Proving that $\mathscr L=\{0^n \big|\text{n is the square of a natural number }\}$ is non regular using the pumping lemma 
I need to prove that the language $\mathscr L=\{0^n
\big|\text{n is the square of a natural number}\}$ is non regular using the pumping lemma

My try:
$\mathscr L=\{\overbrace{\epsilon}^{0^2},\overbrace{0}^{1^2},\overbrace{0000}^{2^2},\overbrace{000000000}^{3^2},\overbrace{0000000000000000}^{4^2},\dots\}$
Suppose that $\mathscr L$ is regular so $\exists$ a word '$x$' with length of at least $n$ 
$|x|\geq n $ such that
$(1)\,\,\,|uv|\leq n$
$(2)\,\,\,|v|\geq 1$
$(3)\,\,\,uv^iw \in \mathscr L$
Now, let as choose the word $\color{blue}{x=0^{n^2}}$  $|x|\geq n$ so we can use $(1)-(3)$
$x=\underbrace{000000\dots}_{n^2\text{ times}}$
$x=uv0^kw=\underbrace{uv0^k}_{\text{n zeros   }}\overbrace{w}^{n^2-n\text{ zeros }}$
Let as choose $\color{blue}{i=??}$ 

I'm difficult choosing the word '$\color{blue}x$' and choosing  $\color{blue}i$ to come to contradiction with the lemma

 A: Start with $x=0^{n^2}$, as you did. Then we can decompose $x$ as $x=uvw$ in such a way that $|uv|\le n$, $|v|\ge 1$, and $uv^kw\in\mathscr{L}$ for all $k\ge 0$. Clearly there are $r\ge 0$ and $s\ge 1$ such that $u=0^r$, $v=0^s$, and $r+s\le n$. Thus, for each $k\ge 0$ we have
$$uv^kw=0^r0^{ks}0^{n^2-r-s}=0^{ks}0^{n^2-s}=0^{n^2+(k-1)s}\;,$$
and we need to show that there is a $k\ge 0$ such that $n^2+(k-1)s$ is not a perfect square.
For $k\ge 0$ let $a_k=n^2+(k-1)s$; the sequence $\langle a_k:k\ge 0\rangle$ is an arithmetic sequence with constant difference $s$, meaning that $a_{k+1}-a_k=s$ for all $k$. On the other hand, the gaps between consecutive squares get bigger and bigger: $(m+1)^2-m^2=2m+1$. Thus, when $k$ is large enough the terms $a_k$ and $a_{k+1}$ are too close together for both to be squares. 
Specifically, if $a_k=m^2$ for some $k$ and $m$, and $s<2m+1$, then $a_{k+1}<(m+1)^2$. Thus, if we choose any $m$ such that $2m+1>s$, and then choose $k$ so that $a_k\ge m^2$, it follows that $a_k$ and $a_{k+1}$ cannot both be squares.
