Actually a DFA cannot be drawn (at least a correct one that is). You're right that the pumping lemma gives a contradiction, it's easy to pump i up or down to make it non-square. If the pumping lemma shows that a language is not regular, then by definition there is no way to draw a DFA to represent that language.
The set of states is infinite, there must be one state to accept a^1, one for a^2, one for a^3 and so on. Then there's all the states between these accepting ones.. it gets messy pretty quickly. Anyways, if the pumping lemma shows a language is not regular (assuming you're applying it properly), then there is no DFA (or regular expression) to represent the language.