If $n^2$ is even $n$ is even I understand that there are already several answers to how to prove this question:
Prove if $n^2$ is even, then $n$ is even.
Prove that if $n^2$ is even then $n$ is even
I am trying to understand why my direct proof is wrong

Given that $a^2$ is even, prove that $a$ is even.
Proof:
We have that $a^2 = 2b$ by the definition of an even number.


*

*$a^2 = 4c^2$ by the fact that an even number times any other number results in an even number.

*$a = 2c$ by the definition of a square root.


Therefore $a$ is even.

My question: If I had validated step (1) by saying that any number squared is an even number would that have been circular logic?
Also, I think this proof is wrong because $c$ might be an irrational number say $\sqrt{2}$.
This makes me think that step (2) was wrong.
Can you please explain why this did not work out?
 A: Maybe you could try something like this if you want to use a similar approach to yours: $a$ has an unique representation in terms of factorization into prime factors: $a = p_1^{n_1} \cdots p_k^{n_k}$. Therefore $a^2 = p_1^{2n_1} \cdots p_k^{2n_k}$. Then if $a$ is even one of the $p_i = 2$.
Hope that makes sense.
A: You could not validate step (1) by saying that any number squared is an even number (e.g., $3^2 = 9$, but 3 is an odd number). As for explicitly why your proof is wrong, the statement that $a^2 = 4c^2$ is a non sequitur; that is, it does not logically follow from what you start out assuming, namely that $a^2 = 2b$. If you really want a direct proof, you could do something as follows, although it does not fall out quite nicely.
Direct proof: Suppose $a^2$ is even. Then $a^2 = 2b$, where $b\in\mathbb{Z}$. Then $\frac{a^2}{2}=b\neq 0 \rightarrow \left[\frac{a^2}{2}\right] = [0]$, where $[0]$ denotes an equivalence class representative of the even integers. This is because there are two congruence classes mod 2, namely the even and odd integers (for which $[0]$ and $[1]$ are equivalence class representatives, respectively). I doubt this is the kind of answer you were looking for though. 
Alternative direct proof: Suppose $a^2$ is even. Then let us represent $a^2$ as follows:
$$
a^2 = a\cdot a = a_1\cdot a_2,
$$
where we know $a_1 = a_2$. Well, when two integers $c$ and $d$ are multiplied together and yield an even integer, then $c$ and $d$ must both be even or one must be even and the other odd (i.e., they cannot both be odd; if you use this fact, then you may need to prove it in case you cannot take it for granted). Well, since $a_1 = a_2$ and $a_1\cdot a_2 = a^2$, we must necessarily have that $a$ is even (this due to the fact that a single number cannot be both even and odd; if $a$ is odd, then we are multiplying two odd numbers, yielding an odd number as a product--this means $a$ can only be even then). This is kind of an ugly proof, but it may be more along the lines of what you are looking for. 
A proof by contraposition seems ideal here though in terms of ease and clarity. 
