Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 2 / 20H / c / Solution

The element $ϵ=3+\sqrt{10}>1$ has norm $-1$. To prove that it is a fundamental unit, let $u=x+y\sqrt{10}$ be a positive real unit with $1<u<ϵ$. Its conjugate is $\pm 1/u$, so $y=(u-\overline{u})/(2\sqrt{10})>0$. For norm $-1$, the increasing function $u+1/u$ gives $y<(ϵ+1/ϵ)/(2\sqrt{10})=1$; for norm $+1$, $y<(ϵ-1/ϵ)/(2\sqrt{10})=3/\sqrt{10}<1$. Either contradicts integer $y$. There is no positive unit between one and $ϵ$. For any positive unit, multiply by a suitable integer power of $ϵ$ to place it in $[1,ϵ)$; it must then equal one. HenceTo 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 ${\mathfrak{p}}_2{\mathfrak{p}}_3=(ϵ+1)$ or ${\mathfrak{p}}_2{\overline{\mathfrak{p}}}_3=(ϵ-1)$. 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:Conversely every displayed element is integral, since $ϵ$ 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 $n=0,1,2,\dots$ 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.