In the following, I shall denote $$\frac{1 + \sqrt{-19}}{2}$$ with $\gamma$. If we consider the ring $R = \mathbb{Z}[\gamma]$, one can show that $R$ is not a Euclidean domain. I am working through a proof that uses the fact that $\{-1, 1\} = R^\times$ (the units in $R$).
This is a result of the fact that $N(xy) = N(x) N(y)$ for all $x, y \in R$ for $N : R \to \mathbb{Z}$ defined by $$N(a + b \gamma) = a^2 + ab + 5b^2 = \left(a + \frac{b}{2}\right)^2 + \frac{19}{4}b^2.$$ I tried working out the details of this assumption (that the equality $N(xy) = N(x) N(y)$ holds for all $x, y \in R$) but I keep getting stuck on the same part.
If we let $x = a + b \gamma$ and $y = c + d\gamma$ for some $a, b, c, d \in \mathbb{Z}$, I get
$$N(xy) = a^2c^2 + a^2cd + abc^2 + 5a^2d^2 + 5b^2c^2 + \gamma^4b^2d^2 + (2\gamma^2 + 10) abcd + \gamma^2abd^2 + \gamma^2b^2cd$$
$$N(x) N(y) = a^2c^2 + a^2cd + abc^2 + 5a^2d^2 + 5b^2c^2 + 25b^2d^2 + abcd + 5abd^2 + 5b^2cd$$
which doesn't seem to correspond. This really doesn't seem that hard but time and time again I can't seem to make the equality hold. Any help with figuring out what I did wrong would be appreciated!
If you consider $R$ as a subring of $\mathbb{C}$, $N$ is just the square of the absolute value, which should make multiplicativity obvious.
I haven't checked the correctness of your expansion of $N(x)N(y)$, but $N(xy)$ looks pretty funny - you shouldn't have $\gamma$ in the expression, since it is an integer. (And it's a mystery how you even got a $\gamma^4$ in there.) If you expand $xy = (a + b\gamma)(c + d\gamma)$ using $\gamma^2 = \gamma - 5$, you should get the right thing.