How is L = a^2n regular if it doesn't pass the pigeon-hole principle test?

Question

I understand that this topic has been discussed, and I have reviewed numerous posts about it on stack overflow. However, my question remains unresolved. Specifically, I am seeking clarification on the use of the Pigeonhole Principle in determining the regularity of a language.

Consider the string a^{12}.

Suppose we have a DFA with 5 states and we process the string a^{12}. By the Pigeonhole Principle, the DFA will revisit a state more than once. If we remove the segment between the revisited states, this could result in an odd number of a's, for instance, a^7

The string a^7 would end in the same accepting state as a^{12}, yet a^7 does not belong to the language L={a^2n∣n≥0}.

My lecturer mentioned that the Pigeonhole Principle is used to prove that languages are non-regular. Based on this principle, I demonstrated how L fails the test, leading to my confusion. So how can L still be classified as a regular language?

This is also another similar proof that it fails:

Answer 1

The simplest DFA has two states S and O, every a switches from S to O and from O to S, and S is accepting. After 2n “a”s we are in state S, and after 2n+1 “a”s we are in the non-accepting state O.

It’s quite obvious that to move from any state to the same state we need exactly 2k “a”s. Your l2 - l1 can never be odd.