# Pell’s equation and side and diagonal numbers

Jordan Bell
October 30, 2016

## 1 Side and diagonal numbers

Heath [17, pp. 117–118]:

 $a_{1}=1,\qquad d_{1}=1.$
 $a_{n}=a_{n-1}+d_{n-1},\qquad d_{n}=2a_{n+1}+d_{n+1}.$
 $\displaystyle d_{n}^{2}-2a_{n}^{2}$ $\displaystyle=4a_{n-1}^{2}+4a_{n-1}d_{n-1}+d_{n-1}^{2}-2(a_{n-1}^{2}+2a_{n-1}d% _{n-1}+d_{n-1}^{2})$ $\displaystyle=4a_{n+1}^{2}+4a_{n+1}d_{n+1}+d_{n+1}^{2}-2a_{n-1}^{2}-4a_{n-1}d_% {n-1}-2d_{n-1}^{2}$ $\displaystyle=2a_{n-1}^{2}-d_{n-1}^{2}$ $\displaystyle=-(d_{n-1}^{2}-2a_{n-1}^{2}).$

As $d_{1}^{2}-2a_{1}^{2}=-1$,

 $d_{n}^{2}-2a_{n}^{2}=(-1)^{n}.$
 $\begin{pmatrix}a_{n}\\ d_{n}\end{pmatrix}=\begin{pmatrix}1&1\\ 2&1\end{pmatrix}\begin{pmatrix}a_{n-1}\\ d_{n-1}\end{pmatrix}=A\begin{pmatrix}a_{n-1}\\ d_{n-1}\end{pmatrix}.$
 $\begin{pmatrix}a_{n+1}\\ d_{n+1}\end{pmatrix}=A^{n}\begin{pmatrix}1\\ 1\end{pmatrix}.$

$A=PDP^{-1}$.

 $D=\begin{pmatrix}1-\sqrt{2}&0\\ 0&1+\sqrt{2}\end{pmatrix},\quad P=\begin{pmatrix}-\frac{1}{\sqrt{2}}&\frac{1}{% \sqrt{2}}\\ 1&1\end{pmatrix},\qquad P^{-1}=\begin{pmatrix}-\frac{1}{\sqrt{2}}&\frac{1}{2}% \\ \frac{1}{\sqrt{2}}&\frac{1}{2}\end{pmatrix}.$
 $A^{n}=\begin{pmatrix}\frac{(1-\sqrt{2})^{n}}{2}+\frac{(1+\sqrt{2})^{n}}{2}&-% \frac{(1-\sqrt{2})^{n}}{2\sqrt{2}}+\frac{(1+\sqrt{2})^{n}}{2\sqrt{2}}\\ -\frac{(1-\sqrt{2})^{n}}{\sqrt{2}}+\frac{(1+\sqrt{2})^{n}}{\sqrt{2}}&\frac{(1-% \sqrt{2})^{n}}{2}+\frac{(1+\sqrt{2})^{n}}{2}.\end{pmatrix}.$
 $\begin{pmatrix}a_{n+1}\\ d_{n+1}\end{pmatrix}=\begin{pmatrix}\frac{1}{2}\left(1-\sqrt{2}\right)^{n}-% \frac{\left(1-\sqrt{2}\right)^{n}}{2\sqrt{2}}+\frac{1}{2}\left(1+\sqrt{2}% \right)^{n}+\frac{\left(1+\sqrt{2}\right)^{n}}{2\sqrt{2}}\\ \frac{1}{2}\left(1-\sqrt{2}\right)^{n}-\frac{\left(1-\sqrt{2}\right)^{n}}{% \sqrt{2}}+\frac{1}{2}\left(1+\sqrt{2}\right)^{n}+\frac{\left(1+\sqrt{2}\right)% ^{n}}{\sqrt{2}}\end{pmatrix}.$

Thus

 $\begin{pmatrix}a_{2}\\ d_{2}\end{pmatrix}=\begin{pmatrix}2\\ 3\end{pmatrix},\qquad\begin{pmatrix}a_{3}\\ d_{3}\end{pmatrix}=\begin{pmatrix}5\\ 7\end{pmatrix},\qquad\begin{pmatrix}a_{4}\\ d_{4}\end{pmatrix}=\begin{pmatrix}12\\ 17\end{pmatrix},\qquad\begin{pmatrix}a_{5}\\ d_{5}\end{pmatrix}=\begin{pmatrix}29\\ 41\end{pmatrix}.$

## 2 Diophantus

If $x^{2}-Ay^{2}=1$ and $y=m(x+1)$ for rational $m$, then $y^{2}=m^{2}(x^{2}+2x+1)$, then $x^{2}-Am^{2}(x^{2}+2x+1)=1$, then $(Am^{2}-1)x^{2}+2Am^{2}x+Am^{2}+1=0$. Write $(x+1)(px+q)=(Am^{2}-1)x^{2}+2Am^{2}x+Am^{2}+1$. Then $px^{2}+(p+q)x+q=(Am^{2}-1)x^{2}+2Am^{2}x+Am^{2}+1$. Then $p=Am^{2}-1$, $q=Am^{2}+1$, and so $p+q=Am^{2}-1+Am^{2}+1=2Am^{2}$. Thus if $x\neq-1$ then $px+q=0$ and hence $(Am^{2}-1)x+Am^{2}+1=0$. Hence $(Am^{2}-1)x=-(Am^{2}+1)$ and so $x=-\frac{Am^{2}+1}{Am^{2}-1}$. Thus

 $y=m(x+1)=m\left(-\frac{Am^{2}+1}{Am^{2}-1}+1\right)\frac{m}{Am^{2}-1}(-Am^{2}-% 1+Am^{2}-1)=\frac{-2m}{Am^{2}-1}.$

Therefore for rational $m$,

 $x=-\frac{Am^{2}+1}{Am^{2}-1},\qquad y=\frac{-2m}{Am^{2}-1}$

satisfy $x^{2}-Ay^{2}=1$. cf. Heath [17, pp. 68–69], Nesselmann [24, p. 331].

Diophantus V.11: $30x^{2}+1=y^{2}$. Say $y=5x+1$. $y^{2}=25x^{2}+10x+1$. Then $5x^{2}-10x=0$, so $x=0$ or $5x-10=0$, i.e. $x=0$ or $x=2$. Hence $x=0,y=1$ and $x=2,y=11$ satisfy $30x^{2}+1=y^{2}$.

Diophantus V.14 [17, pp. 211–212]. $34y^{2}+1=x^{2}$. Say $x=6y-1$. $x^{2}=36y^{2}-12y+1$. Then $2y^{2}-12y=0$, i.e. $y(y-6)=0$ so $y=0$ or $y=6$. Then $x=-1,y=0$ and $x=35,y=6$ satisfy $34y^{2}+1=x^{2}$.

## 3 Fermat

Fermat, February 1657 [31, p. 29]:

Given any number not a square, then there are an infinite number of squares which, when multiplied by the given number, make a square when unity is added.
Example. Given 3, a nonsquare number; this number multiplied by the square number 1, and 1 being added, produces 4, which is a square.
Moreover, the same 3 multiplied by the square 16, with 1 added makes 49, which is a square.
And instead of 1 and 16, an infinite number of squares may be found showing the same property; I demand, however, a general rule, any number being given which is not a square.
It is sought, for example, to find a square which when multiplied into 149, 109, 433, etc., becomes a square when unity is added.

## 4 Wallis

Wallis [1, p. 546].

Stedall [29]

## 5 Brouncker

Weil [32, pp. 92–99].

## 6 Ozanam

Ozanam [25, pp. 503–516], Liv. III, Quest. XXVI.

## 7 Continued fractions

Let

 $[a_{0},a_{1}]=a_{0}+\frac{1}{a_{1}}$

and

 $[a_{0},\ldots,a_{n-1},a_{n}]=\left[a_{0},\ldots,a_{n-2},a_{n-1}+\frac{1}{a_{n}% }\right].$

Then for $1\leq m,

 $[a_{0},\ldots,a_{n}]=[a_{0},\ldots,a_{m-1},[a_{m},\ldots,a_{n}]].$

Define

 $p_{0}=a_{0},\qquad q_{0}=1,\qquad p_{1}=a_{1}a_{0}+1,\qquad q_{1}=a_{1}$

and for $n\geq 2$,

 $p_{n}=a_{n}p_{n-1}+p_{n-2},\qquad q_{n}=a_{n}q_{n-1}+q_{n-2}.$

Hardy and Wright [13, p. 130, Theorem 149]. For $n\geq 0$,

 $[a_{0},\ldots,a_{n}]=\frac{p_{n}}{q_{n}}.$

For $n\geq 1$,

 $p_{n}q_{n-1}-p_{n-1}q_{n}=(-1)^{n-1}.$

For $n\geq 2$,

 $p_{n}q_{n-2}-p_{n-2}q_{n}=(-1)^{n}a_{n}.$

For $n\geq 0$ let

 $a_{n}^{\prime}=[a_{n},a_{n+1},\ldots].$

For $x=[a_{0},a_{1},\ldots]$,

 $x=\frac{a_{n}^{\prime}p_{n-1}+p_{n-2}}{a_{n}^{\prime}q_{n-1}+q_{n-2}},\qquad n% \geq 2.$

Hardy and Wright [13, p. 144, Theorem 176]. A continued fraction $[a_{0},a_{1},\ldots]$ is said to be periodic if there is some $L\geq 0$ and some $k\geq 1$ such that $a_{l+k}=a_{l}$ for all $l\geq L$.

###### Theorem 1.

If $x=[a_{0},a_{1},\ldots]$ is a periodic continued fraction, then $x$ is a quadratic surd.

###### Proof.

Let

 $\overline{a_{L},\ldots,a_{L+k-1}}=[a_{L},a_{L+1},\ldots]=a_{L}^{\prime}.$

Thus

 $\displaystyle[a_{0},\ldots,a_{L-1},a_{L},a_{L+1},\ldots]$ $\displaystyle=[a_{0},\ldots,a_{L-1},a_{L}^{\prime}]$ $\displaystyle=[a_{0},\ldots,a_{L-1},\overline{a_{L},\ldots,a_{L+k-1}}].$

As $a_{L+k}=a_{L},a_{L+k+1}=a_{L+1},\ldots$,

 $\displaystyle a_{L}^{\prime}$ $\displaystyle=[a_{L},a_{L+1},\ldots]$ $\displaystyle=[a_{L},a_{L+1},\ldots,a_{L+k-1},a_{L},a_{L+1},\ldots]$ $\displaystyle=[a_{L},a_{L+1},\ldots,a_{L+k-1},a_{L}^{\prime}].$

Let

 $\frac{p^{\prime}}{q^{\prime}}=[a_{L},a_{L+1},\ldots,a_{L+k-1}],\qquad\frac{p^{% \prime\prime}}{q^{\prime\prime}}=[a_{L},a_{L+1},\ldots,a_{L+k-2}].$

For $t=[a_{L},a_{L+1},\ldots,a_{L+k-1},a_{L}^{\prime}]$,

 $t=\frac{a_{L}^{\prime}p^{\prime}+p^{\prime\prime}}{a_{L}^{\prime}q^{\prime}+q^% {\prime\prime}}=\frac{p^{\prime}t+p^{\prime\prime}}{q^{\prime}t+q^{\prime% \prime}}.$

Hence $q^{\prime}t^{2}+q^{\prime\prime}t=p^{\prime}t+p^{\prime\prime}$, so $q^{\prime}t^{2}+(q^{\prime\prime}-p^{\prime})t-p^{\prime\prime}=0$. For $x=[a_{0},a_{1},\ldots]$,

 $x=\frac{a_{L}^{\prime}p_{L-1}+p_{L-2}}{a_{L}^{\prime}q_{L-1}+q_{L-2}}.$

Then

 $a_{L}^{\prime}=\frac{p_{L-2}-q_{L-2}x}{q_{L-1}x-p_{L-1}}.$

Thus, with $t=a_{L}^{\prime}$,

 $q^{\prime}\left(\frac{p_{L-2}-q_{L-2}x}{q_{L-1}x-p_{L-1}}\right)^{2}+(q^{% \prime\prime}-p^{\prime})\frac{p_{L-2}-q_{L-2}x}{q_{L-1}x-p_{L-1}}-p^{\prime% \prime}=0.$

Then

 $q^{\prime}(p_{L-2}-q_{L-2}x)^{2}+(q^{\prime\prime}-p^{\prime})(p_{L-2}-q_{L-2}% x)(q_{L-1}x-p_{L-1})-p^{\prime\prime}(q_{L-1}x-p_{L-1})^{2}=0.$

Therefore there are integers $a,b,c$ such that

 $ax^{2}+bx+c=0.$

This means that $x$ is a quadratic surd, as $x$ is irrational. ∎

Example. Say $x=[3,2,7,4,\overline{5,1,12}]$. $L=4,k=3$.

 $\frac{p^{\prime}}{q^{\prime}}=[5,1,12]=\frac{77}{13},\qquad\frac{p^{\prime% \prime}}{q^{\prime\prime}}=[5,1]=\frac{6}{1}.$
 $\frac{p_{L-1}}{q_{L-1}}=\frac{p_{3}}{q_{3}}=[3,2,7,4]=\frac{215}{62},\qquad% \frac{p_{L-2}}{q_{L-2}}=\frac{p_{2}}{q_{2}}=[3,2,7]=\frac{52}{15}.$

Then

 $\displaystyle q^{\prime}(p_{L-2}-q_{L-2}x)^{2}+(q^{\prime\prime}-p^{\prime})(p% _{L-2}-q_{L-2}x)(q_{L-1}x-p_{L-1})-p^{\prime\prime}(q_{L-1}x-p_{L-1})^{2}$ $\displaystyle=$ $\displaystyle 13(52-15x)^{2}+(1-77)(52-15x)(62x-215)-6(62x-215)^{2}$ $\displaystyle=$ $\displaystyle 50541x^{2}-350444x+607482.$

Hence $x=[3,2,7,4,\overline{5,1,12}]$ satisfies

 $50541x^{2}-350444x+607482=0.$

In fact,

 $x=\frac{175222+\sqrt{1522}}{50541}.$

Hardy and Wright [13, p. 144, Theorem 177].

###### Theorem 2.

If $x$ is a quadratic surd, then the continued fraction of $x$ is periodic.

Example. Say $x^{2}=218$. $14^{2}=196$.

 $\sqrt{218}=14+\sqrt{218}-14=14+\cfrac{1}{\cfrac{1}{\sqrt{218}-14}}.$
 $(\sqrt{218}-14)(\sqrt{218}+14)=218-196=22,\qquad\frac{1}{\sqrt{218}-14}=\frac{% \sqrt{218}+14}{22}.$

We do not need to compute the decimal expansion of $\sqrt{218}$; we merely have to calculate $\lfloor\frac{\sqrt{218}+14}{22}\rfloor$. Using $14<\sqrt{218}<15$,

 $\frac{1}{\sqrt{218}-14}=1+\frac{\sqrt{218}+14}{22}-1=1+\frac{\sqrt{218}-8}{22}.$

Then

 $\sqrt{218}=14+\cfrac{1}{1+\cfrac{\sqrt{218}-8}{22}}=14+\cfrac{1}{1+\cfrac{1}{% \cfrac{22}{\sqrt{218}-8}}}$
 $(\sqrt{218}-8)(\sqrt{218}+8)=218-64=154,\qquad\frac{1}{\sqrt{218}-8}=\frac{% \sqrt{218}+8}{154}.$

Then

 $\frac{22}{\sqrt{218}-8}=\frac{22\sqrt{218}+176}{154}.$

Using that $14<\sqrt{218}<15$,

 $\frac{22}{\sqrt{218}-8}=3+\frac{22\sqrt{218}+176}{154}-3=3+\cfrac{22\sqrt{218}% -286}{154}.$

Then

 $\sqrt{218}=14+\cfrac{1}{1+\cfrac{1}{3+\cfrac{22\sqrt{218}-286}{154}}}=14+% \cfrac{1}{1+\cfrac{1}{3+\cfrac{1}{\cfrac{154}{22\sqrt{218}-286}}}}.$
 $(22\sqrt{218}-286)(22\sqrt{218}+286)=22^{2}\cdot 218-286^{2}=23716,$
 $\cfrac{1}{22\sqrt{218}-286}=\frac{22\sqrt{218}+286}{23716}.$
 $\cfrac{154}{22\sqrt{218}-286}=\frac{3388\sqrt{218}+44044}{23716}.$

Using $14<\sqrt{218}<15$,

 $\cfrac{154}{22\sqrt{218}-286}=3+\frac{3388\sqrt{218}+44044}{23716}-3=3+\cfrac{% 3388\sqrt{218}-27104}{23716}.$

Then

 $\sqrt{218}=14+\cfrac{1}{1+\cfrac{1}{3+\cfrac{1}{3+\cfrac{3388\sqrt{218}-27104}% {23716}}}}=14+\cfrac{1}{1+\cfrac{1}{3+\cfrac{1}{3+\cfrac{1}{\cfrac{23716}{3388% \sqrt{218}-27104}}}}}.$
 $(3388\sqrt{218}-27104)(3388\sqrt{218}+27104)=3388^{2}\cdot 218-27104^{2}=17676% 95776,$
 $\cfrac{1}{3388\sqrt{218}-27104}=\cfrac{3388\sqrt{218}+27104}{1767695776}.$
 $\cfrac{23716}{3388\sqrt{218}-27104}=23716\cdot\cfrac{3388\sqrt{218}+27104}{176% 7695776}.$

Using $14<\sqrt{218}<15$,

 $\displaystyle\cfrac{23716}{3388\sqrt{218}-27104}$ $\displaystyle=1+23716\cdot\cfrac{3388\sqrt{218}+27104}{1767695776}-1$ $\displaystyle=1+\cfrac{80349808\sqrt{218}-1124897312}{1767695776}.$

Then

 $\displaystyle\sqrt{218}$ $\displaystyle=14+\cfrac{1}{1+\cfrac{1}{3+\cfrac{1}{3+\cfrac{1}{1+\cfrac{803498% 08\sqrt{218}-1124897312}{1767695776}}}}}$ $\displaystyle=14+\cfrac{1}{1+\cfrac{1}{3+\cfrac{1}{3+\cfrac{1}{1+\cfrac{1}{% \cfrac{1767695776}{80349808\sqrt{218}-1124897312}}}}}}.$
 $(80349808\sqrt{218}-1124897312)(80349808\sqrt{218}+1124897312)=142034016204011% 008.$
 $\cfrac{1767695776}{80349808\sqrt{218}-1124897312}=1767695776\cdot\cfrac{803498% 08\sqrt{218}+1124897312}{142034016204011008}.$

Using $14<\sqrt{218}<15$, the floor of the above quantity is 28. Hence

 $\displaystyle\cfrac{1767695776}{80349808\sqrt{218}-1124897312}$ $\displaystyle=28+\cfrac{142034016204011008\sqrt{218}-1988476226856154112}{1420% 34016204011008}$ $\displaystyle=28+\sqrt{218}-14.$

Then

 $\displaystyle\sqrt{218}$ $\displaystyle=14+\cfrac{1}{1+\cfrac{1}{3+\cfrac{1}{3+\cfrac{1}{1+\cfrac{1}{28+% \sqrt{218}-14}}}}}$

Thus for $x=\sqrt{218}$,

 $x-14=\cfrac{1}{1+\cfrac{1}{3+\cfrac{1}{3+\cfrac{1}{1+\cfrac{1}{28+x-14}}}}}.$

Thus for $t=x-14$,

 $t=\cfrac{1}{1+\cfrac{1}{3+\cfrac{1}{3+\cfrac{1}{1+\cfrac{1}{28+t}}}}}.$

Therefore $t=[0,\overline{1,3,3,1,28}]$. Hence $x=14+t=[14,\overline{1,3,3,1,28}]$:

 $\sqrt{218}=[14,\overline{1,3,3,1,28}].$

## 8 Euler

Euler, Algebra [10], Part II, Chapter VII.

## 9 Lagrange

Konen [20, pp. 75–77].

## 10 Chakravala

Hankel [12, pp. 200–203]

Strachey [30, pp. 36–53]. Dickson [5, pp. 349–350].

Colebrooke [3, pp. 170–184]

Colebrooke [3, pp. 363–372]

Datta and Singh [4, II, pp. 93–99]

Datta and Singh [4, II, pp. 146–161]

Datta and Singh [4, II, pp. 161–172]

Suppose that $p_{n},q_{n}$ are relatively prime and

 $Aq_{n}^{2}+s_{n}=p_{n}^{2}.$

If $d$ is a common factor of $q_{n}$ and $s_{n}$ then $d\mid p_{n}^{2}$, so $d$ is a common factor of $p_{n}^{2}$ and $q_{n}^{2}$, which implies that $p_{n}$ and $q_{n}$ have a common factor, a contradiction. Therefore $q_{n}$ and $s_{n}$ are relatively prime. Because $q_{n}$ and $s_{n}$ are relatively prime, by the Kuttaka algorithm there are some $\rho_{n},\rho_{n}^{\prime}$ satisfying $-q_{n}\rho_{n}+s_{n}\rho_{n}^{\prime}=p_{n}$. For $r_{n}=\rho_{n}+k_{n}s_{n}$, $r_{n}^{\prime}=\rho_{n}^{\prime}+k_{n}q_{n}$,

 $\displaystyle-q_{n}r_{n}+s_{n}r_{n}^{\prime}$ $\displaystyle=-q_{n}(\rho_{n}+k_{n}s_{n})+s_{n}(\rho_{n}^{\prime}+k_{n}q_{n})$ $\displaystyle=-q_{n}\rho_{n}-k_{n}q_{n}s_{n}+s_{n}\rho_{n}^{\prime}+k_{n}q_{n}% s_{n}$ $\displaystyle=-q_{n}\rho_{n}+s_{n}\rho_{n}^{\prime}$ $\displaystyle=p_{n}.$

Take $r_{n}<\sqrt{A}. $r_{n}^{\prime}=\frac{p_{n}+q_{n}r_{n}}{s_{n}}$. Let

 $q_{n+1}=r_{n}^{\prime},\qquad p_{n+1}=\frac{p_{n}q_{n+1}-1}{q_{n}},\qquad s_{n% +1}=p_{n+1}^{2}-Aq_{n+1}^{2}.$

Example. $69y^{2}+1=x^{2}$. $A=69$.

$Aq_{0}^{2}+s_{0}=p_{0}^{2}$: $p_{0}=8,q_{0}=1,s_{0}=-5$.

$p_{0}+q_{0}\rho_{0}=\rho_{0}^{\prime}s_{0}$ is equivalent to $8+\rho_{0}=-5\rho_{0}^{\prime}$. It is satisfied by $\rho_{0}=-8,\rho_{0}^{\prime}=0$. Take $r_{0}=-8-5k_{0}=7$. $r_{0}^{\prime}=\frac{p_{0}+q_{0}r_{0}}{s_{0}}=\frac{8+1\cdot 7}{-5}=-3$.

$q_{1}=-3$.

 $p_{1}=\frac{p_{0}q_{1}-1}{q_{0}}=\frac{8\cdot-3-1}{1}=-25.$
 $s_{1}=p_{1}^{2}-Aq_{1}^{2}=4.$

$p_{1}+q_{1}\rho_{1}=\rho_{1}^{\prime}s_{1}$ is equivalent to $-25-3\rho_{1}=4\rho_{1}^{\prime}$. This is satisfied by $\rho_{1}=1,\rho_{1}^{\prime}=-7$. Take $r_{1}=1+4k_{1}=5$. Then $r_{1}^{\prime}=\frac{p_{1}+q_{1}r_{1}}{s_{1}}=\frac{-25-3\cdot 5}{4}=-10$.

$q_{2}=-10$.

 $p_{2}=\frac{p_{1}q_{2}-1}{q_{1}}=\frac{-25\cdot-10-1}{-3}=-83.$
 $s_{2}=p_{2}^{2}-Aq_{2}^{2}=-11.$

$p_{2}+q_{2}\rho_{2}=\rho_{2}^{\prime}s_{2}$ is equivalent to $-83-10\rho_{2}=-11\rho_{2}^{\prime}$. This is satisfied by $\rho_{2}=6,\rho_{2}^{\prime}=13$. Take $r_{2}=6-11k_{2}=6$. Then $r_{2}^{\prime}=\frac{p_{2}+q_{2}r_{2}}{s_{2}}=\frac{-83-10\cdot 6}{-11}=13$.

$q_{3}=13$.

 $p_{3}=\frac{p_{2}q_{3}-1}{q_{2}}=\frac{-83\cdot 13-1}{-10}=108.$
 $s_{3}=p_{3}^{2}-Aq_{3}^{2}=3.$

$p_{3}+q_{3}\rho_{3}=\rho_{3}^{\prime}s_{3}$ is equivalent to $108+13\rho_{3}=3\rho_{3}^{\prime}$. This is satisfied by $\rho_{3}=0,\rho_{3}^{\prime}=36$. Take $r_{3}=36+3k_{3}=6$. Then $r_{3}^{\prime}=\frac{p_{3}+q_{3}r_{3}}{s_{3}}=\frac{108+13\cdot 6}{3}=62$.

$q_{4}=62$.

 $p_{4}=\frac{p_{3}q_{4}-1}{q_{3}}=\frac{108\cdot 62-1}{13}=515.$
 $s_{4}=p_{4}^{2}-Aq_{4}^{2}=-11.$

$p_{4}+q_{4}\rho_{4}=\rho_{4}^{\prime}s_{4}$ is equivalent to $515+62\rho_{4}=-11\rho_{4}^{\prime}$. This is satisfied by $\rho_{4}=5,\rho_{4}^{\prime}=-75$. Take $r_{4}=5-11k_{4}=5$. Then $r_{4}^{\prime}=\frac{p_{4}+q_{4}r_{4}}{s_{4}}=\frac{515+62\cdot 5}{-11}=-75$.

$q_{5}=-75$.

 $p_{5}=\frac{p_{4}q_{5}-1}{q_{4}}=\frac{515\cdot-75-1}{62}=-623.$
 $s_{5}=p_{5}^{2}-Aq_{5}^{2}=4.$

$p_{5}+q_{5}\rho_{5}=\rho_{5}^{\prime}s_{5}$ is equivalent to $-623-75\rho_{5}=4\rho_{5}^{\prime}$. This is satisfied by $\rho_{5}=3,\rho_{5}^{\prime}=-212$. Take $r_{5}=3+4k_{5}=7$. Then $r_{5}^{\prime}=\frac{p_{5}+q_{5}r_{5}}{s_{5}}=\frac{-623-75\cdot 7}{4}=-287$.

$q_{6}=-287$.

 $p_{6}=\frac{p_{5}q_{6}-1}{q_{5}}=\frac{-623\cdot-287-1}{-75}=-2384.$
 $s_{6}=p_{6}^{2}-Aq_{6}^{2}=-5.$

$p_{6}+q_{6}\rho_{6}=\rho_{6}^{\prime}s_{6}$ is equivalent to $-2384-287\rho_{6}=-5\rho_{6}^{\prime}$. This is satisfied by $\rho_{6}=3,\rho_{6}^{\prime}=649$. Take $r_{6}=3-5k=8$. Then $r_{6}^{\prime}=\frac{p_{6}+q_{6}r_{6}}{s_{6}}=\frac{-2384-287\cdot 8}{-5}=936$.

$q_{7}=936$.

 $p_{7}=\frac{p_{6}q_{7}-1}{q_{6}}=\frac{-2384\cdot 936-1}{-287}=7775.$
 $s_{7}=p_{7}^{2}-Aq_{7}^{2}=1.$

Therefore

 $7775^{2}-69\cdot 936^{2}=1.$

Thus $\sqrt{69}\sim\frac{7775}{936}$.

Example. $91y^{2}+1=x^{2}$. $A=91$.

$Aq_{0}^{2}+s_{0}=p_{0}^{2}$: $p_{0}=10$, $q_{0}=1$, $s_{0}=9$.

$p_{0}+q_{0}\rho_{0}=\rho_{0}^{\prime}s_{0}$ is equivalent to $10+\rho_{0}=9\rho_{0}^{\prime}$. This is satisfied by $\rho_{0}=-10$, $\rho_{0}^{\prime}=0$. Take $r_{0}=-10+9k_{0}=8$. Then $r_{0}^{\prime}=\frac{p_{0}+q_{0}r_{0}}{s_{0}}=\frac{10+1\cdot 8}{9}=2$.

$q_{1}=2$.

 $p_{1}=\frac{p_{0}q_{1}-1}{q_{0}}=\frac{10\cdot 2-1}{1}=19.$
 $s_{1}=p_{1}^{2}-Aq_{1}^{2}=-3.$

$p_{1}+q_{1}\rho_{1}=\rho_{1}^{\prime}s_{1}$ is equivalent with $19+2\rho_{1}=-3\rho_{1}^{\prime}$. This is satisfied by $\rho_{1}=1,\rho_{1}^{\prime}=-7$. Take $r_{1}=1-3k_{1}=7$. Then $r_{1}^{\prime}=\frac{p_{1}+q_{1}r_{1}}{s_{1}}=\frac{19+2\cdot 7}{-3}=-11$.

$q_{2}=-11$.

 $p_{2}=\frac{p_{1}q_{2}-1}{q_{1}}=\frac{19\cdot-11-1}{2}=-105.$
 $s_{2}=p_{2}^{2}-Aq_{2}^{2}=14.$

$p_{2}+q_{2}\rho_{2}=\rho_{2}^{\prime}s_{2}$ is equivalent with $-105-11\rho_{2}=14\rho_{2}^{\prime}$. This is satisfied by $\rho_{2}=7,\rho_{2}^{\prime}=-13$. Take $r_{2}=7,r_{2}^{\prime}=-13$.

$q_{3}=-13$.

 $p_{3}=\frac{p_{2}q_{3}-1}{q_{2}}=\frac{-105\cdot-13-1}{-11}=-124.$
 $s_{3}=p_{3}^{2}-Aq_{3}^{2}=-3.$

$p_{3}+q_{3}\rho_{3}=\rho_{3}^{\prime}s_{3}$ is equivalent with $-124-13\rho_{3}=-3\rho_{3}^{\prime}$. This is satisfied by $\rho_{3}=2,\rho_{3}^{\prime}=50$. Take $r_{3}=2-3k_{3}=8$. Then $r_{3}^{\prime}=\frac{p_{3}+q_{3}r_{3}}{s_{3}}=\frac{-124-13\cdot 8}{-3}=76$.

$q_{4}=76$.

 $p_{4}=\frac{p_{3}q_{4}-1}{q_{3}}=\frac{-124\cdot 76-1}{-13}=725.$
 $s_{4}=p_{4}^{2}-Aq_{4}^{2}=9.$

$p_{4}+q_{4}\rho_{4}=\rho_{4}^{\prime}s_{4}$ is equivalent with $725+76\rho_{4}=9\rho_{4}^{\prime}$. This is satisfied by $\rho_{4}=1,\rho_{4}^{\prime}=89$. Take $r_{4}=1$, $r_{4}^{\prime}=89$.

$q_{5}=89$.

 $p_{5}=\frac{p_{4}q_{5}-1}{q_{4}}=\frac{725\cdot 89-1}{76}=849.$
 $s_{5}=p_{5}^{2}-Aq_{5}^{2}=-10.$

$p_{5}+q_{5}\rho_{5}=\rho_{5}^{\prime}s_{5}$ is equivalent with $849+89\rho_{5}=-10\rho_{5}^{\prime}$. This is satisfied by $\rho_{5}=9,\rho_{5}^{\prime}=-165$. Take $r_{5}=9,r_{5}^{\prime}=-165$.

$q_{6}=-165$.

 $p_{6}=\frac{p_{5}q_{6}-1}{q_{5}}=\frac{849\cdot-165-1}{89}=-1574.$
 $s_{6}=p_{6}^{2}-Aq_{6}^{2}=1.$

Therefore

 $1574^{2}-91\cdot 165^{2}=1.$

Thus $\sqrt{91}\sim\frac{1574}{165}$.

Example. $109y^{2}+1=x^{2}$. $A=109$.

$Ay_{0}^{2}+s_{0}=x_{0}^{2}$: $x_{0}=10$, $y_{0}=1$, $s_{0}=-9$.

$x_{0}+y_{0}\rho_{0}=s_{0}\rho_{0}^{\prime}$ is equivalent to $10+\rho_{0}=-9\rho_{0}^{\prime}$. This is satisfied by $\rho_{0}=-10$, $\rho_{0}^{\prime}=0$. Take $r_{0}=-10+9k_{0}=8$. Then $r_{0}^{\prime}=\frac{x_{0}+y_{0}r_{0}}{s_{0}}=\frac{10+1\cdot 8}{-9}=-2$.

$y_{1}=-2$.

 $x_{1}=\frac{x_{0}y_{1}-1}{y_{0}}=\frac{10\cdot-2-1}{1}=-21.$
 $s_{1}=x_{1}^{2}-Ay_{1}^{2}=5.$

$x_{1}+y_{1}\rho_{1}=\rho_{1}^{\prime}s_{1}$ is equivalent with $-21-2\rho_{1}=5\rho_{1}^{\prime}$. This is satisfied by $\rho_{1}=2,\rho_{1}^{\prime}=-5$. Take $r_{1}=2+5k_{1}=7$. Then $r_{1}^{\prime}=\frac{x_{1}+y_{1}r_{1}}{s_{1}}=\frac{-21-2\cdot 7}{5}=-7$.

$y_{2}=-7$.

 $x_{2}=\frac{x_{1}y_{2}-1}{y_{1}}=\frac{-21\cdot-7-1}{-2}=-73.$
 $s_{2}=x_{2}^{2}-Ay_{2}^{2}=-12.$

$x_{2}+y_{2}\rho_{2}=\rho_{2}^{\prime}s_{2}$ is equivalent with $-73-7\rho_{2}=-12\rho_{1}^{\prime}$. This is satisfied by $\rho_{2}=5,\rho_{2}^{\prime}=9$. Take $r_{2}=5,r_{2}^{\prime}=9$.

$y_{3}=9$.

 $x_{3}=\frac{x_{2}y_{3}-1}{y_{2}}=\frac{-73\cdot 9-1}{-7}=94.$
 $s_{3}=x_{3}^{2}-Ay_{3}^{2}=7.$

$x_{3}+y_{3}\rho_{3}=\rho_{3}^{\prime}s_{3}$ is equivalent with $94+9\rho_{3}=7\rho_{3}^{\prime}$. This is satisfied by $\rho_{3}=2,\rho_{3}^{\prime}=16$. Take $r_{3}=2+7k_{3}=9$. Then $r_{3}^{\prime}=\frac{x_{3}+y_{3}r_{3}}{s_{3}}=\frac{94+9\cdot 9}{7}=25$.

$y_{4}=25$.

 $x_{4}=\frac{x_{3}y_{4}-1}{y_{3}}=\frac{94\cdot 25-1}{9}=261.$
 $s_{4}=x_{4}^{2}-Ay_{4}^{2}=-4.$

$x_{4}+y_{4}\rho_{4}=\rho_{4}^{\prime}s_{4}$ is equivalent with $261+25\rho_{4}=-4\rho_{4}^{\prime}$. This is satisfied by $\rho_{4}=3,\rho_{4}^{\prime}=-84$. Take $r_{4}=3-4k_{4}=7$. Then $r_{4}^{\prime}=\frac{x_{4}+y_{4}r_{4}}{s_{4}}=\frac{261+25\cdot 7}{-4}=-109$.

$y_{5}=-109$.

 $x_{5}=\frac{x_{4}y_{5}-1}{y_{4}}=\frac{261\cdot-109-1}{25}=-1138.$
 $s_{5}=x_{5}^{2}-Ay_{5}^{2}=15.$

$x_{5}+y_{5}\rho_{5}=\rho_{5}^{\prime}s_{5}$ is equivalent with $-1138-109\rho_{5}=15\rho_{5}^{\prime}$. This is satisfied by $\rho_{5}=8,\rho_{5}^{\prime}=-134$. Take $r_{5}=8,r_{5}^{\prime}=-134$.

$y_{6}=-134$.

 $x_{6}=\frac{x_{5}y_{6}-1}{y_{5}}=\frac{-1138\cdot-134-1}{-109}=-1399.$
 $s_{6}=x_{6}^{2}-Ay_{6}^{2}=-3.$

$x_{6}+y_{6}\rho_{6}=\rho_{6}^{\prime}s_{6}$ is equivalent with $-1399-134\rho_{6}=-3\rho_{6}^{\prime}$. This is satisfied by $\rho_{6}=1,\rho_{6}^{\prime}=511$. Take $r_{6}=1-3k_{6}=10$. Then $r_{6}^{\prime}=\frac{x_{6}+y_{6}r_{6}}{s_{6}}=\frac{-1399-134\cdot 10}{-3}=913$.

$y_{7}=913$.

 $x_{7}=\frac{x_{6}y_{7}-1}{y_{6}}=\frac{-1399\cdot 913-1}{-134}=9532.$
 $s_{7}=x_{7}^{2}-Ay_{7}^{2}=3.$

$x_{7}+y_{7}\rho_{7}=\rho_{7}^{\prime}s_{7}$ is equivalent with $9532+913\rho_{7}=3\rho_{7}^{\prime}$. This is satisfied by $\rho_{7}=2,\rho_{7}^{\prime}=3786$. Take $r_{7}=2+3k_{7}=8$. Then $r_{7}^{\prime}=\frac{x_{7}+y_{7}r_{7}}{s_{7}}=\frac{9532+913\cdot 8}{3}=5612$.

$y_{8}=5612$.

 $x_{8}=\frac{x_{7}y_{8}-1}{y_{7}}=\frac{9532\cdot 5612-1}{913}=58591.$
 $s_{8}=x_{8}^{2}-Ay_{8}^{2}=-15.$

$x_{8}+y_{8}\rho_{8}=\rho_{8}^{\prime}s_{8}$ is equivalent with $58591+5612\rho_{8}=-15\rho_{8}^{\prime}$. This is satisfied by $\rho_{8}=7,\rho_{7}^{\prime}=-6525$. Take $r_{8}=7,r_{8}^{\prime}=-6525$.

$y_{9}=-6525$.

 $x_{9}=\frac{x_{8}y_{9}-1}{y_{8}}=\frac{58591\cdot-6525-1}{5612}=-68123.$
 $s_{9}=x_{9}^{2}-Ay_{9}^{2}=4.$

$x_{9}+y_{9}\rho_{9}=\rho_{9}^{\prime}s_{9}$ is equivalent with $-68123-6525\rho_{9}=4\rho_{9}^{\prime}$. This is satisfied by $\rho_{9}=1,\rho_{9}^{\prime}=-18662$. Take $r_{9}=1+4k_{9}=9$. Then $r_{9}^{\prime}=\frac{x_{9}+y_{9}r_{9}}{s_{9}}=\frac{-68123-6525\cdot 9}{4}=-31% 712$.

$y_{10}=-31712$.

 $x_{10}=\frac{x_{9}y_{10}-1}{y_{9}}=\frac{-68123\cdot(-31712)-1}{-6525}=-331083.$
 $s_{10}=x_{10}^{2}-Ay_{10}^{2}=-7.$

$x_{10}+y_{10}\rho_{10}=\rho_{10}^{\prime}s_{10}$ is equivalent with $-331083-31712\rho_{10}=-7\rho_{10}^{\prime}$. This is satisfied by $\rho_{10}=5,\rho_{10}^{\prime}=69949$. Take $r_{10}=5,r_{10}^{\prime}=69949$.

$y_{11}=69949$.

 $x_{11}=\frac{x_{10}y_{11}-1}{y_{10}}=\frac{-331083\cdot 69949-1}{-31712}=730289.$
 $s_{11}=x_{11}^{2}-Ay_{11}^{2}=12.$

$x_{11}+y_{11}\rho_{11}=\rho_{11}^{\prime}s_{11}$ is equivalent with $730289+69949\rho_{11}=12\rho_{11}^{\prime}$. This is satisfied by $\rho_{11}=7,\rho_{11}^{\prime}=101661$. Take $r_{11}=5,r_{10}^{\prime}=101661$.

$y_{12}=101661$.

 $x_{12}=\frac{x_{11}y_{12}-1}{y_{11}}=\frac{730289\cdot 101661-1}{69949}=1061372.$
 $s_{12}=x_{12}^{2}-Ay_{12}^{2}=-5.$

$x_{12}+y_{12}\rho_{12}=\rho_{12}^{\prime}s_{12}$ is equivalent with $1061372+101661\rho_{12}=-5\rho_{12}^{\prime}$. This is satisfied by $\rho_{12}=3,\rho_{12}^{\prime}=-273271$. Take $r_{12}=3-5k_{12}=8$. Then $r_{12}^{\prime}=\frac{x_{12}+y_{12}r_{12}}{s_{12}}=\frac{1061372+101661\cdot 8% }{-5}=-374932$.

$y_{13}=-374932$.

 $x_{13}=\frac{x_{12}y_{13}-1}{y_{12}}=\frac{1061372\cdot(-374932)-1}{101661}=-3% 914405.$
 $s_{13}=x_{13}^{2}-Ay_{13}^{2}=9.$

$x_{13}+y_{13}\rho_{13}=\rho_{13}^{\prime}s_{13}$ is equivalent with $-3914405-374932\rho_{13}=9\rho_{13}^{\prime}$. This is satisfied by $\rho_{13}=1,\rho_{13}^{\prime}=-476593$. Take $r_{13}=1+9k_{13}=10$. Then $r_{13}^{\prime}=\frac{x_{13}+y_{13}r_{13}}{s_{13}}=\frac{-3914405-374932\cdot 1% 0}{9}=-851525$.

$y_{14}=-851525$.

 $x_{14}=\frac{x_{13}y_{14}-1}{y_{13}}=\frac{-3914405\cdot(-851525)-1}{-374932}=% -8890182.$
 $s_{14}=x_{14}^{2}-Ay_{14}^{2}=-1.$

$x_{14}+y_{14}\rho_{14}=\rho_{14}^{\prime}s_{14}$ is equivalent with $-8890182-851525\rho_{14}=-\rho_{14}^{\prime}$. This is satisfied by $\rho_{14}=0,\rho_{14}^{\prime}=8890182$. Take $r_{14}=-k_{14}=10$. Then $r_{14}^{\prime}=\frac{x_{14}+y_{14}r_{14}}{s_{14}}=\frac{-8890182-851525\cdot 1% 0}{-1}=17405432$.

$y_{15}=17405432$.

 $x_{15}=\frac{x_{14}y_{15}-1}{y_{14}}=\frac{-8890182\cdot 17405432-1}{-851525}=% 181718045.$
 $s_{15}=x_{15}^{2}-Ay_{15}^{2}=9.$

$r_{15}=8,r_{15}^{\prime}=35662389$.

$y_{16}=35662389$.

 $x_{16}=\frac{x_{15}y_{16}-1}{y_{15}}=\frac{35662389\cdot 35662389-1}{17405432}% =372326272.$
 $s_{16}=x_{16}^{2}-Ay_{16}^{2}=-5.$

$\rho_{16}=2,\rho_{2}^{\prime}=-88730210$. $r_{16}=2-5k_{16}=7$. Then $r_{16}^{\prime}=-124392599$.

$y_{17}=-124392599$.

 $x_{17}=-1298696861.$
 $s_{17}=12.$

$r_{18}=5,r_{18}^{\prime}=-160054988$.

$y_{18}=-160054988$.

 $x_{18}=-1671023133.$
 $s_{18}=-7.$

$\rho_{19}=2,\rho_{19}^{\prime}=284447587$. Take $r_{19}=2-7k_{19}=9$. Then $r_{19}^{\prime}=444502575$.

$y_{19}=444502575$.

 $x_{19}=4640743127.$
 $s_{19}=4.$

$\rho_{20}=3,\rho_{20}^{\prime}=1493562713$. Take $r_{20}=3+4k_{20}=7$. Then $r_{20}^{\prime}=1938065288$.

$y_{20}=1938065288$.

 $x_{20}=20233995641.$
 $s_{20}=-15.$

$r_{21}=8,r_{21}^{\prime}=-2382567863$.

$y_{21}=-2382567863$.

 $x_{21}=-24874738768.$
 $s_{21}=3.$

$\rho_{22}=1,\rho_{22}^{\prime}=-9085768877$. Take $r_{22}=1+3k_{22}=10$. Then $r_{22}^{\prime}=-16233472466$.

$y_{22}=-16233472466$.

 $x_{22}=.-169482428249.$
 $s_{22}=-3.$

$\rho_{23}=2,\rho_{23}^{\prime}=67316457727$. Take $r_{23}=2-3k_{23}=8$. Then $r_{23}^{\prime}=99783402659$.

$y_{23}=99783402659$.

 $x_{23}=1041769308262.$
 $s_{23}=15.$

$r_{24}=7,r_{24}^{\prime}=116016875125$.

$y_{24}=116016875125$.

 $x_{24}=1211251736511.$
 $s_{24}=-4.$

$\rho_{25}=1,\rho_{25}^{\prime}=-331817152909$. Take $r_{25}=1-4k_{25}=9$. Then $r_{25}^{\prime}=-563850903159$.

$y_{25}=-563850903159$.

 $x_{25}=.-5886776254306.$
 $s_{25}=7.$

$r_{26}=5,r_{26}^{\prime}=-1243718681443$.

$y_{26}=-1243718681443$.

 $x_{26}=-12984804245123.$
 $s_{26}=.-12.$

$r_{27}=7,r_{27}^{\prime}=1807569584602$.

$y_{27}=1807569584602$.

 $x_{27}=18871580499429.$
 $s_{27}=5.$

$\rho_{28}=3,\rho_{28}^{\prime}=4858857850647$. Take $r_{28}=3+5k_{28}=8$. Then $r_{28}^{\prime}=6666427435249$.

$y_{28}=6666427435249$.

 $x_{28}=69599545743410.$
 $s_{28}=-9.$

$\rho_{29}=1,\rho_{29}^{\prime}=-8473997019851$. Take $r_{29}=1+9k_{29}=10$. Then $r_{29}^{\prime}=-15140424455100$.

$y_{29}=-15140424455100$.

 $x_{29}=-158070671986249.$
 $s_{29}=1.$

Therefore

 $158070671986249^{2}-109\cdot 15140424455100^{2}=1.$

Thus $\sqrt{109}\sim\frac{158070671986249}{15140424455100}$.

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