For any b>0 exists k>0 such that for any finite set A of integers either|kA|>|A|^b or|A^k|>|A|^b

You need to know: Notation ${\mathbb Z}$ for the set of integers, notation $|A|$ for the size of set A.

Background: For $A,B\subset {\mathbb Z}$ , denote $A+B=\{a+b, \, a\in A, b\in B\}$ and $A\times B=\{a\cdot b, \, a\in A, b\in B\}$ . For $A\subset {\mathbb Z}$ and positive integer k, denote $kA= A + A + \dots + A$ and $A^k = A \times A \times \dots \times A$ , where A in the sum and in the product is repeated k times.

The Theorem: On 3rd September 2003, Jean Bourgain and Mei-Chu Chang submitted to arxiv a paper in which they proved that for any integer $b>0$ there is an integer $k = k(b) > 0$ such that $\max(|kA|,|A^k|)\geq|A|^b$ holds for any non-empty finite set $A \subset {\mathbb Z}$ .

Short context: In 1983, Erdős and Szemerédi proved the existence of positive constants c and $\epsilon$ such that inequality $\max(|A+A|, |A \cdot A|) \geq c|A|^{1+\epsilon}$ holds for any non-empty finite set $A \subset {\mathbb Z}$ (see here for a finite field analogue of this result). The Theorem states that if we consider k-fold sums and products for sufficiently large k, then exponent $1+\epsilon$ in the Erdős-Szemerédi theorem can be replaced by an arbitrary large constant.