First, let me just say that this language is regular, and I understand why. But before I understood that, I tried proving that L isn't regular with pumping lemma. I just can't figure what is wrong with this prove:
let it be N>0 from the lemma, w=$a^m=xyz$, m>N.
We will mark |y|=k, |xz|=r ($k>0$, $r≥0$).
so w=$a^{k+r}$ and we also know that $a^{nk+r}$∈L, for all n≥0.
case 1- k mod 3=1, r mod 3=0: we will choose n=3 and get a contradiction to the lemma.
case 2- k mod 3=1, r mod 3=1: we will choose n=2 and get a contradiction.
case 3- k mod 3=2, r mod 3=0: we will choose n=3 and get a contradiction.
we got contradiction for every possible case of m mod 3=(k+r) mod 3≠0, and therefore condradicted the lemma.
To see that $L$ is regular, consider the DFA $M=(Q,\Sigma, \delta, q_0, F)$ where $Q=\{0,1,2\}$, $\Sigma=\{a\}$, $q_0=0$, $F=\{1,2\}$, and transition function $$\delta(n,a) = \delta(n+1,a),$$ with $n\bmod 3$, or equivalently, the regular expression $$(a\cup aa)(aaa)^*.$$