Let $A$ be a non empty finite sequence of $n$ distinct integers $a_1<a_2<...<a_n$. Define $|A+A|=\{a_i+a_j | 1 \le i,j \le n\}$ e.g. for $A=\{1,4\}$, $A+A=\{2,5,8\}$
Prove that $|A+A|=2n-1$ if and only if the sequence $A$ is an AP.
My approach is for any fixed number $k\in [2,2n], \ a_i+a_{k-i}$ is constant. So, if choose an AP it is easy to show that $A+A$ will have $2n-1$ elements. My query is how to prove the converse.
It is easy to find $2n - 1$ different elements in $A + A$: $$a_1 + a_1 < a_1 + a_2 < a_1 + a_3 < \dots < a_1 + a_n < a_2 + a_n < a_3 + a_n < \dots < a_n + a_n.$$
Thus if $|A + A| = 2n - 1$, then these must be all elements of $A + A$.
Now what happens to the other sums?
E.g. we may also create the following $2n - 1$ elements: $$a_1 + a_1 < a_1 + a_2 < a_2 + a_2 < a_2 + a_3 < \dots < a_2 + a_n < a_3 + a_n < \dots < a_n + a_n.$$