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Suppose $A\subseteq \mathbb{Z}$ has positive upper density, that is $$ \limsup_{n\to \infty} \frac{|A\cap\{1,2,\dots,n\}|}{n}>0. $$ I would like to know whether the following statement for $A$ can be proven without applying Szemerédi's theorem on arithmetic progression:

For any integer $n$, there exists an integer $b$ such that $\{b, 2b,\dots, nb\}\subseteq A-A$, where $A-A=\{a_1-a_2: a_1, a_2\in A\}$.

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  • $\begingroup$ Isn't $A-A$ empty? $\endgroup$
    – Mike Stay
    Commented Jan 26, 2017 at 19:51
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    $\begingroup$ $A - A = \{x \in A : x = y - z \text{ for some } y, z \in A\}$. $\endgroup$
    – Peter Humphries
    Commented Jan 26, 2017 at 20:02
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    $\begingroup$ I would expect that it should be possible to modify appropriately the argument of Croot, Ruzsa, and Schoen (people.math.gatech.edu/~ecroot/kterm.pdf) who proved that if $A\subset[1,n]$ has size $|A|>(3n)^{1-1/(k-1)}$, then the sumset $A+A$ contains an arithmetic progression of length at least $k$. $\endgroup$
    – Seva
    Commented Jan 26, 2017 at 20:13

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