= Solution
By the <Plünnecke-Ruzsa inequality>,
$$
|4A|\leq K^4|A|.
$$
Apply the <Ruzsa covering lemma> to $3A$ and $A$. Since $A=-A$, there is a set $X\subseteq3A$ with $|X|\leq K^4$ such that
$$
3A\subseteq X+A-A=X+2A.
$$
Adding $A$ and reusing this inclusion inductively gives
$$
\boxed{mA\subseteq(m-2)X+2A\qquad(m\geq3).}
$$
A sum of $m-2$ members of the fixed set $X$ depends only on the multiplicity of each member. The number of possible multiplicity vectors is at most $m^{|X|}$, and therefore
$$
|(m-2)X|\leq m^{|X|}\leq m^{K^4}.
$$
Since $|2A|\leq K|A|\leq K^m|A|$, it follows that
$$
\boxed{|mA|\leq K^m m^{K^4}|A|.}
$$
For fixed $K$, this differs from the Plünnecke–Ruzsa bound $K^m|A|$ only by a <polynomial> factor in $m$, so both have the same leading <exponential function> factor $K^m$.
Back to article page
