Solution
= Solution
Form the <graph> whose <vertices> are the points $x_i$, joining two when their <Euclidean distance> is one. Two distinct points have at most two common unit-distance neighbours, because two <unit circles> intersect in at most two points. The graph is therefore $K_{3,2}$-free. Part (c) bounds its <edges>[unordered edges] by $2n^{3/2}$, so the number of <ordered pairs>[ordered] unit-distance pairs is at most
$$
\boxed{4n^{3/2}}.
$$