= Solution
For $K=\mathbb Q(\sqrt{10})$, the <ring of integers of a quadratic field> is
$$
\mathcal O_K=\mathbb Z[\sqrt{10}].
$$
The element
$$
\varepsilon=3+\sqrt{10}
$$
has <field norm> $9-10=-1$, so it is a <unit from norm one or minus one>[unit]. We claim that
$$
\mathcal O_K^\times
=\{\mathord\pm\varepsilon^n:n\in\mathbb Z\}.
$$
Let $u$ be any unit. Its norm is $\pm1$. After changing its sign and replacing it by its inverse if necessary, its first real embedding satisfies $u\geq1$. Choose $n\in\mathbb Z$ so that
$$
1\leq v=u\varepsilon^{-n}<\varepsilon.
$$
Write $v=a+b\sqrt{10}$ with $a,b\in\mathbb Z$. Its conjugate is $v'=\pm v^{-1}$, so $|v'|\leq1$. Therefore
$$
|a|=\frac{|v+v'|}{2}<\frac{\varepsilon+1}{2}<4,
\qquad
|b|=\frac{|v-v'|}{2\sqrt{10}}
<\frac{\varepsilon+1}{2\sqrt{10}}<2.
$$
Thus $|a|\leq3$ and $|b|\leq1$. The equation $a^2-10b^2=\pm1$ now shows directly that the only value in $[1,\varepsilon)$ is $v=1$: if $b=0$ then $a=\pm1$, while if $|b|=1$ then $a=\pm3$ and the positive possibilities are either below $1$ or equal to $\varepsilon$. Hence $u=\varepsilon^n$ after normalization, proving the claimed description of the <units of Q of square root ten>.
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