मलाही या समस्येची लाज वाटेल.

   शीर्षकाचे भाषांतर असे आहे: I would be ashamed of this problem too"

Someone, signed "ANONYMOUS", commented on yesterday's post. I wanted to delete the comment, usually insulting, but I researched it and the guy is right. He, in the comment, refers to this problem:

In translation:

                      "10.     Show that if   $a,\;b,\;c \in \mathbb{R}$  and  $ab+bc+ca=0$ , then

$2\sqrt{a^2+b^2+c^2}\geqslant 3\sqrt[3]{|abc|}.$

C. Ionescu-Țiu "

Indeed, the problem is a bit strange. Unless it is a typographical error, it insults the intelligence of a solver with minimal knowledge. I won't bother with it any further, I'll just say this:

 it is true under any conditions for any numbers  $a,\;b,\;c$  not just those that satisfy the relation $ab+bc+ca=0$. In addition, the  $=$  sign only occurs in the case of  $a=b=c=0$.

          Proof  CiP  For non-negative numbers $x,\;y,\;z$ , holds the inequality

  AM-GM :   $\frac{x+y+z}{3}\geqslant \sqrt[3]{xyz}$ 

 so

$x+y+z\geqslant 3\cdot \sqrt[3]{xyz}$

Replacing above $x=a^2,\;y=b^2,\;z=c^2$, where the numbers  $a,\;b,\;c$  are arbitrary, not bound by any condition, we obtain

$a^2+b^2+c^2\geqslant 3\cdot \sqrt[3]{a^2b^2c^2}$

and taking the square root of both sides we have

$$2\cdot \sqrt{a^2+b^2+c^2}\geqslant 2\sqrt{3}\cdot \sqrt[3]{|abc|}$$

But, since  $2\sqrt{3}=\sqrt{12}>\sqrt{9}=3$, the above results in

$$2\sqrt{a^2+b^2+c^2}>3\cdot \sqrt[3]{|abc|}\;.$$

$\blacksquare$

More in Joke, more in Serious : A Problem Close to Logic

 I don't expect C. Ionescu-Țiu to show much logic in His Problems. Here is Problem E:6061 from the magazine in the picture.

I have published more about this issue of the Magazine elsewhere.

In translation:

                        "E:6061*. Consider the real and positive numbers  $a,\;b,\;c,\;d$  such

                        that $a+b=c+d$.  Show that:

                          1). If  $ab>cd$  then  $a^2+b^2<c^2+d^2$  and the converse.

                          2). If  $ab>cd$  then  $|a-b|<|c-d|$.

                          3). If  $a^2+b^2<c^2+d^2$  then  $|a-b|<|c-d|$  and the converse."

          Solution CiP (an improvised solution, at the school level)

               Let us remember that everywhere in what follows holds the equation:

$a+b=c+d \tag{1}$

               1). Direct implication:         $ab>cd\;\Rightarrow\;a^2+b^2<c^2+d^2$

(1)$\;\Rightarrow\;\;(a+b)^2=(c+d)^2$

$\Rightarrow\;\;a^2+b^2+2\cdot ab=c^2+d^2+2\cdot cd$

$\Rightarrow\;\;a^2+b^2-c^2-d^2=2\cdot (cd-ab)$

and from hypothesis  $ab>cd$  it follows  $cd-ab<0$,  so  $a^2+b^2-c^2-d^2<0$, or equivalently : $a^2+b^2<c^2+d^2$.

qed

                    Converse implication :          $a^2+b^2<c^2+d^2\;\Rightarrow\;ab>cd$ 

              $a^2+b^2<c^2+d^2\;\;\Rightarrow\;\;-a^2-b^2>-c^2-d^2\;\;\underset{(1)}{\Rightarrow}$

$\Rightarrow\;\;(a+b)^2-a^2-b^2>(c+d)^2-c^2-d^2\;\;\Leftrightarrow\;\;2\cdot ab>2\cdot cd\;\;\Leftrightarrow\;\;ab>cd$

qed

          Remark CiP  Based on the equation

$a^2+b^2-c^2-d^2=2\cdot (cd-ab)$

we have the logical equivalence

$a^2+b^2-c^2-d^2<0\;\;\;\Leftrightarrow\;\;\;cd-ab<0$

and the statement  "1)" is obtained immediately.

<end Rem>

               2). Implication:   $ab>cd\;\;\Rightarrow\;\;|a-b|<|c-d|$

$ab>cd\;\;\Rightarrow\;\;-4\cdot ab<-4\cdot cd\;\;\Rightarrow\;\;(a-b)^2-(a+b)^2<(c-d)^2-(c+d)^2\;\;\Rightarrow$

$\;\underset{(1)}{\Rightarrow}\;\;(a-b)^2<(c-d)^2\;\;\Rightarrow\;\;\sqrt{(a-b)^2}<\sqrt{(c-d)^2}\;\;\Leftrightarrow\;\;|a-b|<|c-d|$.

qed

          Remark CiP

                 a)  The converse is also valid:  $|a-b|<|c-d|\;\;\Rightarrow\;\;ab>cd$

Because  $|a-b|<c-d|\;\;\Rightarrow\;\;(a-b)^2<(c-d)^2\;\;\Rightarrow\;\;-(a-b)^2>-(c-d)^2\;\;\Rightarrow$

$\underset{(1)}{\Rightarrow}\;\;(a+b)^2-(a-b)^2>(c+d)^2-(c-d)^2\;\;\Leftrightarrow\;\;4\cdot ab>4\cdot cd$, &c...

                 b) As in the Remark from point 1), we can prove point 2) together with its converse, based on the equation

$(a-b)^2-(c-d)^2=4\cdot(cd-ab)$

<end Rem>

               3).  Direct implication:     $a^2+b^2<c^2+d^2\;\;\Rightarrow\;\;|a-b|<|c-d|$

From  $2a^2+2b^2<2c^2+2d^2\;\;\Rightarrow\;\;(a+b)^2+(a-b)^2<(c+d)^2+(c+d)^2\;\;\Rightarrow$

$\underset{(1)}{\Rightarrow}\;\;(a-b)^2<(c-d)^2\;\;\Rightarrow\;\;\sqrt{(a-b)^2}<\sqrt{(c-d)^2}\;\;\Leftrightarrow\;\;|a-b|<|c-d|$

qed

                     Converse implication:     $|a-b|<|c-d|\;\;\Rightarrow\;\;a^2+b^2<c^2+d^2$

From the hypothesis  $|a-b|<|c-d|$  follows immediately the inequality  

$(a-b)^2<(c-d)^2 \tag{2}$

$\Rightarrow\;\;2a^2+2b^2-4ab<2c^2+2d^2-4cd\;\;\Rightarrow\;\;2(a^2+b^2-c^2-^2)<4ab-4cd=$

$=(a+b)^2-(a-b)^2-[(c+d)^2-(c-d)^2]\underset{(1)}{=}(c-d)^2-(a-b)^2\underset{(2)}{<}0.$

qed

               Remark CiP   Both implications, direct and converse, result at once from the equation

$2(a^2+b^2-c^2-d^2)=(c-d)^2-(a-b)^2$

<end Rem>

With this we solved the exercise, and something extra.

                    Final Remarks CiP

               1.  For the logical propositions:

$p :  ab>cd$

$q :  a^2+b^2<c^2+d^2$

$r :   |a-b|<|c-d|$

we have the logical equivalents

$p\;\;\Leftrightarrow\;\;q\;\;\Leftrightarrow\;\;r$

               2.  Due to the symmetry in $a$ and $b$ on the one hand and in $c$ and $d$ on the other hand, we could assume from the beginning that  $a\leqslant b$ and $c\leqslant d$. With this, the proposition $p$ is true  $\Leftrightarrow\;\;ab\underset{(1)}{>}c(a+b-c)\;\;\Leftrightarrow\;\;c^2-c(a+b)+ab>0\;\;\Leftrightarrow\;\;(c-a)(c-b)>0\;\;\Leftrightarrow\;\;c\in (0,\;a)\cup (b,\;+\infty)$.

and similar for $d$, so ultimately we have on the real axis the ordering $0<c<a\leqslant b<d$, to which is added the condition that the segments with ends $a$ and $b$, respectively $c$ and $d$ have the same midpoint.

$\blacksquare$

Nicușor DAN - A Day in the Life of a President // Nicușor DAN - O Zi din Viața unui Președinte

 Your doctoral thesis is inaccessible to the public. In its place we put a work from our youth...

His work is not too extensive either...

Diophantine Equation $x^2+y^2=2\cdot z^2$ in terms of the correct solution of the Pythagorean Equation

 In this post we will solve in integers the equation

$$x^2+y^2=2\cdot z^2 \tag{E}$$

Noting that  $(x,y,z)=(\pm a,\pm b,\pm c)$ are solutions, once one of them is, we will limit ourselves to solutions in positive integers. 

We will see that the solution of equation (E) depends on the general solution of the Pythagorean equation $x^2+y^2=z^2$. But I have explained this in more detail here.

          If the integers $(x,y,z)$ verify the equation (E) then, since  $2\cdot z^2$ is an even number, $x$ and  $y$  must necessarily have the same parity. So $\frac{x+y}{2}$ and  $\frac{x-y}{2}$ are integers. It can be seen that (E) is equivalent to

$$\left ( \frac{x+y}{2}\right )^2+\left ( \frac{x-y}{2} \right )^2=z^2 \tag{P}$$

which shows that  $\left (\frac{x+y}{2},\frac{x-y}{2},z \right )$  are solutions of the Pythagorean equation.

     Returning to the previously mentioned Post, the positive integers solutions (not only the primitive ones) of the equation  (P) are given by the families with parameters $s,\; t$ and $d$ ($d$ being a multiplicity factor):

$\begin {cases}\frac{x+y}{2}=2dst\\\frac{x-y}{2}=d(s^2-t^2) \tag{1}\\z=d(s^2+t^2) \end{cases}$

and

$\begin{cases}\frac{x+y}{2}=d(s^2-t^2)\\\frac{x-y}{2}=2dst \tag{2}\\z=d(s^2+t^2)\end{cases}$

where $d\in \mathbb{Z}$ and

$s>t>0$ are two coprime integers of opposite parity.                        (st)

          Then the general solution in positive integers of equation (E) will be, solving for (1) and (2):

$x=d\cdot (s^2+2st-t^2),\;\;y=d\cdot (-s^2+2st+t^2),\;\;z=d\cdot (s^2+t^2) \tag{3.1}$

and

$x=d\cdot (s^2+2st-t^2),\;\;y=d\cdot (s^2-2st-t^2),\;\;z=d\cdot (s^2+t^2) \tag{3.2}$

We notice that the value of  $x$  in the two formulas is the same. And the value of  $y$  is given by two formulas differing only in sign: for $s>\sqrt{2}\cdot t$  we have $y_1<0$ and for $s<\sqrt{2}\cdot t$  we have  $y_2<0$. Since we are only interested in positive values, we can express the final result in the form

          The integer and positive solutions of the equation (E) are

$x=d\cdot (r^2+2st-t^2),\;\;y=d\cdot |s^2-2st-t^2|,\;\;z=d\cdot (s^2+t^2) \tag{DST}$

where $d\in \mathbb{N}$  and  $s,\;t$  check the condition (st) above.

          Remark CiP  In the table below we show some solutions for $d=1$ (i.e. primitive solutions, gcd(x,y,z)=1)

ATTENTION! The values ​​$s=9,\; t=7$  do not satisfy the condition (st). The solution $(x,y,z)=(158,94,130)=2\cdot (79,47,65)$ is the multiple of another solution...

Karmaşık görünen 29 071 numaralı sorunun çok basit olduğu ortaya çıktı // The Problem 29 071 that seemed complicated turned out to be too trivial

The author, Mihály Bencze, is famous, which is perhaps why the Magazine GM-B accepted this issue.

In tranlation

                          "29071.   Let the function  $f\;:\;\mathbb{N} \to \mathbb{N}$  be defined by

$f(n)=\left [ \frac{n}{3}\right ]+\left [ \frac{n+1}{5} \right ]+\left [ \frac{n+2}{7}\right ]$ ,  for all $n\in\mathbb{N}.$

                                  Prove that the function is neither injective nor surjective."

ANSWER CiP

$f(0)=f(1)=0$  , so the function is NOT injective;

$f(8)=4,f(9)=6$ and the function $f$, which is increasing, never takes the value $5.$

                Solution CiP

               It was not said in the statement, as many other times, that  $[\alpha]$  denotes the integer part of the real number  $\alpha$. By definition, $[\alpha]\in\mathbb{Z}$  is the (uniquely determined) integer that satisfies one of the conditions below:

$$[\alpha]\leqslant \alpha <[\alpha]+1,\quad \quad \alpha -1<[\alpha]\leqslant \alpha\;. \tag{][}$$

 

          The case of injectivity:

$f(0)=\left [ \frac{0}{3}\right ]+\left [ \frac{1}{5}\right ]+\left [\frac{2}{7}\right ]=0+0+0=0:$

$f(1)=\left [\frac{1}{3}\right ]+\left [\frac{2}{5}\right ]+\left [\frac{3}{7}\right ]=0+0+0=0.$

So the function $f$ is NOT injective.

          The case of surjectivity:

     Let's first note that $f$ is increasing, because function $[x]$ is: $x<y\Rightarrow [x]\leqslant [y]$

        (The pedantic solver would do this:

        let $x<y;$ if $x\leqslant [y]$ then, because $[x]\leqslant x$ we deduce from the transitivity of the inequalities that $[x]\leqslant [y];$

                           if $[y]<x$ then, because $x<y\overset{(][)}{<}[y]+1$ we deduce from transitivity that $[y]<x<[y]+1\overset{(][)}{\Rightarrow} [x]=[y]$.)

          We see that $f(8)=\left [2\frac{2}{3}\right ]+\left [1\frac{4}{5}\right ]+\left [1\frac{3}{7}\right ]=2+1+1=4$,

$f(9)=\left [\frac{9}{3}\right ]+\left [\frac{10}{5} \right ]+\left [1\frac{4}{7}\right ]=3+2+1=6$

so $f(n)$ takes the values ​​4 and 6 for the consecutive numbers 8 and 9, so, being increasing, it cannot take the value 5 for any natural number $n$. (The pedant would say so: $n\leqslant 8\Rightarrow f(n)\leqslant 4;\;\;n\geqslant 9\Rightarrow f(n)\geqslant 6$  and there are no other possibilities.  )

The function $f$ is not surjective.

$\blacksquare$

            Remark CiP

          I suspected from the beginning that the values ​​of the function $f$ make "jumps". I looked for values ​​of $n$ for which each of the three fractions would be a natural number. For this we set the conditions:

$$\begin {cases}n=3\cdot k\;\;\;\;\;\;\;\;(1)\\n+1=5\cdot l\;\;\;(2)\\n+2=7\cdot m\;\;(3)\end{cases}$$

We substitute (1) into (2) and we obtain

$$3\cdot k+1=5\cdot l\;\;\Leftrightarrow\;\;5\cdot l-3\cdot k=1 \tag{kl}$$ 

which is a linear Diophantine equation. We see the solution $k=3,\;l=2$ of (kl) so it has the general solution

$$k=5\cdot p+3,\;\;l=3\cdot p+2 \;,\;\;p\in\mathbb{N}\tag{4}$$

Then  $n\underset{(1)}{=}3\cdot k\underset{(4)}{=}3(5p+3)=15\cdot p+9$ which substituted into (3) gives us

$$(15\cdot p+9)+2=7\cdot m\;\;\Leftrightarrow\;\;7\cdot m-15\cdot p=11 \tag{mp}$$

Here we see an solution  $m=-7,\;p=-4$, so the general solution of  (mp) is

$$m=15\cdot t-7,\;\;p=7\cdot t-4\;,\;\;t\in\mathbb{N} \tag{5}$$

Finally  $n=15p+9\underset{(5)}{=}15(7\cdot t-4)+9=105\cdot t-51.$

     The first value of  $n$ we are looking for is $n=105-51=54.$ Calculating, we get

$f(53)=\left [\frac{53}{3}\right ]+\left [ \frac{54}{5}\right ]+\left [\frac{55}{7}\right ]=\left [17\frac{2}{3}\right ]+\left [10\frac{4}{5}\right ]+\left [ 7\frac{6}{7}\right ]=17+10+7=34,$

$f(54)=\left [\frac{54}{3}\right ]+\left [\frac{55}{5}\right ]+\left [\frac{56}{7}\right ]=18+11+8=37,$

$f(55)=\left [\frac{55}{3}\right ]+\left [\frac{56}{5}\right ]+\left [\frac{57}{7}\right]=\left[18\frac{1}{3}\right]+\left [11\frac{1}{5}\right]+\left[8\frac{1}{7}\right]=18+11+8=37.$

From here it is immediately clear that the function  $f$  is neither injective nor surjective.

<end Rem>

          A PEDANT's remark  Instead of (1)-(3) we could use congruences according to different moduli. 

$n\equiv 0\;(mod\;3),\;\;n+1\equiv 0\;(mod\;5)\Leftrightarrow n\equiv  -1\;(mod\;5),\;\;n\equiv -2\;(mod\;7).$

 Now,  $3\cdot k \equiv -1\;(mod\;5)\Rightarrow 6\cdot k \equiv -2\;(mod\;5)\Leftrightarrow k\equiv -2\;(mod\;5).$ So  $k=5\cdot p-2,\;p\in\mathbb{Z}$  and further  $n=3\cdot k=15\cdot p-6\equiv -2\;(mod\;7)\Rightarrow 15\cdot p\equiv 4\;(mod\;7)\Rightarrow p\equiv 4\equiv -3\;(mod\;7).$ Thus  $p=7\cdot t-3,\;t\in\mathbb{Z}$, so $n=15p-6=15(7t-3)-6=105\cdot t-51.$

<end rem>

İkisi Bir Arada: Yağ ve su gibi karışmazlar // Two in One : they don't mix, like oil and water

 Strange juxtaposition of two issues in SGMB 3/2025

In translation:

     "Let  $a,\;b,\;c$  be positive real numbers. Show that

$\frac{\sqrt{a}}{b+c}=\frac{\sqrt{b}}{a+c}=\frac{\sqrt{c}}{a+b} \tag{E}$

                                  if and only if  $a=b=c.$"

                                        Solution CiP

            Let's assume that  $a\neq b$. Then, from the first two of the equalities (E), we get

$0<\frac{\sqrt{a}}{b+c}=\frac{\sqrt{b}}{a+c}=\frac{\sqrt{a}-\sqrt{b}}{(b+c)-(a+c)}=\frac{\sqrt{a}-\sqrt{b}}{b-a}=-\frac{\sqrt{a}-\sqrt{b}}{a-b}=$

$=-\frac{\sqrt{a}-\sqrt{b}}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}=-\frac{1}{\sqrt{a}+\sqrt{b}}<0$

but this is ABSURD. 

          So $a=b$, and therefore from the equalities (E) there remains

$$\frac{\sqrt{a}}{a+c}=\frac{\sqrt{c}}{2\cdot a}$$

Assuming now that  $c\neq a$, the above results in

$0<\frac{\sqrt{a}}{a+c}=\frac{\sqrt{a}-\sqrt{c}}{(a+c)-2a}=-\frac{\sqrt{a}-\sqrt{c}}{a-c}=-\frac{1}{\sqrt{a}+\sqrt{c}}<0$

this is again ABSURD. So  $c=a=b$.

QED $\blacksquare$

Pell's Equations $\;\;x^2-D\cdot y^2=\pm1$

           Let Peel's Equation be

$$x^2-D\cdot y^2=1.\tag{P}$$

where $D$ is a given positive nonsquare integer.

     In solving (P) we must find the fundamental solution, i.e. a pair of positive integers $(x_1,y_1)$ wich satisfie them,and $x_1$ is minimal. An algorithm that determines this, found by me in the book "Dr. Paul RADOVICI-MĂRCULESCU, Probleme de Teoria Elementară a Numerelor" (Ed. Tehnică, București, 1986), at page 178, uses Continued Fraction.

      We develop $\sqrt{D}$ in continued fraction (their shape is well known):

$\sqrt{D}=[a_0,\; \overline{a_1,\;\dots,\;a_{k-1},\;2a_0}] \tag{D}$

                  $\blacklozenge$ if $k$=odd, $k=2t+1$ then the fundamental solution is  $x_1=A_{4t+1},\;y_1=B_{4t+1}; \tag{O}$

                   $\blacklozenge$ if $k$=even, $k=2t$ then the fundamental solution is $x_1=A_{2t-1},\;y_1=B_{2t-1}.\tag{E}$

I wrote $A_k$ the numerator and $B_k$ the denominator of the $k$th convergent of (D), i.e.

$[a_0,\;a_1,\;\dots, a_k]=\frac{A_k}{B_k}.$

I found their calculation formulas in the book ""I.M. VINOGRADOV, Bazele Teoriei Numerelor"(Ed. Academiei, 1957), at page 17

$$A_i=a_i\cdot A_{i-1}+A_{i-2},\;\;B_i=a_i\cdot B_{i-1}+B_{i-2},\;\;i\geqslant 1$$

with initial values  $A_{-1}=1,\;B_{-1}=0;\,\,\,A_0=a_0,\;B_0=1$

where A_n is the numerator and B_n is the denominator, called continuants,^[1][2] of the nth convergent.

       Schematically, obtaining $A_i$ can be represented as follows

$\begin{cases}\;\;\;\;\;\;\;\;\times \swarrow a_i\\\underset{+}{\underbrace{A_{i-2}\;A_{i-1}}}\;\fbox{Ai}\tag{S}\end{cases}$

And similar for $B_i$.

          Example 1        $x^2-13\cdot y^2=1$

         We have the continued fraction expansion

$\sqrt{13}=[3,\;\overline{1,1,1,1,6}]$   so $k=5$.

We apply the scheme (S), the calculated elements being written in color, and we obtain the table:

\begin{array}{|c|c}\hline\\\;i&-1&0&1&2&3&4&5&6&7&8&\color{Green}9&10\\\hline \\a_i&\;&3&1&1&1&1&6&1&1&1&1&6\\\hline \\A_i&1&3&\color{Red}4&\color{Red}7&\color{Red}{11}&\color{Red}{18}&\color{Red}{119}&\color{Red}{137}&\color{Red}{256}&\color{Red}{393}&\color{Green}{649}&\color{Red}{4287}\\\hline\\B_i&0&1&\color{Red}1&\color{Red}2&\color{Red}3&\color{Red}5&\color{Red}{33}&\color{Red}{38}&\color{Red}{71}&\color{Red}{109}&\color{Green}{180}&\color{Red}{1189}\\\hline \end{array}

We are in the case (O),  $k=5=2\cdot t+1\;\Rightarrow t=2$ and the minimal solution is (colored green)

$x_1=A_9=649,\;\;y_1=B_9=180.$

          Remark CiP  The result is in agreement with that given in the Table on page 127 of the book

ANDREESCU Titu, ANDRICA Dorin, CUCUREZEANU Ion - An Introduction to Diophantine Equations : A Problem-Based Approach (Birkhäuser Springer, New York Dordrecht Heidelberg London, 2010)

<end Rem>

          Example 2         $x^2-31\cdot y^2=1$

          We have the continued fraction expansion

$\sqrt{31}=[5,\;\overline{1,1,3,5,3,1,1,10}]$    so  $k=8.$

Now we are in the case (E):  $8=k=2\cdot t\;\Rightarrow\;t=4$  and the minimal solution is, according to the calculations in the table below

$x_1=A_7=1520,\;\;y_1=B_7=273.$

(Compare with the Table mentioned in the previous Remark.)

\begin{array}{|c|c|}\hline\\i&-1&0&1&2&3&4&5&6&\color{Green}7\\\hline \\a_i&\;&5&1&1&3&5&3&1&\color{Green}1\\\hline\\A_i&1&5&6&11&39&206&657&863&\color{Green}{1520}\\\hline\\B_i&0&1&1&2&7&37&118&155&\color{Green}{273}\\\hline\end{array}

          As for ALL the solutions in natural numbers of the equation (P), they are given by the formulas (O) - wiyh $t=0,\,1,\,2,\dots$ and (E) - with $t=1,\;2,\dots$. However, the further calculation of $n$th convergent  is too laborious.

          A simpler approach involves writing a system of recurrence relations.

          Suppose, for the equation (P) that we have found the fundamental solution $(x_1,y_1)$. Then, using the matrix

$P=P_D=\begin{pmatrix}x_1&D\cdot y_1\\y_1&x_1\\\end{pmatrix} \tag{M}$

we write the linear recurrence relation in the form 

$\begin{pmatrix}x_{n+1}\\y_{n+1} \end{pmatrix}=P \cdot \begin{pmatrix}x_n\\y_n \end{pmatrix}=\begin{pmatrix}x_1&D\cdot y_1\\y_1&x_1 \end{pmatrix} \cdot \begin{pmatrix}x_n\\y_n \end{pmatrix}\;,n=0,\;1,\dots\tag{MX}$

with  $(x_0,y_0)=(1,0)$ which is the universal solution for the equations (P),

that is

$\begin{cases}x_{n+1}=x_1\cdot x_n+(Dy_1)\cdot y_n\\y_{n+1}=y_1\cdot x_n+x_1 \cdot y_n \end{cases},\;\;x_0=1,\;y_0=0,\;\;\;n=0,\;1,\dots \tag{XY}$

     The two-recurrence system (XY) can be transformed into second-order recurrence relations, as shown in another post.

        Example 3             $x^2-3\cdot y^2=1$

        Fundamental solution of this equation is $(x_1,y_1)=(2,1)$. So  $P=\begin{pmatrix}2&3\cdot 1\\1&2\end{pmatrix}$  and the recurrence relations are written

$\begin{cases}x_{n+1}=2\cdot x_n+3\cdot y_n\\y_{n+1}=x_n+2 \cdot y_n\end{cases},\;x_0=1,\;y_0=0\;\;\;(n=0,\;1,\dots).$

The first few solutions appear in the Table below:

\begin{array}{c|c}n&0&1&2&3&4&5&6\\\hline \\x_n&1&2&7&26&97&362&1351\\\hline \\y_n&0&1&4&15&56&209&780 \end{array}

       Or we can write the second-order recurrence relations, as mentioned above, with the same results:

$x_{n+2}=4\cdot x_{n+1}-x_n,\;\;x_0=1,\;x_1=2\;\;\;\;;\;\;\;\;y_{n+2}=4\cdot y_{n+1}-y_n,\;\;y_0=0,\;y_1=1\;\;(n=0,\;1,\dots)$.

          Now let the Negative Pell (or "minus Pell") equation be 

$x^2-D\cdot y^2=-1 \tag{-P}$

Not all of these equations have solutions. But when they do, they solve similarly. I first encountered it in an article in GM-P, no. 4/1989, pages 175-178: Gheorghe UDREA - "Asupra Formei Solutiilor Ecuatiilor  $x^2-D\cdot y^2=\pm1$". We quote from here what interests us.

         Let's go back to (D).

$\quad\blacklozenge$ if $k$=odd the solutions of the equation (-P) are

$x_n=A_{k(2n-1)-1},\;\;\;y_n=B_{k(2n-1)-1},\;\;\;n=1,\;2,\;3,\dots ;\tag{-O}$

$\quad\blacklozenge$ if $k$=even the equation (-P) has NO solutions.

So, the smallest positive solution, when it exists (so $k$ is odd) is

$x_1=A_{k-1}\,,\,\,\,y_1=B_{k-1} \tag{-F}$

Furthermore, we can also determine the smallest positive solution(the fundamental one) of the equation (P) by the formula 

$x_1'+y_1'=(x_1+y_1\cdot \sqrt{D})^2 \tag{+P}$

          Example 4    $x^2-13\cdot y^2=-1$

         We have (see Example 1) $\sqrt{13}=[3,\overline{1,1,1,1,6}]$  so $k=5$. Minimal solution is $x_1=A_4=18,\;y_1=B_4=5.$ Further we have

$(18+5\cdot \sqrt{13})^2=324+25\cdot 13+180\cdot \sqrt{13}=649+180 \cdot \sqrt{13}$

so minimai solution of $x^2-13\cdot y^2=1$ is $x_1'=649,\;y_1'=180$ as we have already found.

Images of the older manuscript and some drafts.