# Consider $P\geq 3$ points in the plane. We count exactly $n$ different pairwise distances between distinct points. Prove that $P \leq (n+1)^2$

With the usual euclidean distance, i can't figure out how to solve this... I tried to draw the equality case, and then adding another point at one of the distances already in but of course couldn't find any correct placement

Assume that we have $$M=(n+1)^2+1$$ points in the plane $$P_1,P_2,\ldots,P_M$$, labeled in such a way that $$i implies that the abscissa of $$P_i$$ is less than the abscissa of $$P_j$$.
Let $$y_1,y_2,\ldots,y_M$$ the sequence of the ordinates of $$P_1,P_2,\ldots,P_M$$. By Erdos-Szekeres this sequence admits a weakly monotonic subsequence $$y_{\sigma(1)},y_{\sigma(2)},\ldots,y_{\sigma(n+2)}$$, so by considering the distances between $$P_{\sigma(1)}$$ and $$P_{\sigma(2)},\ldots,P_{\sigma(n+2)}$$ we have at least $$n+1$$ different pairwise distances.
Let $$g(N)$$ denote the minimal number of distinct distances between $$N$$ points in the plane. In his 1946 paper, Erdős proved the estimates $$\sqrt{N-3/4}-1/2\leq g(N)\leq c N/\sqrt{\log N}.$$ The lower bound is essentially the question of the OP.
After a long string of improvements, Larry Guth and Nets Katz proved in 2015 the lower bound $$g(N) \ge cN/ {\log N}$$ that almost matches Erdos' upper bound. See [1] for more details and references.