Unable to understand an inequality in an application of the pumping lemma for context-free languages The problem

Prove that the language

$\qquad L = \{a^n b^j \mid n = j^2\}$ 

is not context free using pumping lemma.

Approach taken by the book

To prove such statements, the book takes the approach of a game played against an opponent who strives to fail our effort to prove that the language is not context free. The steps involved in the game are as follows:


The opponent chooses m such that for all w $\in$ L, |w| >= m
We choose the string w
The opponent decomposes w in uvxyz such that |vxy| &lt;= m and |vy| >= 1
We pump v and y i-times to get the string uv$^i$xy$^i$z. 

Now, if for any i = 0,1,2,...

uv$^i$xy$^i$z $\notin$ L

the language L is not context free.


The solution to the above problem


Opponent chooses m
We choose w = a$^{m^2}$b$^m$   
The opponent decomposes w in uvxyz as follows:


Pumping v and y i-times yields string with m$^2$+(i-1)k$_1$ a's and m+(i-1)k$_2$ b's.
If opponent takes k$_1$ $\ne$ 0 and k$_2$ $\ne$ 0, we can take i = 0, such that 

(m-k$_2$)$^2 \leq$  (m-1)$^2$             ... since k$_2\ne$ 0 making minimum value of k$_2$ is 1

= m$^2$-2m+1

&lt; m$^2$ - k$_1$    

Q. This last line I did not understand. How is -2m+1 &lt; -k$_1$? Especially because I can find the below decomposition uvxyz for which -2m+1 > -k$_1$.




I must be missing some stupid algebra here.

The solution further continues saying that the pumped resultant string does not belong to L.


Similar argument can be done if user select k$_1$ = 0 and k$_2$ $\ne$ 0  or k$_1$ $\ne$ 0 and k$_2$ = 0
The Pumping Lemma requires $|vxy| \le m$ so that the pumped parts are of bounded length. Thus $k_1+k_2\le m$. With this it is easy.

As $k_2 \neq 0$, we must have $k_1&lt;m$. Also $m&gt;0$ as the pumping constant must be positive (but in fact we can assume it to be as large as we want, of course).

Now $2m-1 \ge m &gt; k_1$.

That's all.