Proving that something is a multiplicative norm

Question

Obviously, we need to show that the 2 conditions for a multiplicative norm are satisfied for $N$. Here is the first:

If $N(\alpha)=0$, it means that $a^2+2b^2=0$ for $a, b\in\mathbb{Z}$, which is only possible when both $a=0$ and $b=0$. Thus, $N(\alpha)=0$ iff $\alpha=0$. Not too bad.

The problem I am having is with the second condition, which is to show that $N(\alpha\beta)=N(\alpha)\cdot N(\beta)$

Answer 1

$$N(a+ib\sqrt 2)N(c+id\sqrt 2)=(a^2+2b^2)(c^2+2d^2)=(ac)^2+4(bd)^2+2(bc)^2+2(ad)^2,$$ and $$N((a+ib\sqrt 2)(c+id\sqrt 2))=N((ac-2bd)+i(ad+bc)\sqrt 2)=(ac-2bd)^2+2(ad+bc)^2=N(a+ib\sqrt 2)N(c+id\sqrt 2) $$

Answer 2

This norm is the same as the square of the modulus of $a+ib\sqrt2$ as a complex number,

which is known to be multiplicative.

$|a+ib\sqrt2|^2=(a+ib\sqrt2)(a-ib\sqrt2)=a^2+2b^2=N(a+ib\sqrt2)$, and

so $N(\alpha)\cdot N(\beta)=|\alpha|^2|\beta|^2=|\alpha\beta|^2=N(\alpha)N(\beta).$