Coprime of $x+\sqrt{-2}$ and $x-\sqrt{-2}$

Question

Let $A = \mathbb{Z}[\sqrt{-2}]= \{a+b\sqrt{-2} \ : a, b \in \mathbb{Z}\}$. Show that if $x, y \in \mathbb{N}$ such that $x^2+ 2 = y^3$, then $x+\sqrt{-2}$ and $x-\sqrt{-2}$ are coprime.

I think we can proceed in using the norm $N(a+b\sqrt{-2})=a^2+2b^2$. So far I know that $y^3=(x+\sqrt{-2}\,)(x-\sqrt{-2}\,)$ and $\Bbb Z[\sqrt{-2}\,]$ has a unique factorization.

Defintion : $d=\gcd(a,b)$ if $(i)$ $d | a$ and $d|b$ and $(ii)$ if $e|a$ and $e|b$, then $e|d$.

Is there a faster way to do this question instead of using the definition? Otherwise, how could I do this question in using the definition?

Answer 1

It's easy to see that $\sqrt{-2}$ is prime: if $\sqrt{-2}\mid(a+b\sqrt{-2})(c+d\sqrt{-2})$ then $2\mid(a^2+2b^2)(c^2+2d^2)$ so either $a$ or $c$ is even.

Suppose the elements are not coprime. If $d$ is a prime such that $d\mid(x+\sqrt{-2})$ and $d\mid(x-\sqrt{-2})$, then $d\mid 2\sqrt{-2}=-(\sqrt{-2})^3$, so $d=\sqrt{-2}$ (up to associates; here we use the UFD property).

Therefore $\sqrt{-2}\mid x$; from $x=(a+b\sqrt{-2})\sqrt{-2}$ we get $a=0$ and $x=-2b$, so $x$ is even. This easily leads to a contradiction.

Answer 2

Let $d|x+\sqrt{-2}$ and $d|x-\sqrt{-2}$. By adding and subtracting, we obtain $d|2x$ and $d|2\sqrt{-2}$, hence $d|\operatorname{gcd}(2x,2\sqrt{-2})=2$.

The factorization of $2$ is $2 = -1 \cdot \sqrt{-2} \cdot \sqrt{-2}$, hence we are left with the options (up to units) $d \in \{1,\sqrt{-2},2\}$.

Finish the proof by showing that $\sqrt{-2} \nmid x+\sqrt{-2}$ (of course you need $x$ odd for this, so you should show that $x$ is odd, using elementary number theory). Then $d=1$ is the only possible option left.