1. For a set of real numbers $A$, let $A-A$ be the set of distinct pairwise differences of $A$; that is,
$$
A-A:=\{a-b: a, b \in A\}
$$

If $|A-A|=25$, find the sum of all possible values of $|A|$.
1. For a set of real numbers $A$, let $A-A$ be the set of distinct pairwise differences of $A$; that is,
$$
A-A:=\{a-b: a, b \in A\}
$$

If $|A-A|=25$, find the sum of all possible values of $|A|$.
Answer: 76
Solution. First, we show that $6 \leq|A| \leq 13$. To do this, we show that if $|A|=n$, then $2 n-1 \leq|A-A| \leq n(n-1)+1$. The upper bound is easily attained; the number of distinct differences is bounded above by the number of ordered pairs $(a, b)$ of distinct elements of $A$ plus 1 , to account for the remaining pairs $(a, a)$ which give a difference of 0 ; this is exactly $n(n-1)+1$. For the lower bound, suppose that $A=\left\{a_{1}, a_{2}, \ldots, a_{n}\right\}$ with $a_{1}<a_{2}<\cdots<a_{n}$. Then the $n-1$ differences $a_{n}-a_{1}, a_{n}-a_{2}, \ldots, a_{n}-a_{n-1}$ are all positive and distinct and appear in $A-A$; their additive inverses are all negative and distinct and also appear in $A-A$. Also, $0 \in A-A$ and 0 is not among those differences previously listed. This gives us at least $2(n-1)+1=2 n-1$ differences in $A-A$. These bounds show us that if $|A-A|=25$, we must have $6 \leq|A| \leq 13$ as desired.
We then show that for any $n$ with $6 \leq n \leq 13$, there exists a set $A$ such that $|A|=n$ and $|A-A|=25$. Consider now the sets $B=\{1,2,4,8,12,13\}$ and $C=\{1,2, \ldots, 13\}$. Then we have $|B|=6,|C|=13, B \subseteq C$, and $B-B=C-C=\{-12,-11, \ldots, 0, \ldots, 11,12\}$. Now, for any set $A$ with $B \subseteq A \subseteq C$, we have $B-B \subseteq A-A \subseteq C-C$, and so we must have $A-A=B-B=C-C$. For any $n$ with $6 \leq n \leq 13$, it is always possible to choose $A$ such that $|A|=n$ and $B \subseteq A \subseteq C$, and this will give us the desired $A$.
Thus, the sum of all possible values of $|A|$ is $6+7+\cdots+13=76$.