Regular language not accepted by DFA having at most three states
The pumping lemma can be stated to take into account the number of states in the DFA. Every language $L$ accepted by a DFA with $p$ states satisfies the following pumping lemma:


  Each word $w$ of length at least $p$ can be broken up as $w=xyz$, where $|xy| \leq p$ and $|y| \geq 1$, such that $xy^iz \in L$ for all $i \geq 0$.


You can use this characterization to prove that the language $\{0^p\}$ requires $p+1$ states.

Another method is to use the Myhill--Nerode theorem. Two words $x,y$ are inequivalent (with respect to some language $L$) if for some word $z$, either $xz \in L$ and $yz \notin L$ or the other way around. The Myhill--Nerode theorem states that if there are $p$ pairwise inequivalent words, then every DFA for $L$ has at least $p$ states. For the example $L = \{0^p\}$, you can find $p+1$ pairwise inequivalent words, namely $\epsilon,0,\ldots,0^p$.