= Solution
The element $\epsilon=3+\sqrt{10}>1$ has norm $-1$. To prove that it is a <fundamental unit>, let $u=x+y\sqrt{10}$ be a positive real unit with $1<u<\epsilon$. Its conjugate is $\pm1/u$, so $y=(u-\overline u)/(2\sqrt{10})>0$. For norm $-1$, the increasing function $u+1/u$ gives $y<(\epsilon+1/\epsilon)/(2\sqrt{10})=1$; for norm $+1$, $y<(\epsilon-1/\epsilon)/(2\sqrt{10})=3/\sqrt{10}<1$. Either contradicts integer $y$. There is no positive unit between one and $\epsilon$. For any positive unit, multiply by a suitable integer power of $\epsilon$ to place it in $[1,\epsilon)$; it must then equal one. Hence
$$
\boxed{\mathcal O_K^*=\{\pm\epsilon^n:n\in\mathbb Z\}.}
$$
To classify the norm-six elements, a principal ideal generated by such an element has norm 6. Unique prime-ideal factorization and part (b) force it to be either $\mathfrak p_2\mathfrak p_3=(\epsilon+1)$ or $\mathfrak p_2\overline{\mathfrak p}_3=(\epsilon-1)$. Therefore its generator is an associate of one of these two elements. Since their norms are respectively 6 and $-6$, the unit multiplying the first must have even exponent and the unit multiplying the second must have odd exponent. Combining the two cases gives all integer solutions:
$$
\boxed{x+y\sqrt{10}=\pm\epsilon^n(\epsilon+(-1)^n),\qquad n\in\mathbb Z.}
$$
Conversely every displayed element is integral, since $\epsilon$ is a unit, and has norm 6. This proves completeness of the <norm-six equation in the quadratic field of square root ten>.
The PDF's restriction to $n=0,1,2,\ldots$ misses conjugate solutions. In particular $(x,y)=(4,-1)$ satisfies the equation but is not in that restricted family; it is the $n=-1$ member. A nonnegative-index version becomes correct by adding the conjugates of every displayed element, with both signs. The restriction as printed cannot describe all signed integer pairs.
Back to article page
