$a^{2n}$ where $n\geq1$ is not Regular using Pumping Lemma I'm having trouble understanding Pumping Lemma for $RL$ , i know that $0^{2n}$ is regular and we can design a FSM for it.Here's my pumping Lemma statement

Let $p$ be the pumping length and my string is $0^{2p}$
1. $z=|0^{2p}|=&gt; 2p &gt; p $ (Satisfied)
2. $z=uvw$ such that $|uv|\geq p$, I'm choosing $|u|=n-p$ and $|v|=p$ so that $|uv|\geq p$ (Satisfied)
3.Now $u(v^i)w$, I choose $i=3$ then $|uvw|=2(n+p)$ which is even 
But if $n=2$ then $|uvw|=2n+p$, how do i know if this could be even/odd?
You can't prove that a language is regular by applying the pumping lemma.

Regular Languages are a subset of Languages that satisfy the pumping lemma. So you can only proof irregularity with the lemma. (By assuming that the language is regular and coming to a contradiction)

One way to show that a language is regular, is e.g. constructing a Regular Expression, NFA or DFA (all of these being equivalent).