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As the question asks, is it possible for a graph to have a chromatic number larger than three without it having a 4 vertice complete graph as a sub-graph?

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    In general, there's no "easy" characterization of 3-colorability. Determining whether a graph is 3-colorable is NP-complete, so no characterization can be in P (such as checking whether it contains $K_4$, which is $O(n^4)$).2012-10-31

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Here's a simple counterexample with 6 vertices. Take a 5-cycle $C_5$. Add a new vertex $A$ and connect it by an edge with each vertex of $C_5$. The resulting graph is not 3-colorable. It it were 3-colorable, then $C_5$ would be 2-colorable, which it isn't.

Wheel graph

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    yes, thank you. I think i mixed up the concept of subgraph vs subdivision.2016-02-15
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I think this is the example referred to by Jeremy Hurwitz (the "squash" corresponding to vertices 4,5,6,7)

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A non-constructive counter-example comes from a theorem of Erdős: For any $\chi$ and $g$, there exists graphs of chromatic number at least $\chi$ and girth at least $g$. If we pick $g=\chi = 4$ then there exists a graph which is not $3$-colourable and is triangle free and therefore cannot possible contain a $K_4$ sub-graph.

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Let $G$ be a square with one diagonal. Note that the opposite corners must have the same color in any 3-coloring. We can therefore use this gadget to replace any vertex of a graph without changing its 3-coloarability.

So, for example, we can take $K_4$ and replace one vertex with $G$ (connect two of the edges to one corner and the third edge to the other corner).

There's probably a smaller counter example.

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    Sorry! That should have said "square with one diagonal". I've fixed it.2012-11-01