# Need a proofreading why all the units are satisfied $a^2-2b^2 =\pm1$ for $\mathbf{Z}$

All the units are satisfied Pell’s equation $a^2-2b^2=\pm1$ for $\mathbf{Z}[\sqrt{2}]$, $a,b\in\mathbf{Z}$.

Here is my proof:

Let $a+b\sqrt{2}$ be a unit $\in\mathbf{Z}[\sqrt{2}]$. This implies
$$(a+b\sqrt{2})(c+d\sqrt{2})=1, c+d\sqrt{2}\in\mathbf{Z}\sqrt{2}$$
$$\implies \mathrm{norm}((a+b\sqrt{2})(c+d\sqrt{2}))=1$$
$$\implies \mathrm{norm}(a+b\sqrt{2})\cdot \mathrm{norm}(c+d\sqrt{2})=1$$
$$\implies (a+b\sqrt{2})(a-b\sqrt{2})(c+d\sqrt{2})(c-d\sqrt{2})=1$$
$$\implies (a^2-2b^2)(c^2-2d^2)=1$$
$$\implies \left[ a^2-2b^2=1\text{ and } {c}^2-2{d}^2=1\right]\text{ or } \left[ a^2-2{b}^2=-1\text{ and }{c}^2-2{d}^2=-1\right]$$
$$\implies {a}^2-2{b}^2=\pm1$$