There are $N>1$ points in a plane. Consider the set of all distances between pairs of the points. Let $n$ be the number of elements of this set. We know that $$n \leq {N \choose 2}.$$ What is the lower bound estimate on $n$?
For the sake of clearity, if points are $A_1, A_2, \dots, A_N$, and $S = \{d(A_i, A_j)| 1\leq i < j \leq N \}$, then $n = |S|$.
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The abstract of this article says:
A classical problem in combinatorial geometry is that of determining the minimum number $f(n)$ of different distances determined by $n$ points in the Euclidean plane. In 1952, L. Moser proved that $f(n) > \frac{n^{2/3}}{2\cdot3^{1/3}} - 1$ and this has remained for 30 years as the best lower bound known for f(n). It is shown that $f(n) > cn^{5/7}$ for some fixed constant c.