Show that $L=\{a^{n^2} | n \geq 0\}$ is not regularHey guys. I'm taking a CS class and this stuff is really new to me so bear with me. I tried to look if I get some contradiction by using the pumping lemma for regular languages and I worked it out like this:Suppose $L$ is regular. Then there must be a natural number $m$ for all words $z$ in $L$ with length $|z| \geq m$ and there exists a decomposition $z = uvw, |uv| \leq m, |v| > 0$, so that $u(v^i)w$ is in the language for any $i \geq 0$.Consider the string $a^{m^2}$. Then $uv = a^{k^2} = a^{x+y}$, for some $k \leq m$ and $x = (k-1)^2$.  Then $v = a^y = a^{2k-1}$.Let $i = 2$. Then $u(v^2)w = a^{x+2y}$. But $\sqrt{x+2y}$ is not necessarily a natural number -> Contradiction! Hence, $L$ can not be regular.Well, I know that this way is unnecessarily complicated and you can prove it differently (I already know the most simple solution). But my question here is: Is my proof valid as well or does it contain any flaw? Is it formally correct? I appreciate any feedback! Thanks!
Proof that $a^{n^2}$ is not regular
regular languages;pumping lemma