How does this actually solve this simple proof?

I'm having a really hard time learning proofs, and cannot pick up on why this is actually proved. This isn't a homework problem, it's straight from the solutions manual (it just doesn't tell me why the proof is actually solved).

The Proof:

If $x$ is an even integer, then $x^2$ is even.

Proof. Suppose $x$ is even. Thus $x=2a$ for some $a\in\mathbb Z$.
Consequently $x^2=(2a)^2=4a^2=2(2a^2)$.
Therefore $x^2=2b$, where $b$ is the integer $2a^2$.
Thus $x^2$ is even by definition of an even number.

I'm confused on the third line. I understand that $x^2=2b$, but why do we need to know "where $b=2a^2$, and why do we need to know it? How did we even get that value?

Shoot, I think I just got it as I typed it out... What it's really saying is that $x^2=2(2a^2)$, and since $x^2=(2a)^2$ factors down to $2(2a^2)$, $b$ is a valid even integer $2a^2$ when it's value is for '$b$' in $x^2=2b$.

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I think you miss opined the proof or it is wrong. $4a^2=2(2a^2)$ is the correct step. – Thomas Andrews Jan 10 '13 at 5:43
you're right, updated. – user56763 Jan 10 '13 at 5:55
– A.Schulz Jan 10 '13 at 6:42
Many times you'll find in such "collections of examples" steps that are obvious (and that you'd probably just omit), for completeness and clarity. For homework/exams you'll have to find out what level of detail is appropiate (you´ll have read of the 100+ page proof that 2 + 2 = 4...) – vonbrand Jan 23 '13 at 2:15

To show $x^2$ even, you need to exhibit some integer $b$ with $x^2=2b$. The preceding calculation shows that the (obviously integer) expression $2a^2$ can be chosen as $b$.