Suppose $L = \{0^k \mid \text{$k$ is composite}\}$. Prove that this language is not regular.
What bugs me in this lemma is that when I choose a string in $L$ and try to consider all cases of dividing it into three parts so that in each case it violates lemma, I always find one case that does not violate it. A bit of help will be appreciated.
Thanks in advance.
My attempt:
Suppose that $L$ is regular.
Choosing string $x = 0^{2k}$ where $k$ is prime ($2k$ pumping constant)
We can divide $x$ into three parts $u, v, w$ such that: $$|uv| \le 2k \qquad |v| > 0\qquad uv^iw \in L \text{ for $i \ge 0$}$$
If $u$ and $w$ are empty, all conditions are met.
It is the same when I change $2$ for any other number. Maybe I'm choosing wrong.
