All integral Pell solutions are $\pm {\epsilon }_D^k$

The element $\alpha$ is nonzero because $N_D(\alpha )=1$. If $\alpha =1$ or $\alpha =-1$, then already $\alpha =\pm {\epsilon }_D^0\mathrm{.}$ Assume now that $\alpha \mathrm{\ne }\pm 1$. If $\alpha >1$, then ${\alpha }^{-1}=1\mathrm{/}\alpha$ satisfies $0<{\alpha }^{-1}<1$. Writing $\alpha =x+y\sqrt{D}$, one gets $x=\frac{\alpha +{\alpha }^{-1}}{2}>0,y=\frac{\alpha -{\alpha }^{-1}}{2\sqrt{D}}>0,$ so $\alpha$ is a positive Pell solution. Hence [F2] gives $\alpha ={\epsilon }_D^k$ for a unique $k\ge 1$. If $0<\alpha <1$, then ${\alpha }^{-1}>1$, so the previous argument shows that ${\alpha }^{-1}$ is a positive Pell solution; [F1] and [F2] therefore give ${\alpha }^{-1}={\epsilon }_D^k$ for a unique $k\ge 1$, hence $\alpha ={\epsilon }_D^{-k}$. If $\alpha <0$, apply the previous positive cases to $-\alpha$, whose norm is still $1$. Therefore $\alpha =\pm {\epsilon }_D^m$ for some integer $m$.

The representation is unique. Indeed, if $\sigma {\epsilon }_D^m=\tau {\epsilon }_D^n,\sigma ,\tau \in \{\pm 1\},$ then the powers ${\epsilon }_D^m$ and ${\epsilon }_D^n$ are positive real numbers, so equality forces $\sigma =\tau$. After cancelling the common sign, suppose for contradiction that $m<n$. Then [F1] gives $1={\epsilon }_D^{n-m}\mathrm{.}$ Multiplying by ${\epsilon }_D$ yields ${\epsilon }_D={\epsilon }_D^{n-m+1}\mathrm{.}$ Both sides are representations of the positive Pell solution ${\epsilon }_D$, so [F2] forces $n-m+1=1$, impossible because $m<n$. The case $n<m$ is symmetric. Hence $m=n$.