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If a simple graph has the same chromatic polynomial as the Turán graph $T_r(n)$, it has the same number of vertices and edges, and the same chromatic number $r$. It is therefore $K_{r+1}$-free and attains equality in Turán's theorem. The equality case forces it to be isomorphic to $T_r(n)$.

The complete bipartite graph $K_{3,3}$ has vertex sets $A=\{a_1,a_2,a_3\}$ and $B=\{b_1,b_2,b_3\}$, with every edge between $A$ and $B$ and none inside either part. Its six vertices support $3\cdot 3=9$ edges and no triangles in a graph: among any three vertices, two belong to the same independent set and have no joining edge.

Add the single edge $a_1{a}_2$. Each $b_j$ is a common neighbour of its endpoints, producing the triangle in a graph $a_1{a}_2{b}_j$. These are exactly the three new triangles in a graph, because every triangle in a graph must use the only edge inside a part. The new count is ten edges; nine was the last possible count without a triangle in a graph on six vertices.

In general, every simple graph with a finite set of $n\ge 0$ vertices that is triangle-free has at mostedges. This is the $r=2$ case of Turán's theorem, with forbidden clique $K_3$. Equality holds exactly for the balanced complete bipartite graph $K_{\lfloor n/2\rfloor ,\lceil n/2\rceil }$, up to graph isomorphism; empty parts are allowed for $n<2$.^[ref]

The floor function gives the exact integer bound. Splitting $n$ vertices into two independent sets of sizes $a$ and $n-a$, and putting in every cross edge, givesThe count is largest when $a$ is an integer nearest $n/2$. For $n=2k$ it is $k^2$; for $n=2k+1$ it is $k(k+1)$. Both equal $\lfloor n^2/4\rfloor$. This proves attainability for every $n$, including the empty graph.

A short proof of the upper bound uses degrees of vertices. In a triangle-free graph, the endpoints $u,v$ of an edge have disjoint vertex neighbourhoods: a common neighbour would complete a triangle in a graph. Thus $d(u)+d(v)\le n$. Summing over edges counts $d(v)$ once for each of the $d(v)$ edges incident to $v$, givingFor $n>0$, the Cauchy-Schwarz inequality and handshaking lemma yieldIf $m>0$, division gives $m\le n^2/4$; $m=0$ is immediate. Since $m$ is an integer, apply the floor function. The case $n=0$ is immediate too. This proves the bound; the full equality characterization follows from Turán's theorem.

The useful reversal is that more than $\lfloor n^2/4\rfloor$ edges force a triangle in a graph, whatever their placement. A global count certifies a local pattern. The extremizer is bipartite, but a triangle-free graph need not be bipartite: a five-vertex graph cycle has five edges, below the bound of six. Historically, Willem Mantel posed the triangle-exclusion question in 1907.^[ref]

The chromatic polynomial $p_G(t)$ counts proper vertex colorings with $t$ labeled colors at positive integer $t$. For an edge $e$, the deletion-contraction recurrence for the chromatic polynomial is $p_G=p_{G-e}-p_{G/e}$, where parallel edges in the contracted graph are merged. Induction on edges, starting from the edgeless polynomial $t^n$, shows that the coefficients have alternating signs: the contracted graph has one fewer vertex, so subtracting its alternating polynomial reinforces the same signs. The leading coefficient stays one. The coefficient of $t^{n-1}$ decreases by one per added edge, hence it is $-m$. Thus $p_G(t)={\sum }_{i=0}^n(-1{)}^{n-i}{a}_i{t}^i$, with $a_n=1$, $a_{n-1}=m$ and all $a_i\ge 0$.

For a tree, choose a root, give it any of $t$ colors, and give every other vertex any of the $t-1$ colors different from its parent. This gives $p_G(t)=t(t-1{)}^{n-1}$. Conversely, a graph with this polynomial has $m=n-1$. The chromatic polynomial of a disjoint union is the product of those of its connected components, each divisible by $t$. The given polynomial has a simple zero at zero, forcing a single connected component. A connected graph with $n-1$ edges is a tree. The converse holds.

Finally the polynomial determines the number of vertices, the number of edges and the chromatic number. If it equals that of the Turán graph $T_r(n)$, with $1\le r\le n$, then $G$ has chromatic number $r$, so has no $K_{r+1}$, and has the maximum possible number of edges for such a graph. The equality case in Turán's theorem makes $G$ isomorphic to $T_r(n)$. This is the chromatic polynomial determines a Turán graph property. If $r>n$, the graph is simply complete and its edge count already determines it.

The Turán graph $T_r(n)$ is a complete multipartite graph with $r$ independent sets of sizes differing by at most one, with every cross edge present. Empty parts are allowed when $r>n$.^[ref]

In this example, the independent sets are $A=\{a_1,a_2,a_3\}$, $B=\{b_1,b_2,b_3\}$ and $C=\{c_1,c_2\}$. Count the edges between each pair of parts:Any four vertices put at least two in the same independent set, by the pigeonhole principle, so they cannot form $K_4$. Conversely, every choice of one vertex from each part is a triangle in a graph, giving $3\cdot 3\cdot 2=18$ triangles in a graph. Clique exclusion here does not mean triangle-free.

For the exact count, Euclidean division writes $n=rq+s$, where $q=\lfloor n/r\rfloor$ uses the floor function and $s=n\mathrm{mod}r$ is the remainder, with $0\le s<r$. There are $s$ parts of size $q+1$ and $r-s$ of size $q$. For any part sizes $a_1,\dots ,a_r$ summing to $n$, count cross edges by expanding the square:ThereforeFor $n=8,r=3$, we have $q=2,s=2$, giving $(64-2\cdot 9-1\cdot 4)/2=21$. The Quadratic Turan edge bound gives $64/3$ instead: the exact correction subtracts $1/3$.

Balancing is an integer optimization with a simple reason. If two part sizes satisfy $a\ge b+2$, moving one vertex from the larger part to the smaller decreases their contribution to the sum of squares byso it increases the edge count by $a-b-1$. For eight vertices, parts $4,2,2$ give $20$ edges; moving one vertex produces $3,3,2$ and $21$ edges. The most symmetric permitted sizes pack the most cross edges, which explains the sharp construction in Turán's theorem.