The following expression can be obtained by converting the continued fraction of quadratic irrationals to single fraction. $$ \sqrt{N} = \frac{b\sqrt{N}+aN}{a\sqrt{N}+b} $$ The equation holds for any values of $a$ and $b$. However, by calculating from the regular continued fraction, $a$ and $b$ are restricted to unique pair for $N$. While investigating how to infer these $a$ and $b$ from $N$, I found the following properties under limited conditions. (not proven)
Let $N$ be expressed as follows. $$ N=p^2\pm q \ \ \ \ (p,q \in \mathbb{N}, 1 < q<p) $$ When $q$ divides $2p$, we obtain $a$ and $b$ as follows. $$ a = \frac{2p}{q},\ b=ap\pm1 $$
I would like to know why this is happening. If it is already known, please point it out. Thank you.
concrete example
The regular continued fraction of $\sqrt{7}$ is
$$ \sqrt{7} = 2 + \frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{2+\sqrt{7}}}}} $$
Return to the single fraction. $$ \sqrt{7} = \frac{8\sqrt{7}+21}{3\sqrt{7}+8} $$
$(a,b)=(3,8)$ is uniquely determined for $N=7$.
On the other hand, $7$ can be expressed as follows.
$$ 7=p^2-q=3^2-2 $$
Since $q=2$ divides $2p=6$, we can obtain $a$ and $b$ as follows.
$$ a = \frac{6}{2} = 3,\ \ b = 3\cdot3-1 = 8 $$
your matrix is the generator of the (oriented) automorphism group of the quadratic form $x^2 - N y^2.$ Given the smallest positive (nonzero) $u,v$ solving the Pell equation $u^2 - N v^2 = 1,$ your matrix is $$ A = \left( \begin{array}{cc} u & Nv \\ v & u \end{array} \right) $$ and your fraction is $$ \frac{u \sqrt N + v N}{v \sqrt N + u} $$
Without matrices, taking integers $x,y,$ we can confirm that $$ (ux + Nv y)^2 - N(vx + uy)^2 = x^2 - N y^2 $$ This identity is the generalization of "Vieta Jumping," which is the phrase used in contests for automorphisms of any form $f(x,y)=x^2 - kxy + y^2 .$ Let's see, for the general indefinite form $g(x,y) = a x^2 + bxy + c y^2$ with $ \Delta = b^2 - 4ac > 0$ but $\Delta$ not a square, there is an expression for the generating automorphism using nontrivial solutions to $ \sigma^2 - \Delta \tau^2 = 4.$ Also, some forms have automorphisms of determinant $-1;$ in your $x^2 - N y^2 $ such a thing is $(x,y) \mapsto (-x,y) $