Show that this ideal is not principal

Question

I've been working on a task dealing with the real quadratic number field $K = \mathbb{Q}(\sqrt{42})$, where $\mathcal{O}_K$ is its ring of integers and $p$ is a prime ideal over $2$ and $q$ a prime ideal over $3$. I was then able to show that $pq$ is a principal ideal (as $pq= (6+ \sqrt{42}) \mathcal{O}_K$) and that $p$ is not a principal ideal:

I argued that if $p$ were a principal ideal, there would be an element $a=x+y\sqrt{42} \in \mathcal{O}_K=\mathbb{Z}[\sqrt{42}]$ such that $2=N(p)=|N_{K/ \mathbb{Q}}(a)|= |x^2 -42y^2|$. Then by looking at the equation $x^2-42y^2= \pm 2$ modulo $3$ and $7$, I was able to show that there are no solutions $(x,y) \in \mathbb{Z}^2$.

Based on this approach, I wanted to show that also $q$ cannot be a principal ideal. I know that it cannnot be one as otherwise a contradiction would arise when looking at the classes as $1=[pq]=[p][q]$, $[p]=[q]^{-1}$ and $[p] \neq 1$.

It would be great if anyone could tell me how to prove that there's no solution for $ \pm 3 = x^2-42y^2$ using a modulo argument. By looking at the equation modulo $7$, I was able to reduce it to the case $x^2-42y^2=-3$ but was not able to show that this last equation has no solution in $\mathbb{Z}$.

Thank you!

Answer 1

Mod $8$ clinches it.

Suppose $x^2-42y^2=\pm 3$.

Then $x^2 - 2y^2 \equiv \pm 3 \pmod 8$.

Clearly, $x$ must be odd, hence $x^2\equiv 1 \pmod 8$.

If y is also odd, then

$$x^2 - 2y^2 \equiv \pm 3 \pmod 8 \implies -1 \equiv \pm 3 \pmod 8$$

contradiction.

Hence y must be even. But then $2y^2 \equiv 0 \pmod 8$, so

$$x^2 - 2y^2 \equiv \pm 3 \pmod 8 \implies 1 \equiv \pm 3 \pmod 8$$

contradiction.

Therefore the equation $x^2-42y^2=\pm 3$ has no integer solutions.