Valid Alternative Proof to an Elementary Number Theory question in congruences? So, I've recently started teaching myself elementary number theory (since it does not require any specific mathematical background and it seems like a good way to keep my brain in shape until my freshman college year) using Elementary Number Theory by Underwood Dudley. I've run across the following question:

Show that every prime (except $2$ or $3$) is congruent to $1$ or $3$ (mod $4$)

Now I am aware of one method of proof, which is looking at all the residues of $p$ (mod $4$) and eliminating the improbable ones. But before that, I found another possible way of proving it.
Since $p$ must be odd: $p$  $\equiv$ $1$ (mod $2$)
We can write this as: 2$|$$(p-1)$
But since $p$ is odd we can also say: 2$|$$(p-3)$
If $a|b$ and $c|d$ then $ac|bd$, then it follows that:
$4|(p-1)(p-3)$
The $3$ possibilites then are:
$1.$ $4|(p-1)(p-3)$
$2.$ $4|(p-1)$
$3.$ $4|(p-3)$
Thus, by the definition of congruence, we have the 3 possibilites:
$1.$ $p \equiv 1$ (mod $4$)
$2.$ $p \equiv 3$ (mod $4$)
$3.$ $4|(p-1)(p-3) = 4|p^2-4p+3$ therefore $p^2-4p+3 = 4m$
Then $p^2+3 = 4m +4p$. 
Set $m+p=z$
Then from $p^2+3 = 4z$ it follows that $p^2 \equiv -3$ (mod $4$) (Is this correct?)
Can anyone please tell me if this is a valid proof? Thank you in advance.
EDIT: Taken the first possibility into account. I also realize that there are much simpler proofs but I was curious as to whether this approach also works.

I don't think there is anything wrong with your proof, but put it more simply, the congruence
$$
p \equiv 1 \pmod{2}
$$ 
is equivalent to
$$
\text{either}\quad
p \equiv 1 \pmod{4},
\quad\text{or}\quad
p \equiv 3 \pmod{4}.
$$
This is simply because if $p \equiv 0, 2 \pmod{4}$, then $p \equiv 0 \pmod{2}$. 
An then, note that the prime $3$ does fit in, $3 \equiv 3 \pmod{4}$.
