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?
$4n-2 = (a/b)^2$ so $b$ divides $a$.

But $\operatorname{gcd}(a,b) = 1$
so $b = 1$.

So now $2$ divides $a$
so write $a = 2k$
then by substitution, we get that
$2n-1 = 2k^2$

Left side is odd but the right side is even. Contradiction!
Suppose $n$ is a natural number. If $n$ is even, then $n^2 = (2k)^2 = 4k^2$ is divisible by 4. If $n$ is odd, then $n^2 = (2k-1)^2 = 4k^2-4k+1 = 1\ \textrm{mod}\ 4$.

Hence, if $n$ is a natural number, then $n^2$ is either 0 or 1 mod 4.

Thus, if $m$ is a natural number and is 2 or 3 mod 4, then $\sqrt{m}$ is not a natural number, and hence must be irrational, since $\sqrt{m}$ is either irrational or natural.

Hence, as a corollary, since $4n-2$ is 2 mod 4, then $\sqrt{4n-2}$ is irrational.

(This shows that $\sqrt{4n-3}$ is also irrational.)