Prove that the sum of the squares of two odd integers cannot be a perfect square.

The author wrote the follow proof:

Assume, to the contrary, that there exist odd integers $\displaystyle x$ and $\displaystyle y $such that $\displaystyle x^2 + y^2 = z^2$, where $\displaystyle z \in \mathbb{Z}$. Then $\displaystyle x = 2a+1$ and $\displaystyle y = 2b+1$, where $\displaystyle a,b \in \mathbb{Z}$. Thus

$\displaystyle x^2+y^2 = (2a+1)^2+(2b+1)^2$

$\displaystyle =4a^2+4a+1+4b^2+4b+1 $

$\displaystyle =4(a^2+1+b^2+b)+2=2[2(a^2+a+b^2+b)+1]=2s$

where $\displaystyle s = 2(a^2+a+b^2+b)+1$ is odd integer. If $\displaystyle z$ is even, then $\displaystyle z=2c$ for some integer $\displaystyle c$ and so $\displaystyle z^2 = 2(2c)$, where $\displaystyle 2c^2$ is an even integers; while $\displaystyle z$ is odd, than $\displaystyle z^2$ is odd. Produce a contradiction in each case.

Remark: I just don't see how z can be anything else be even. To me it's obvious $\displaystyle z^2=2s $ explicitly implied that $\displaystyle z^2$ is even, which by some theorem, $\displaystyle z$ is also even.

Question: Which part of the proof gave the author the reason to believe that it's odd? Further how in the case of it being even be a contradiction to $\displaystyle z^2$ being a perfect square?

I just can't see it. Could someone please show me the author's intent?