Square Root Norm Product Identities: (a^2+db^2)(x^2+dy^2)

Recall how we defined the norm of $\mathbb{Z}[\sqrt{d}]$ to be. $N(\alpha)=N(a+b\sqrt{d})=\alpha\overline{\alpha}=a^2-db^2$. We also showed that norms were multiplicative with the following proof: $$\begin{align}N(\alpha\beta) &= \alpha\beta\overline{\alpha\beta}\\ &= \alpha\overline{\alpha}\beta\overline\beta\\ &= N(\alpha)N(\beta).\end{align}$$

For convenience, let's work in $\mathbb{Z}[\sqrt{-d}]$ so that $N(a+b\sqrt{d})=a^2+db^2$. Let $\alpha=a+b\sqrt{-d}$ and $\beta=x+y\sqrt{-d}$. Then, we know that $$\alpha\beta=(ax-bdy)+(ay+bx)\sqrt{-d}.$$ Thus, $$\begin{align} N(\alpha\beta)&=N(\alpha)N(\beta)\\ (ax-bdy)^2+d(ay+bx)^2&=(a^2+db^2)(x^2+dy^2). \end{align}$$ If we let $d=1$, then we get the well-known identity, $$(ax-by)^2+(ay+bx)^2=(a^2+b^2)(x^2+y^2).$$ This tells us that if two integers are the sum of two squares, then their product is also a sum of two squares.