polynomial progressions, under the condition that $P_1(0)=\dots=P_k(0)=0$ . The Theorem removes this condition, and also generalises the result to the multivariate case, but introduces a new condition that “ $P_i-P_j$ has degree exactly d”. Proving any result of this kind without any additional condition is difficult, because the case $k=2$ with $P_1-P_2=2$ would imply famous twin prime conjecture.

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

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The Markoff-Hurwitz equation has (c+o(1))(log R)^b integer solutions up to R

You need to know: Set ${\mathbb Z}$ of integers, perfect square, o(1) notation.

Background: For integer parameters $n\geq 3$ , $a\geq 1$ , and $k\in {\mathbb Z}$ consider the equation $x_1^2+x_2^2+\dots+x_n^2=ax_1x_2\dots x_n+k$ . This is called generalized Markoff-Hurwitz equation (the Markoff-Hurwitz equation corresponds to the case $k=0$ ). For $R>0$ , let $M_{n,a,k}(R)$ be the number of integer solutions $(x_1, \dots, x_n)$ to this equation with $|x_i|\leq R$ for $1\leq i \leq n$ .

The Theorem: On 20th March 2016, Alex Gamburd, Michael Magee, and Ryan Ronan submitted to arxiv a paper in which they proved that if $n\geq 3$ , $a\geq 1$ , and $k\in {\mathbb Z}$ are such that $k-n+2$ and $k-n-1$ are not perfect squares, then either $\lim\limits_{R\to\infty}M_{n,a,k}(R)<\infty$ or $M_{n,a,k}(R)=(c+o(1))(\log R)^\beta$ , where $c=c(n,a,k)>0$ and $\beta=\beta(n)>0$ are positive constants.

Short context: The generalized Markoff-Hurwitz equation contains many important and well-studied equations as special cases. For example, with $n=a=3$ and $k=0$ , it reduces to the equation $x_1^2+x_2^2+x_3^2=3x_1x_2x_3$ , whose solutions in positive integers are known as Markoff triples, and have applications in number theory, group theory, and geometry. The Theorem establishes an asymptotic formula for the number of integer solutions of this equation.

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

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A problem posed by Erdős and Szekeres in 1950 has a positive answer

You need to know: Rational and irrational numbers, complex numbers, notation $i$ for $\sqrt{-1}$ , absolute value $|z|$ of complex number $z$ , limit inferior $\liminf$ .

Background: For any real $x$ , by $e^{ix}$ we mean $\cos(x)+i\sin(x)$ .

The Theorem: On 16th February 2016, Artur Avila, Svetlana Jitomirskaya, and Christoph Marx submitted to arxiv a paper in which they proved, among other results, that for all irrational $\alpha$ , one has $\liminf\limits_{n\to\infty}\max\limits_{|z|=1}\prod\limits_{k=1}^n\left|z-e^{2\pi i k \alpha}\right|<\infty$ .

Short context: The Theorem gives a positive answer to a long-standing open problem posed by Erdős and Szekeres in 1950. Interestingly, the authors needed to solve this elementary-looking number theoretic problem on a way to deriving deep results related to spectral theory of so-called Harper’s model used in statistical mechanics.

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

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The main conjecture in Vinogradov’s Mean Value Theorem is true

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

Background: For positive integers s,k, and N, let $J_{s,k}(N)$ 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 $1\leq x_i,y_i \leq N$ for $1\leq i \leq s$ .

The Theorem: On 4th December 2015, Jean Bourgain, Ciprian Demeter, and Larry Guth submitted to arxiv and the Annals of Mathematics a paper in which they proved that for any natural numbers $s\geq 1$ and $k\geq 2$ , and any real $\epsilon>0$ , there exist a constant $C=C(s,k,\epsilon)$ such that $J_{s,k}(N) \leq C N^\epsilon(N^s+N^{2s-\frac{1}{2}k(k+1)})$ for all $N\geq 2$ .

Short context: The Theorem confirms a famous conjecture, known as the main conjecture in Vinogradov’s Mean Value Theorem. The estimate is optimal up to constant and $N^\epsilon$ factors. Before 2015, the conjecture was proved if $s \geq k(k + 1)$ , and, more recently, for $k\leq 3$ . The Theorem confirms the conjecture in general.

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

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