Is this an acceptable congruency proof? I have a past exam question that I proved as follows:
$$(\forall n\in \Bbb Z)\bigl((3n^2-5\equiv 2 \pmod 4)\lor(3n^2-5\equiv 3 \pmod 4)\bigr)$$
If odd:
$$3n^2 - 7 = k4,k\in \mathbb Z$$
$$3(2l+1)^2 - 7 = k4, l\in \mathbb Z$$
$$12l^2+12l-4 = 4k$$
$$3l^2+3l-1=k$$
If even
$$3n^2 - 8 = k4$$
$$3(2l)^2 - 8 = k4, l\in \mathbb Z$$
$$12l^2 -8 = 4k$$
$$3l^2-2=k$$
$$\therefore \forall n \in \mathbb Z,   3n^2 - 5 \equiv (2\mod 4) \lor (3 \mod 4)$$
Is this a sufficient proof for this question? Any tips to improve?
Edit: Also I feel like I got lucky with the whole even or odd thing, and I don't want to rely on that in the exams, how do people determine how they are going to attack this sort of problem? Or is it a matter of lots of practice?
 A: First of all, the last line does not mean anything. You cannot put "$\vee$" between two non-propositional expressions. 
Second of all, if you did not write any comment to this proof, it's not understandable at all. I did not understand what you assumed and what you reached. But I'm assuming that you first assumed $n$ is odd, and tried to showed something that I could not understand. You assume that proposition is true, and tried to show that there is no contradiction, I suppose. Then, did the same thing with the assumption $n$ is even. But this is not the way of proving things. Showing that there is no contradiction between two propositions does not implies that they are both true. What if they're both false? Or is there a contradiction between "Today is rainy in Istanbul" and "Today is not rainy in New York". No, but when you prove that "Today is rainy in Istanbul", can you say that "Today is not rainy in New York"? Of course not. 
What you had to do was to assume that the proposition was false. That is:
$$ \neg( \forall n\in \mathbb{Z}, ((3n^2−5≡2(mod4))∨(3n^2−5≡3(mod4))) $$
$$ \implies \exists n \in \mathbb{Z}, \neg ((3n^2−5≡2(mod4))∨(3n^2−5≡3(mod4)) $$ 
Start your proof with that assumption and show that there is a contradiction. 
A: If that is the requested proof style, probably you proved the thing; but my opinion is “that's cryptography, rather than mathematics“. You probably have better to say that you're going to find $k$, assuming that $n$ is odd or even.
But before plunging in algebraic substitutions, it's better to simplify the problem. You'll probably see how the exercise was conceived.
The statement is equivalent to showing that $3n^2\equiv 2+5\pmod{4}$ or $3n^2\equiv3+5\pmod{4}$ that's the same as
$$
3n^2\equiv 3\pmod{4}\quad\text{or}\quad 3n^2\equiv0\pmod 4
$$
Since $3\cdot3=9\equiv 1\pmod{4}$, the statement becomes
$$
n^2\equiv 1\pmod{4}\quad\text{or}\quad n^2\equiv0\pmod 4
$$
Now it should be easier.
