If $n$ is any positive integer, prove that $sqrt{4n-2}$ is irrational

If $n$ is any positive integer, prove that $\sqrt{4n-2}$ is irrational.

I've tried proving by contradiction but I'm stuck, here is my work so far:

Suppose that $\sqrt{4n-2}$ is rational. Then we have $\sqrt{4n-2}$ = $\frac{p}{q}$, where $p,q \in \mathbb{Z}$ and $q \neq 0$.

From $\sqrt{4n-2}$ = $\frac{p}{q}$, I just rearrange it to:

$n=\frac{p^2+2q^2}{4q^2}$. I'm having troubles from here, $n$ is obviously positive but I need to prove that it isn't an integer.

Any corrections, advice on my progress and what I should do next?

Answer

The number $\sqrt{4n-2}$ is rational iff $4n-2 = a^2$ reduction mod 4 shows that this is impossible.

Here is a proof of the general fact that $\sqrt{k}$ is irrational unless $k$ is a square: Suppose $\frac{u}{v}$ is a solution to $x^2 - k = 0$, then it is an integer $i$ by Gauss lemma, but then $k = i^2$.