Characterising $(aa)^*$ in first order logic
Short answer. There is no such first-order formula, you need a monadic second order formula.

Details. This can be proved directly using an Ehrenfeucht-FraÏssé games argument if you want to stay inside logic, but the real answer to your question is the conjunction of three results.  

[1] Büchi (1960): A language is monadic second order expressible iff it is regular.
[2] McNaughton-Pappert (1971): A language is first-order expressible iff it is star-free.
Counter-free Automata. Research Monograph 65. With an appendix by William Henneman. MIT Press. p. 48. ISBN 0-262-13076-9.
[3] Schützenberger (1965): A regular language is star-free if and only if its syntactic monoid is aperiodic.  On finite monoids having only trivial subgroups 

[3] gives an algorithm to decide whether a given regular language is star-free (and hence, first-order expressible, by [2]). Now, the syntactic monoid of $(aa)^*$ is the cyclic group of order $2$, which is not aperiodic. To get a monadic second order formula, you need first to have first order "macros" to express $\min$, $\max$ and $+1$ and then the following second order formula will do the job 
$$
  \exists X\ \forall x\ (\min \in X \wedge \max \notin X) \wedge (x \in X \leftrightarrow x+1 \notin X)
$$