Confirmation of Proof: For some $v\in\mathbb{Z}$, $12v^2 - 4v + 3$ is never a square.

Question

I am trying to prove that $\sqrt{3}$ is irrational. I can use modulo, and the Fundamental Theorem of Algebra (FTOA), but I decided to approach it a little bit differently this time: I decided to use some quadratics.

It is quite long, but it is unfinished, as there is this one part I need to prove:

That for some $v\in\mathbb{Z}$, namely the set of all integers, $12v^2 - 4v + 3$ is never a square.

I have no idea how to approach this. What are some useful skills or knowledge I should know about square numbers or similar (quadratic) problems? I would really appreciate a hint, but full answers are acceptable.

Thank you in advance.

Answer 1

$$12v^2-4v+3\equiv-1\pmod4$$

But for any integer $a,$ $$a^2\equiv0,1\pmod4$$

Answer 2

$$ 12v^2-4v+3=n^2\Rightarrow v = \frac{1}{24}\left(4\pm\sqrt{16-16 \times 3(3-n^2)}\right) $$

but

$$ 1-3(3-n^2) \ge 0\Rightarrow n^2 \le \frac{8}{3}\Rightarrow n^2\le 2\Rightarrow n \in \{0, \pm 1\} $$