Last day on quadratic rings
$\Z[\sqrt{2}]$ with norm $N(a+b\sqrt{2})$ given by the absolute value of $a^22b^2$.

units in this ring come from $a^22b^2=\pm 1$. Compute $\pm 1+2b^2$ and look for squares. Or notice that $1+\sqrt{2}$ has norm $1$ and consider all powers.

the division algorithm for $\alpha$ and $\beta$ is found by taking \(\frac{\alpha}{\beta}=\frac{\alpha\overline{\beta}}{N(\beta)}=\frac{x}{N(\beta)}+\frac{y}{N(\beta)}\sqrt{2}\) where $x$ and $y$ are integers. Then choose the closest integers $u$ and $v$ to $x/N(\beta)$ and $y/N(\beta)$ respectively, so we can write \(\frac{x}{N(\beta)}+\frac{y}{N(\beta)}\sqrt{2}=u+v\sqrt{2} + (r+s\sqrt{2})\) where $r$ and $s$ have absolute value at most $1/2$. Therefore the norm of $r+s\sqrt{2}$ is at most $3/4$.
Multiplying the expression for $\alpha/\beta$ through by $\beta$ gives
\(\alpha=\beta(u+v\sqrt{2})+(r+s\sqrt{2})N(\beta)\) and the remainder term has norm at most $3/4N(\beta)$.
 The prime $p$ remains prime in $\Z[\sqrt{2}]$ if $x^22$ is irreducible mod $p$. This happens when 2 is a quadratic nonresidue mod $p$. The “supplement” to the law of quadratic reciprocity says this happens when $p$ is not congruent to $\pm 1$ mod $8$. So for example $7$ is not prime, it satisfies $7=(3\sqrt{2})(3+\sqrt{2})$ but $11$ is prime.
Remark: The ring $\Z[\sqrt{3}]$ is trickier; if you use the approach above you end up with $u+v\sqrt{3}$ with $u$ and $v$ at most $1/2$ in absolute value; but the norm of $1/2+\sqrt{3}/2$ is 1 which does not yield the necessary estimate, at least unless we are a bit more careful. In fact the remainder term is the absolute value of \(N(\beta)((\frac{x}{N\beta}u)^23(\frac{y}{N\beta}v)^2).\) The second term is the absolute value of the difference of two squares $A^23B^2$, which is at most the maximum of $A^2$ or $3B^2$ and is therefore at most $3/4$.
Finally we look at the imaginary ring $\Z[\rho]$ where $\rho=e^{2\pi i/3}$. Here there are finitely many units (6) and the key geometric observation is that the integer lattice is “small enough”.