Prove the following:
If $r = a + b\sqrt{N}$ produces a solution of Pell’s equation $x^2 − Ny^2 = 1$ then so does $r^n = (a + b\sqrt{N})^n$, where $n$ is any integer.
There are three cases for $n$, which are $n>0$, $n=0$ and $n<0$. Let's consider each case in turn.
Case $n>0$
We proved this case in the previous Exercise (8.3).12.
Case $n=0$
We have
$$ r^n = (a+b\sqrt{N})^0 = 1 = 1 + 0\sqrt{N}$$
This produces the solution $(x,y)=(1,0)$ which is indeed a solution to $x^2-Ny^2=1$.
Case $n<0$
We interpret negative indices to mean the multiplicative inverse as follows
$$ r^{-m} \times r^m = 1 $$
for some natural number $m$.
Writing $r^m=a+b\sqrt{N}$ and $r^{-m}=\alpha + \beta \sqrt{N}$, we have
$$ \begin{align} 1 & = (a+b\sqrt{N}) \times (\alpha + \beta \sqrt{N}) \\ \\ & = a \alpha + a \beta \sqrt{N} + b \alpha \sqrt{N} + b \beta N^2 \\ \\ 1 + 0 \sqrt{N} & = (a \alpha + b \beta N^2) + (a \beta + b \alpha)\sqrt{N} \end{align}$$
This means
$$ a \alpha + b \beta N^2 = 1 \tag{i}$$
$$ a \beta + b \alpha = 0 \tag{ii}$$
Since $r^m$ produces a solution to Pell's equation, we also have
$$ a^2 -Nb^2 = 1 \tag{iii}$$
We aim to show that $\alpha^2 - N \beta^2 =1$.
We start by multiplying (i) by $b$,
$$ \begin{align} a \alpha b + b^2 \beta N^2 & = b \\ \\ \beta(a^2 - b^2 N^2) &= -b \tag{using (ii)} \\ \\ \beta &= - b \tag{(iv) using (iii)}\end{align} $$
We proceed by multiplying (i) by $\beta$,
$$ \begin{align} a \alpha \beta + b \beta^2 N^2 & = \beta \\ \\ b (\beta^2 N^2 - \alpha^2) & = \beta \tag{using (ii)} \\ \\ \beta ( \alpha^2 - \beta^2 N^2) & = \beta \tag{using (iv)}\\ \\ \alpha^2 - \beta^2 N^2 & = 1 \end{align} $$
And so we have shown that $r^{n}$, where $n<0$, produces a solution to $x^2 -Ny^2=1$.
By showing that all the cases $n>0, n=0, n<0$ lead to $r^n$ producing a solution to $x^2-Ny^2=1$, where $r$ produces a solution, we have shown $r^n$ does for any integer.