The degree of a <chromatic polynomial> is the number of vertices and the negative of its next coefficient is the number of edges. Equality with $p_{K_n}$ therefore forces $n$ vertices and $\binom n2$ edges. A simple <graph> on those vertices with that many edges contains every possible edge, so $\boxed{G\cong K_n}$.

Solution

Past exam of the graph Theory and Probabilistic Combinatorics course of the University of Oxford / 2008 / A / 2 / d / Solution

 Solution (source code)