Let $n\in \mathbb{Z}$, $n\geq3$ and $d=n^2-2$. I want to show that $n^2-1+n\sqrt{d}$ is the fundamental unit in $\mathbb{Z}[\sqrt{d}]$.
Substituting $n=3,4,5$ gives the elements $8+3\sqrt{7}$, $15+4\sqrt{14}$ and $24+5\sqrt{23}$ respectively, which, by inspection, are the fundamental units of $\mathbb{Z}[\sqrt{7}]$, $\mathbb{Z}[\sqrt{14}]$ and $\mathbb{Z}[\sqrt{23}]$ respectively. So by empirical observation, the statement seems to be true, at least for the first few values of $n$.
My ideas so far on how to go about proving this have been the following: We know that any unit in $\mathbb{Z}[\sqrt{d}]$ must be expressible as some power of the fundamental unit, or the additive inverse of some power of the fundamental unit, and furthermore, we know that only the fundamental unit has this property. So if we can show that any unit in $\mathbb{Z}[\sqrt{d}]$ must be expressible in the form $\pm(n^2-1+n\sqrt{d})^r$ for some $r\in \mathbb{Z}$, this would prove that $n^2-1+n\sqrt{d}$ is the fundamental unit. However, I seem to be at a loss as to how to prove this, and I suspect that there may be a simpler proof of the statement.
All help or input would, as always, be highly appreciated.
Update: In response to a highly relevant comment, I add the following: If $d$ is not square free, the statement seems to fail. $\textit{E.g.}$ if we take $n=10$. Then $d=10^2-2=98+2*7^2$, and we get the element $99+10\sqrt{98}$, but the fundamental unit in $\mathbb{Z}[\sqrt{98}]$ is demonstrably $1+\sqrt{2}$.
In fact, the fundamental unit is only defined for $\mathbb{Z}[\sqrt{d}]$, when $\mathbb{Z}[\sqrt{d}]$ is the ring of integers of some quadratic number field. This is only the case when $d$ is square-free. We observe that as $d \not\equiv 1\ (\textrm{mod}\ 4)$ for all $n\geq3$, $\mathbb{Z}[\sqrt{d}]$ does indeed constitute the ring of integers of some quadratic number field whenever $d$ is square-free.
But I do still suspect that the above statement is true for $d$ square-free.
Best Answer
You wrote "the fundamental unit in $\mathbf Z[\sqrt{98}]$ is demonstrably $1 + \sqrt{2}$" but this is wrong because $1+\sqrt{2}$ is not even an element of $\mathbf Z[\sqrt{98}] = \mathbf Z[7\sqrt{2}]$. When $A$ is a ring that is a finite free $\mathbf Z$-module (such as $\mathbf Z[\sqrt{m}]$ for all integers $m$ that are not perfect squares, even $m$ that have square factors bigger than $1$ such as $98 = 2 \cdot 49$), the unit group $A^\times$ is finitely generated and a set of multiplicatively independent generators of $A^\times$ modulo its roots of unity is called a set of fundamental units for $A$. If $A^\times$ is finite, such as when $A = \mathbf Z[i]$, there are no fundamental units.
In any case, a real quadratic ring $\mathbf Z[\sqrt{m}]$ turns out to have a unit group of the form $\pm u^\mathbf Z$ for a unique unit $u > 1$ and we call $u$ a fundamental unit for $\mathbf Z[\sqrt{m}]$. To show $n^2 - 1 + n\sqrt{n^2-2}$ is the fundamental unit of $\mathbf Z[\sqrt{n^2-2}]$ when $n \geq 3$, show it is the smallest unit greater than $1$. Start by showing no unit $a+b\sqrt{n^2-2}$ in $\mathbf Z[\sqrt{n^2-2}]$ can have $|b| < n$.