I tried the following attempt :
It exists $p$ such that for all word $w\in L,|w|\ge p$. The lemma's conditions are satisfied. This is true for all $w$, therefore this is true in particular
- for $w=aaa\in L$ (But I don't know if I generalize enough, yet the first one which can generalize a formul for prime numbers is said to have found the one million dollar prize of the century...).
- for $w=a^p$, saying then that $p$ has a prime length.
Thanks to the pumping lemma we can write that
- $|y|\ge 1$
- $|xy|\le p$
- $xy^iz\in L, \forall i\ge 0$
$|xy|\le p \Rightarrow |xy|$ can be or can't be of prime number length.
It's not sure that $|xy^iz|$ has a prime number length.
Let's assume that $L$ is a regular language, then it satisfies the Puping Lemma. Let's take $w\in L$ such that $|w|$ is a Prime number.
We then take $w=xyz$ with $|xy|\le p$ and $|y|>$ (all of that to satisfy the Pumping Lemman).
Let's take $i=|xz|$ then $|xy^iz|=|xz|^2$ which isn't a prime number as far as it is the square of something. Therefore $|xy^iz|\not\in L$ for any $i$.
Yet, I don't feel like usin the pumping lemma here...
Can you help me formalize this proof ?