Proving the Language is not regular for $(a^n)^n$

Question

I am trying to solve a pumping Lemma Problem where $L= \{(a^n)^n ; n\ge 0\}$ I am having a lot of trouble with the pumping lemma and understanding using it with different languages. Here is what I have so far and would appreciate any input on it.

Assume for contradiction that $L$ is a regular language. Since $L$ is infinite we can apply the pumping Lemma. Let $p$ be the critical length for $L$.

Let $w = (a^p)^p$ and length of $|w|>=p$. From the pumping lemma we can write $w = (a^p)^p = xyz$. With $|xy| \le p$ and $|y| \ge 1$. Thus $y = a^k, 1\le k\le p$. So for $xy^iz \in L$ we can take $xy^0z = xz = (a^{p-k})^p$ which should be contained in $L$ but since it is not we have a contradiction thus the language is not regular.

Any help or recommendations is appreciated, thank you.

Answer 1

You're on the right track. Here are a few missing details. First, note that $(a^n)^n=a^{n^2}$, so you want to prove that $L=\{a^{n^2}\mid n\ge 0\}$ isn't regular.

Assume $L$ is regular. Since $L$ is infinite, there's an integer $p$ for which the PL applies, so choose $s=a^{p^2}$ and write $s=xyz$ with $|xy|\le p$ and $|y|\ge 1$. Hence, $y=a^k$ with $1\le k\le p$. Then the PL implies that $xz\in L$. Thus we know that $|xz| =p^2-k$, [You goofed here.] and since $k\le p$, we have $$ |xz| =p^2-k\ge p^2-p >p^2-2p+1=(p-1)^2 $$ and so $(p-1)^2< |xz|<p^2$ and so $xz\notin L$, since $xz$ lies strictly between two adjacent (lengthwise) elements in $L$ and so can't be in $L$, a contradiction.

[The only way this could fail is if $-p\le -2p+1$ and that only holds when $p\le 1$ which we know must be true, since $p$ is the number of states of a FA that accepts $L$ and clearly a one-state FA won't recognize $L$.]