Pumping Lemma: 1^n^2 is not regular
Your proof is correct, but you could elaborate your arguments if you wish.

We have $m\gt 0$ (typo in your question I think, you have used $\ge$) and $l+m\le p$. As $l\ge0$, $0\le p-m\implies m\le p$

$$m\gt0\implies p^2+m\gt p$$
$$\begin{align}m\le p&amp;\implies p^2+m\le p^2+p\lt p^2+2p+1=(p+1)^2\\
&amp;\implies p^2+m\lt (p+1)^2\end{align}$$

$p\lt 2p+1$ is true because $p\ge1$.

Thus, $p^2+m$ lies strictly between two consecutive squares, and hence cannot be a perfect square.