Note: this is very similar to this problem Norm on integral domain, but I’m attempting the proof for extra practice. Idk, it might even be equivalent to the first proposition but just worded differently.
Let $d>1$ be a square free integer. Consider the set $\mathbb{Z}[\sqrt{d}]= \{a+b\sqrt{d} : a, b\in \mathbb{Z}\}$.
Prove that every element in $\mathbb{Z}[\sqrt{d}]$ can be uniquely written in the form $a+b\sqrt{d}$.
Proof. Let $x \in \mathbb{Z}[\sqrt{d}]$ where $d$ is a square free integer. Then $x = a + b \sqrt{d}$. Next, suppose that $x = n+m\sqrt{d}$ where $a \neq n$ and $b \neq m$. Then $a + b \sqrt{d} = n + m \sqrt{d}$, however, $\sqrt{d} \notin \mathbb{Z}$, so $b \sqrt{d} = m \sqrt{d}$, and since this is an integral domain, $b=m$. So, $a + b \sqrt{d} = n + b \sqrt{d}$, which means that $a = n$, so we have a contradiction. Therefore, every element in $\mathbb{Z}[\sqrt{d}]$ can be uniquely written as $a + b \sqrt{d}$.