Problem 5

Prove that for every natural number

m

, there exists a finite set

S

of points in a plane such that every point

A

S

has exactly

m

points in

S

at unit distance from

A

Step 4 of 6: Retain the two families of expected neighbors

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|a+e^{i\theta}b-(a_i+e^{i\theta}b)|=1,\quad |a+e^{i\theta}b-(a+e^{i\theta}b_j)|=1

Detailed analysis

If $a_i$ is one of the $m$ unit neighbors of $a$ in $S$ , then $a_i+e^{i\theta}b$ is a unit neighbor of $a+e^{i\theta}b$ . If $b_j$ is one of the $n$ unit neighbors of $b$ in $T$ , then $a+e^{i\theta}b_j$ is also a unit neighbor. Distinctness makes these $m+n$ points distinct.