Regarding the Problem 16 667

 Here, page 47, the solution was published. I haven't found the Statement yet.

     We select the property in the photo.

 ANSWER CiP

We will show that one of the numbers  $a_k$  is equal to 1, 

and the others are zero. Then  $b=k$.

                              Solution CiP

               We will treat the case  $n=4$  in detail, so for

$a_1,\;\;a_2,\;\;a_3,\;\;a_4\geqslant 0 \tag{1}$

we have equations

\begin{cases}a_1+a_2+a_3+a_4=1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(2)\\1\cdot a_1+2\cdot a_2+3\cdot a_3+4\cdot a_4=b\;\;\;\;\;\;\;\;\;\;\;\;(3)\\1^2\cdot a_1+2^2\cdot a_2+3^2\cdot a_3+4^2\cdot a_4=b^2\;\;\;\;\;(4)\end{cases}

Subtracting (2) from (3) we have  $b-1=a_2+2\cdot a_3+3\cdot a_4\geqslant 0$,  so  $b\geqslant 1.$

          $\textbf{If}\;\;\bf{b\geqslant 4}$  then 

$b^2-4\cdot b\;\underset{(4)-4\times (3)}{=}(a_1+4a_2+9a_3+16a_4)-4\cdot (a_1+2a_2+3a_3+4a_4)$,  so

$0\leqslant b^2-4b=-3a_1-4a_2-3a_3\overset{(1)}{\leqslant} 0 \tag{5}$

Therefore, in (5) we have everywhere the sign  $"="$. In particular

$b=4,\;\;a_1=a_2=a_3=0,\;\;and\;\;a_4\overset{(2)}{=}1 \tag{6}$

        $\textbf{Let}\;\;\bf{b\in[i,i+1)}$  for some  $i\in\{1,\;2,\;3\}$. We calculate first

$b^2-b\cdot i\;\overset{(4)}{\underset{i\times (3)}{=}}(1a_1+4a_2+9a_3+16a_4)-(i\cdot a_1+2i\cdot a_2+3i\cdot a_3+4i\cdot a_4)$  so

$b^2-bi=(1-i)a_1+2(2-i)a_2+3(3-i)a_3+4(4-i)a_4 \tag{7}$

On the other hand

$b-i\cdot 1\overset{(3)}{\underset{i\times (2)}{=}}(a_1+2a_2+3a_3+4a_4)-(i\cdot a_1+i\cdot a_2+i\cdot a_3+i\cdot a_4)=$

$=(1-i)a_1+(2-i)a_2+(3-i)a_3+(4-i)a_4$

which, multiplied by  $b$  gives us

$b^2-bi=b(1-i)a_1+b(2-i)a_2+b(3-i)a_3+b(4-i)a_4 \tag{8}$

Equating (7) and (8)

$(1-i)a_1+2(2-i)a_2+3(3-i)a_3+4(4-i)a_4=b(1-i)a_1+b(2-i)a_2+b(3-i)a_3+b(4-i)a_4$

we obtain

$(b-1)(i-1)a_1+(b-2)(i-2)a_2+(b-3)(i-3)a_3+(b-4)(i-4)a_4=0 \tag{9}$

               For  $i=1$, (so  $1\leqslant b<2$), the first term cancels out, and all coefficients of   $a_2,\;a_3,\;a_4$  are  $>0$.

Therefore  $a_2=a_3=a_4=0$,  so (cf. (2))  $a_1=1$  and then  $b\overset{(3)}{=}1$.

                For  $i=2$, (so  $2\leqslant b<3$), then the second term cancels out, and all coefficients of  $a_1,\;a_3,\;a_4$  are  $>0$.

Therefore  $a_1=a_3=a_4=0$,  so (cf.(2))  $a_2=1$  and then  $b\overset{(3)}{=}2$.

                For  $i=3$, (so  $3\leqslant b <4$), then the third term cancels out, and all coefficients of  $a_1,\;a_2,\;a_4$  are  $>0$.

Therefore  $a_1=a_2=a_4=0$, so (cf.(2))  $a_3=1$  and then  $b\overset{(3)}{=}3$.

The last three conclusions, together with (6), validate the answer in this particular case.

               The general case extends what is shown in case  $n=4$. So let be the numbers

$a_1,\;a_2,\;\dots,a_n\;\geqslant 0 \tag{11}$

that verify the relations

$$a_1+a_2+\dots+a_n=\sum_{k=1}^na_k=1 \tag{12}$$

$$a_1+2\cdot a_2+\dots+n\cdot a_n=\sum_{k=1}^nk\cdot a_k=b \tag{13}$$

$$a_1+4\cdot a_2+\dots +n^2\cdot a_n=\sum_{k=1}^nk^2\cdot a_k=b^2 \tag{14}$$

We have

$$b-1\overset{(13),(12)}{=}\sum_{k=1}^nk\cdot a_k-\sum_{k=1}^na_k=\sum_{k=1}^n(k-1)\cdot a_k=\sum_{k=2}^n(k-1)a_k\geqslant 0$$

so  $b\geqslant 1$.

          If  $b\geqslant n$  then

$$0\leqslant b^2-n\cdot b\overset{(14),(13)}{=}\sum_{k=1}^nk^2\cdot a_k-n \sum_{k=1}^nk\cdot a_k=\sum_{k=1}^n(k^2-nk)a_k=\sum_{k=1}^{n-1}k(k-n)a_k\leqslant 0$$

and according to (11) it result  $a_1=\dots=a_{n-1}=0,\;b=n,\;a_n\overset{(12)}{=}1$.

          If  $b\in[i,i+1)$  for some  $i\in\{1,\dots,n-1\}$, then, on the one hand

$$b^2-i\cdot b\;\overset{(14),(13)}{=}\sum_{k=1}^nk^2a_k-i\cdot \sum_{k=1}^nka_k=\sum_{k=1}^nk(k-i)a_k \tag{15}$$

On the other hand

$$b-i\;\overset{(13),(12)}{=}\sum_{k=1}^nka_k-i\cdot \sum_{k=1}^na_k=\sum_{k=1}^n(k-i)a_k$$

which, multiplied by  $b$  gives us

$$b^2-ib=\sum_{k=1}^nb(k-i)a_k \tag{16}$$

Equating (15) and (16)  we obtain

$$\sum_{k=1}^n(b-k)(i-k)a_k=0$$

or, excluding the term  $a_i$ wich has coefficient zero

$(b-1)(i-1)a_1+\dots+(b-i+1)a_{i-1}+(b-i-1)(-1)a_{i+1}+\dots (b-n)(i-n)a_n=0$.

Let's analyze the coefficients of the numbers  $a_k$  in the equation above :

          for  $k<i\;\Rightarrow\;(b-k)(i-k)\;\overset{k<i\leqslant b}{>}0;$

          for  $k>i\;\Rightarrow\;(b-k)(i-k)=(k-b)(k-i)\overset{k\geqslant i+1>b}{>}0$

Then it turns out that  $a_1=\dots=a_{i-1}=0=a_{i+1}=\dots =a_n$,  so  $a_i\overset{(12)}{=}1$  and  $b\underset{(13)}{=}i.$

$\blacksquare$

PLM inseamna o trivialitate.... PLV este prescurtarea oficiala

 Am postat in Aug/01/2024 Decizia de Recalculare a pensiei.

Un comentator anonim, cam sugubat, s-a interesat de Decizia de trecere la Pensia pentru Limita de Varsta. Prescurtat PLV. I-am raspuns ... la fel de trivial. Scuze cititorilor mai sensibili...

Aici sunt cele 3 pagini din Decizie...

Talonul de pensie pe luna AUGUST 2025, insemnat cu compararile din "Marea Recalculare"

Am incasat si niste diferente pntru lunile MAI-JUL, cum se vede mai jos

Voi posta si Talonul pe SEPtembrie, care asa va fi de aici incolo....cat voi trai.

Edit Sep 8, 2025 : Iata Talonul ....final:

A Problem in "GHEBA"

 The author was a celebrity in the second half of the 20th century for his Mathematical Problem Collections for middle school students.

          Ironically, his name has come up in a controversy. See here and here.

          I solved it from his 1973 collection. The 1975 edition, slightly modified, is here. A list of exercises is selected here and here.

          

         I liked the following problem which seems like Elementary Arithmetic :

               "Un obiect se vinde cu 39 lei, castigandu-se atat la suta cat a costat

                 obiectul. Care a fost costul obiectului ?"

In translation : 

"An object is sold for 39 lei, earning a percentage of the cost of the object.

             What was the cost of the object ?"    

              The answer is 30 lei.  That is, an object that costs 30 lei was sold for 30% more. That is, a commercial addition of  $30\cdot \frac{30}{100}=9$ lei. So the selling price is

30 lei  +  9 lei = 39 lei.

          In my personal edition the problem appears on page 136, Problem #5. In the 1975 edition the problem appears on page 201, Problem #5. Among more modern editions, the problem appears on page 192, Problem  #22.

               How can this problem be solved arithmetically ?

I don't know the answer, but I solved it using algebra. In fact, the problem is included in the Chapter on 2nd Grade Equations.

          Solution CiP

          Let  $x$  be the initial price of the object. The object is being sold for  $x\text{%}$  more. The commercial markup is therefore  $x\cdot \frac{x}{100}=\frac{x^2}{100}$. The selling price of the object will be  $x+\frac{x^2}{100}$. 

     So we have the quadratic equation

$x+\frac{x^2}{100}=39$

     In real numbers the equation has two solutions  $x_1=30,\;\;x_2=-130$. Of these, only the first has significance for our problem.

$\blacksquare$

A Problem That Has a Chance of Becoming a Theorem // Problema, kuri gali tapti teorema

          I heard the expression in the title from a somewhat megalomaniac fellow mathematician.

          It does, and we will try to prove the following formula :

                    Let be the nonzero numbers  $a_1,\;\dots a_m$. The following equation holds :

$a_1\cdot \left ( \frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+\dots +\frac{1}{a_m}\right )+(a_2-a_1)\cdot \left ( \frac{1}{a_2}+\frac{1}{a_3}+\dots +\frac{1}{a_m}\right )+$

$+(a_3-a_2)\cdot \left (\frac{1}{a_3}+\dots +\frac{1}{a_m}\right )+\dots+(a_{m-1}-a_{m-2})\cdot \left ( \frac{1}{a_{m-1}}+\frac{1}{a_m}\right )+$

$+(a_m-a_{m-1})\cdot \frac{1}{a_m}=m \tag{1}$

I seem to see that a "proof without words" is imminent.

A Mathematical Olympiad in the middle of nowhere // O Olimpiadă de Matematică la dracu-n praznic

 The International Mathematical Olympiad "Tuymaada" ...

          Even students from Romania participated there. We read on Andrei Alex ECKSTEIN's blog : "The Tuymaada International Multidisciplinary Olympiad is a competition held annually in Yakutsk, Sakha Republic (Russian Federation). The competition has sections: mathematics, computer science, physics and chemistry. There are two days of competition. Participation is numerically small, for example in 2013 150 students from 6 countries participated, including Romania. Although very far away, participation in the competition is justified by the exceptional quality of the problems. Since 2000, the competition has had a section dedicated to juniors."

       Follow the path : 

HOME$\rightarrow$PROBLEME  DIVERSE$\rightarrow$CONCURSURI$\rightarrow$"TUYMAADA"

     I was interested in "INTERNATIONAL OLYMPIAD "TUYMAADA-2025" (mathematics) Second day" Problems 6 from both the Seniors and Juniors :

     Senior League 6. In a sequence $(x_n)$, the number  $x_1$ is positive and

                             rational, and

 $x_{n+1} = \frac{\{nx_n\}}{n}$    for $n\geqslant 1$ 

                             ($\{a\}$ denotes the fractional part of  $a$). Prove that this   

                             sequence contains only finitely many non-zero terms 

                              and their sum is an integer. 

(V. Kolezhuk, O, Tarakanov )

     Junior League 6. In a sequence $(x_n)$, the first number  $x_1$ is positive,

                              and

 $x_{n+1} =\frac{\{ nx_n\}}{ n}$   for $n\geqslant 1$

                            ($\{a\}$ denotes the fractional part of  $a$). Prove that the

                               sequence does not contain zeroes if and only if  $x_1$ is

                                 irrational.

 (V. Kolezhuk, O, Tarakanov )

                     $\blacklozenge$CiP Comments 

We will refer to these problems by the notations  S6, J6 respectively.

           $\blacklozenge$Problem J6 has a logical aspect

$$\forall n\;(x_n\neq 0)\;\Leftrightarrow\; x_1\notin\mathbb{Q}$$

The statement

$x_1\notin \mathbb{Q}\;\Rightarrow\;\forall n\;(x_n\neq 0)$

is almost trivial: from $\{nx_n\}=nx_n-[nx_n]$ we have  $x_{n+1}=\color{Red}{x_n}-\frac{[nx_n]}{n}$, so

$x_n\notin \mathbb{Q}\;\Rightarrow\;x_{n+1}\notin \mathbb{Q}$,  hence $x_{n+1}\neq 0$. Thus we have  $\forall n\;(x_n \neq 0).$

For statement

$\forall n\;(x_n \neq 0)\;\Rightarrow\;x_1\notin \mathbb{Q}$

we prove its converse instead

$x_1 \in \mathbb{Q}\;\Rightarrow\;\exists n\;(x_n=0) \tag{SJ_1}$

that is a common requirement for both problems J6, S6.

           $\blacklozenge$Let's look at some examples.

          Example 1  $x_1=\frac{2}{3}$

                    $x_2=\frac{\{x_1\}}{1}=\left\{\frac{2}{3}\right \}=\frac{2}{3}\;;\;x_3=\frac{\{2x_2\}}{2}=\frac{\left \{\frac{4}{3}\right \}}{2}=\frac{\frac{1}{3}}{2}=\frac{1}{6}\;;\;x_4=\frac{\{3x_3\}}{3}=\frac{\left \{\frac{3}{6}\right \}}{3}=\frac{\frac{3}{6}}{3}=\frac{1}{6}$

                    $x_5=\frac{\{4x_4\}}{4}=\frac{\left \{\frac{4}{6}\right \}}{4}=\frac{\frac{4}{6}}{4}=\frac{1}{6}\;;\;x_6=\frac{\{5x_5\}}{5}=\frac{\left \{\frac{5}{6}\right \}}{5}=\frac{\frac{5}{6}}{5}=\frac{1}{6}$

                    $x_7=\frac{\{6x_6\}}{6}=\frac{\{1\}}{6}=0$  and from here on out, all  $x_n=0,\;n\geqslant 8$.

                    The sum of the nonzero terms is

$$x_1+x_2+x_3+x_4+x_5+x_6=\frac{2}{3}+\frac{2}{3}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=2 \tag{S_Ex_1}$$

                    We will see later the connection with Egyptian writing

$$\frac{2}{3}=\frac{1}{2}+\frac{1}{6} \tag{E_Ex_1}$$

           Example 2.   $x_1=\frac{3}{4}$

                    $x_2=\frac{\{x_1\}}{1}=\left\{\frac{3}{4}\right\}=\frac{3}{4}\;;\;x_3=\frac{\{2x_2\}}{2}=\frac{\left\{\frac{6}{4}\right \}}{2}=\frac{\frac{2}{4}}{2}=\frac{1}{4}\;;\;x_4=\frac{\{3x_3\}}{3}=\frac{\left\{\frac{3}{4}\right \}}{3}=\frac{\frac{3}{4}}{3}=\frac{1}{4}$

                    $x_5=\frac{\{4x_4\}}{4}=\frac{\{1\}}{4}=0$   and from here on out, all  $x_n=0,\;n\geqslant 6$.

                    The sum of the nonzero terms is

$$x_1+x_2+x_3+x_4=\frac{3}{4}+\frac{3}{4}+\frac{1}{4}+\frac{1}{4}=2 \tag{S_Ex_2}$$

                    and the representation as a sum of Egyptian fractions of  $x_1$  is

$$\frac{3}{4}=\frac{1}{2}+\frac{1}{4} \tag{E_Ex_2}$$

         Example 3.   $x_1=\frac{17}{5}$

                   $x_2=\frac{\{x_1\}}{1}=\left\{ \frac{17}{5}\right \}=\frac{2}{5}\;;\;x_3=\frac{\{2x_2\}}{2}=\frac{\left\{\frac{4}{5}\right\}}{2}=\frac{\frac{4}{5}}{2}=\frac{2}{5}\;;\;x_4=\frac{\{3x_3\}}{3}=\frac{\left\{\frac{6}{5}\right\}}{3}=\frac{\frac{1}{5}}{3}=\frac{1}{15}$

                    $x_5=\frac{\{4x_4\}}{4}=\frac{\left\{\frac{4}{15}\right\}}{4}=\frac{1}{15}\;;\;x_6=\frac{\{5x_5\}}{5}=\frac{\left\{\frac{5}{15}\right\}}{5}=\frac{\frac{5}{15}}{5}=\frac{1}{15}\;;\;x_7=\frac{\{6x_6\}}{6}=\frac{\left\{\frac{6}{15}\right\}}{6}=\frac{\frac{6}{15}}{6}=\frac{1}{15}$

                    $x_8=\frac{\{7x_7\}}{7}=\frac{\left\{\frac{7}{15}\right\}}{7}=\frac{\frac{7}{15}}{7}=\frac{1}{15}\;;\;x_9=\frac{\{8x_8\}}{8}=\frac{\left\{\frac{8}{15}\right\}}{8}=\frac{\frac{8}{15}}{8}=\frac{1}{15}\;;\;x_{10}=\frac{\{9x_9\}}{9}=\frac{\left\{\frac{9}{15}\right\}}{9}=\frac{\frac{9}{15}}{9}=\frac{1}{15}$

                    and so on...  $x_{11}=\frac{1}{15}=x_{12}=x_{13}=x_{14}=x_{15}\left (=\frac{\{14x_{14}\}}{14}=\frac{\left\{\frac{14}{15}\right\}}{14}=\frac{\frac{14}{15}}{14}=\frac{1}{15}\right )$

                   but(!)  $x_{16}=\frac{\{15x_{15}\}}{15}=\frac{\{1\}}{15}=0$  and from here on out,  $x_n=0,\;n\geqslant 17$.

                   The sum of the nonzero terms is

$$x_1+x_2+x_3+x_4+\dots+x_{15}=\frac{17}{5}+2\cdot\frac{2}{5}+12\cdot \frac{1}{15}=5\tag{S_Ex_3}$$

                    and the representation as a sum of Egyptian fractions of  $x_1$  is

$$\frac{17}{5}=3+\frac{1}{3}+\frac{1}{15} \tag{E_Ex_3}$$

          Example 4.  $x_1=\frac{7}{6}$   

                    $x_2=\frac{\{x_1\}}{1}=\left \{\frac{7}{6}\right\}=\frac{1}{6};$ without further calculation, so in the previous examples, we have

                    $x_3=x_4=x_5=x_6=\frac{1}{6},\;x_7=\frac{\{6x_6\}}{6}=\frac{\{1\}}{6}=0$, and furher  $x_n=0,\;n\geqslant 8$.

                    The sum of the nonzero term is

$$x_1+x_2+\dots+x_6=\frac{7}{6}+5\cdot \frac{1}{6}=2 \tag{S_Ex_4}$$

                    and  $x_1$  has the representation as the sum of Egyptian fractions

$$\frac{7}{6}=1+\frac{1}{6} \tag {E_Ex_4}$$

          Example 5.   $x_1=\frac{3}{7}$

                     We quickly see that  $x_3=x_2=\{x_1\}=\frac{3}{7}$;  then that  $x_{11}=x_{10}=\dots=x_4=\frac{\{3x_3\}}{3}=\frac{\left\{\frac{9}{7}\right\}}{3}=\frac{\frac{2}{7}}{3}=\frac{2}{21}$;

                    and the eye, increasingly experienced, sees that  $x_{231}=x_{230}=\dots =x_{12}=\frac{\{11 x_{11}\}}{11}=\frac{\left\{\frac{22}{21}\right\}}{11}=\frac{\frac{1}{21}}{11}=\frac{1}{231}$. 

                    We're almost done, because  $x_{232}=\frac{\{231x_{231}\}}{231}=\frac{\{1\}}{231}=0$  and  $x_n=0,\;n\geqslant 232$.

                    The sum of the nonzero terms is

$$x_1+x_2+x_3+(x_4+x_5+\dots+x_{11})+(x_{12}+x_{13}+\dots+x_{231})=3\cdot \frac{3}{7}+8\cdot \frac{2}{21}+220\cdot \frac{1}{231}=3 \tag{S_Ex_5}$$

                   and  $x_1$  has the representation as the sum of Egyptian fractions

$$\frac{3}{7}=\frac{1}{3}+\frac{1}{11}+\frac{1}{231} \tag{E_Ex_5}$$

 

           $\blacklozenge$The hard truth about the sequence $(x_n)$ is, this sequence being the protocol of a slow and sad decomposition of a rational number  $x_1=\frac{a}{b}$ to the form - pay attention to colors -

$\frac{a}{b}=\color{Blue}s+\frac{1}{k_1}+\dots+\frac{1}{k_{\color{Red}m}} \tag{E_1}$

where

$s=[x_1],\;\;k_1<k_2<\dots<k_m \tag{E_2}$

$k_i,\;i=1,\dots m,$  are positive integers, and

$x_n=\frac{1}{k_i}+\dots +\frac{1}{k_m}\;\;for\;\;k_{i-1}<n\leqslant k_i  \tag{E_3}$

          Moreover, the sum of the nonzero terms of the sequence is equal to (!!colors!!)

$x_1+x_2+\dots+x_{k_m}=\color{Blue}s+\color{Red}m \tag{S}$

     Returning to the previous examples  

          In Example 1,  $x_7=0,\;\;x_6=x_5=x_4=x_3=\frac{1}{6},\;\;x_2=x_1=\frac{2}{3}=\frac{1}{2}+\frac{1}{6}.$ So  $\color{Blue}{s=0},\;\color{Red}{m=2},\;k_1=2,\;k_2=6.$

           In Example 2,  $x_5=0,\;\;x_4=x_3=\frac{1}{4},\;\;x_2=x_1=\frac{3}{4}=\frac{1}{2}+\frac{1}{4}.$ So $\color{Blue}{s=0},\;\;\color{Red}{m=2},\;k_1=2,\;k_2=4.$

           In Example 3,  $x_{16}=0,\;\;x_{15}=x_{14}=\dots=x_4=\frac{1}{15},\;\;x_3=x_2=\frac{2}{5}=\frac{1}{3}+\frac{1}{15}.$ So  $\color{Blue}{s=3},\;\;\color{Red}{m=2},\;k_1=3,\;k_2=15.$

           In Example 4,  $x_7=0,\;\;x_6=x_5=\dots=x_2=\frac{1}{6}$. So  $\color{Blue}{s=1},\;\;\color{Red}{m=1},\;k_1=6.$

           In Example 5,  $x_{232}=0,\;\;x_{231}=x_{230}=\dots=x_{12}=\frac{1}{231},\;\;x_{11}=x_{10}=\dots=x_4=\frac{2}{21}=\frac{1}{11}+\frac{1}{231},\;\;$

            $x_3=x_2=\frac{3}{7}=\frac{1}{3}+\frac{1}{11}+\frac{1}{231}.$ So  $\color{Blue}{s=0},\;\;\color{Red}{m=3},\;k_1=3,\;k_2=11,\;k_3=231.$

            All relations (E_1), (E_3), (S) are verified.

          $\blacklozenge$A final example, treated more expeditiously:

              Let  $x_1=\frac{7}{8}$. The calculations show that

$\color{Blue}{s=0},\;\;k_1=2,\;\;k_2=3,\;\;k_3=24,\;\;\color{Red}{m=3}$

$x_{25}=0,\;\;x_{24}=\dots=x_4=\frac{1}{24},\;\;x_3=\frac{3}{8}=\frac{1}{3}+\frac{1}{24},\;\;x_2=x_1=\frac{7}{8}=\frac{1}{2}+\frac{1}{3}+\frac{1}{24}$

and

$x_1+\dots +x_{24}=2\cdot \frac{7}{8}+\frac{1}{3}+21\cdot \frac{1}{24}=3$                  

<end CiP comments>$\blacklozenge$

                    Proving that the sequence contains

only a finite number of nonzero terms

Let's prove (SJ_1) that is, if   $x_1$ is a rational number then the sequence contains a term (and from here on all terms) equal to zero.

          If  $x_1\in\mathbb{Z}$, then  $\{x_1\}=0$  and statement (SJ_1) is true.

          Otherwise,  $x_2\;\overset{def}{=}\;\frac{\{1\cdot x_1\}}{1}=\{x_1\} \in (0;1)$  and we will clarify this case in the following. 

     So, let

$x_2=\frac{a}{b},\;\;a<b \tag{1}$

 and  $a\; and\; b$  are positive coprime integers.

We choose  $k$-the smallest natural number such that 

$x_2 \geqslant \frac{1}{k} \tag{2}$

Obviously  $k\geqslant 2$.

               If  $k=2$, that is  $\frac{a}{b}=x_2\geqslant 2$ (so $2a-b\geqslant 0$ we'll use that later)

we have  $2x_2-1\geqslant 0$ and, as  $x_2<1$, we also have  $2x_2-1<1$. Using the periodicity of the function  $\{\cdot\}\;(i.e.\;\{\alpha\}=\{\alpha-1\})$  we find

$x_3=\frac{\{2x_2\}}{2}=\frac{\{2x_2-1\}}{2}\underset{0\leqslant 2x_2-1<1}{===}\frac{2x_2-1}{2}=x_2-\frac{1}{2} \tag{3}$

With (1) this is written

$x_3=\frac{2a-b}{2b},\;\;\;0\leqslant 2a-b \;\overset{\color{Red}!a<b}{<}\;a \tag{4}$

     So, if  $x_3$  is not eqal zero, then the numerator of  $x_3$  in irreducible form, is  $\leqslant 2a-b$  hence (cf. (4)) less than the numerator of  $x_2$.

<end case $k=2$>

              If  $k>2$  we will prove the following facts :

(i)    $2\leqslant i \leqslant k\;\;\Rightarrow\;\;x_i=x_2$

(ii)   $x_{k+1}=x_2-\frac{1}{k} \tag{5}$

          We show (i) one by one, up close and close.

         Obviously (i) holds for  $i=2$. 

We assume the statement is true for some  $i$, with  $i\geqslant 2\;\;and\;i+1\leqslant k$. We have  $i\leqslant k-1$  and, considering how we defined the number  $k$, it results (cf. (2))  $x_2<\frac{1}{i}.$  But then

$x_{i+1}=\frac{\{i \cdot x_i\}}{i}\underset{\overbrace{x_i=x_2}}{==}\frac{\{i \cdot x_2\}}{i}\overset{\underbrace{ix_2<1}}{==}\frac{i\cdot x_2}{i}=x_2$

 so  (i)  is also true for  $i+1$.(It's like a kind of Induction.)

             For  (ii) , since from the way we chose the number $k,\;\;k-1$ does not verify  (2) - that is  $x_2<\frac{1}{k-1}$-, we have the inequalities

$$\frac{1}{k}\overset{(2)}{\leqslant} x_2\underset{(1)}{=}\frac{a}{b}<\frac{1}{k-1}\overset{k>2}{<}\frac{2}{k} \tag{6}$$

Then  $b\leqslant k\cdot a<2b$, so  $0\leqslant ka-b<b\;\left ( \Leftrightarrow 0\leqslant \frac{ka-b}{b}<1\right )$,  and

  $x_{k+1}=\frac{\{k\cdot x_k\}}{k}=\frac{\{k\cdot x_2\}}{k}=\frac{\left\{\frac{ka}{b}\right\}}{k}=\frac{\left \{\frac{ka}{b}-1\right \}}{k}=\frac{\left \{\frac{ka-b}{b}\right \}}{k}=\frac{\frac{ka-b}{b}}{k}=\frac{ka-b}{kb}=x_2-\frac{1}{k} \tag {7}$

     From  $(k-1)a<b$ (cf.(6)) we have  $ka-b<a$  and from  $x_{k+1}=\frac{ka-b}{kb}$  we see that the numerator of  $x_{k+1}$  in irreducible form, is  $\leqslant ka-b$  hence less than the numerator  $a$  of  $x_2$.

<end case $k>2$>

(Notice that  (3)  is embedded in  (7) for  $k=2$.)

          "As the numerators cannot decrease infinitely, at some moment the next term will be  0." (We have reproduced the words of the authors of the Official Solution.)

<<end of Proof (SJ_1)>> $\blacksquare$

Before moving on, let's try to find out the 

Structure of  the Sequence  $(x_n)_{n\geqslant 1}$  

and its Connection with the Egyptian Fractions

Let  $k_m$  be the index of the last nonzero term. From  $x_{k_m+1}=0$, applying the analogous of (5)

$x_{k_m+1}=x_{k_m}-\frac{1}{k_m} \tag{8}$ 

it result

$x_{k_m}=\frac{1}{k_m} \tag{9'}$

(The same thing can be obtained from  $x_{k_m+1}=\frac{\{k_m\cdot x_{k_m}\}}{k_m}=0\;\;\overset{\underbrace{x_n<\frac{1}{n-1}}}{\Rightarrow}\;\;k_m\cdot x_{k_m}=1\;\;etc$  )

     Let  $k_{m-1}$  be the largest index  $<k_m$  for wich  $x_{k_{m-1}}\neq x_{k_m}.$  We have

$\frac{1}{k_m}=x_{k_m}=x_{k_m-1}=\dots =x_{k_{m-1}+1} \tag{10'}$

abd from  (8)  written for  $k_{m-1}\;:$

$x_{k_{m-1}+1}=x_{k_{m-1}}-\frac{1}{k_{m-1}}\overset{(10')}{\Leftrightarrow}\frac{1}{k_m}=x_{k_{m-1}}-\frac{1}{k_{m-1}}$

it result

$x_{k_{m-1}}=\frac{1}{k_{m-1}}+\frac{1}{k_m} \tag{9''}$

     If  $k_{m-2}$  is the largest index $<k_{m-1}$  for wich  $x_{k_{m-2}}\neq x_{k_{m-1}}$, that is

$\frac{1}{k_{m-1}}+\frac{1}{k_m}=x_{k_{m-1}}=x_{k_{m-1}+1}=\dots=x_{k_{m-2}+1} \tag{10''}$

then we get

$\frac{1}{k_{m-1}}+\frac{1}{k_m}\;\overset{(10'')}{=}$$x_{k_{m-2}+1}\overset{(8)}{\underset{for\;k_{m-2}}{=}}\;x_{k_{m-2}}-\frac{1}{k_{m-2}}$

and it result

$x_{k_{m-2}}=\frac{1}{k_{m-2}}+\frac{1}{k_{m-1}}+\frac{1}{k_m} \tag{9'''}$

(9'), (9''), (9''')  as well as  (10'), (10'')  confirm the equations  (E_3)  from CiP_Comments above.

          Further, in the words of the Authors of the Official Solutions :

 "Restoring the sequence backwards in this way, we arrive at the desired formula."

It's about the formula  (E_1), but the statement doesn't seem very convincing.

          Let's say it's still good. But I hope it doesn't create confusion that I used the fraction  $\frac{a}{b}$  in (1), in a different context. 

          I will now present one last example, to show that we can start the algorithm with any real number. The reader is advised to treat Examples 1-5 and the final one in my comments at the beginning in the same way. Let's choose a favorite number of Archimedes:

$x_1=-\frac{22}{7}$

$\color{Blue}s=[x_1]=\left [-4+\frac{6}{7}\right ]=\color{Blue}{-4}$

$x_2=\{x_1\}=\frac{6}{7}\;;$

we have  $\frac{6}{7}\geqslant \frac{1}{2}$  so we choose in  (2)  $k_1=2$. The next term will therefore be different from  $x_2$. Let's find it :

$x_3=\frac{\{2x_2\}}{2}=\frac{\left \{\frac{12}{7}\right \}}{2}=\frac{\left \{\frac{5}{7}\right \}}{2}=\frac{\frac{5}{7}}{2}=\frac{5}{14}\;;$

the calculation is also in agreement with  (3). We have  $\frac{5}{14}\geqslant 3$ so we choose in  (2)  $k_2=3$. Because  $k_2-k_1=3-2=1$, this value appears only once in the string. Let's move on.

$x_4=\frac{\{3x_3\}}{3}=\frac{\left\{\frac{15}{14}\right \}}{3}=\frac{\frac{1}{14}}{3}=\frac{1}{42}$

and obvously  $x_4\geqslant \frac{1}{42}$  so  $k_3=42$. All of the following  $k_3-k_2=42-3=39$  terms have the same value :

$x_4=x_5=\dots=x_{42}=\frac{1}{42}$

and finally  $x_{43}=0.$  So  $\color{Red}{m=3}$.

          The calculation of the sum of nonzero terms has, in the official solution, the aspect

$x_1+x_2+\dots +x_{k_m}=s+k_1 \cdot \frac{1}{k_1}+k_2\cdot \frac{1}{k_2}+\dots +k_m\cdot \frac{1}{k_m}=s+m$

This did not appear obviously in the calculations of the previous examples, although the result was correct. That this is indeed the case follows if we write the terms in the form :

$x_4=\color{Violet}{\frac{1}{42}}\;\;\;\;\}\;k_3-k_2=39\;values$

$x_3=\color{Teal}{\frac{1}{3}}+\color{Violet}{\frac{1}{42}}\;\;\;\}\;k_2-k_1=1\;value$

$x_2=\color{Red}{\frac{1}{2}}+\color{Teal}{\frac{1}{3}}+\color{Violet}{\frac{1}{42}}\;\;\}\;k_1-1=1\;value$

$x_1=\color{Blue}{-4}+\color{Red}{\frac{1}{2}}+\color{Teal}{\frac{1}{3}}+\color{Violet}{\frac{1}{42}}\;\;\}\;1\;value$

We see that each value $\frac{1}{k_i}$  appears  $k_i$ times. So  this sum is  $\color{Blue}s+\color{Red}m=-4+3=-1$.

$\blacksquare\;\blacksquare$

Problem #1 from JBMO TEAM SELECTION TEST 2025 - GREECE

 Obtained from here.

         "Problem 1.

           (a) Let the positive integers  $p,\;q$  be prime numbers and let  $a$  be a positive

           integer. If  $a$  divides the product  $p\cdot q$ , and it holds that  $a>p$  and  $a>q$ , 

           prove that  $a=pq$.

          (b) Determine all pair  $(p,q)$  of prime numbers such that  $p^2+3pq+q^2$ 

           eqals a perfect square."

ANSWER CiP

(b)  $(3,\;7)$  and  $(7,\;3)$

Solution CiP

(a)  $a\mid p\cdot q\;\Rightarrow\;\;p\cdot q=a\cdot b$  for a certain  $b\in \mathbb{N}$. Noting that  $p$  divides the product $a\cdot b$ then, since it is prime, it follows

$$p\mid a\;\;\;or\;\;\;p\mid b.$$

      If  $p\mid a$, then  $a=p\cdot c$  for a certain  $c\in \mathbb{N}$. Then, from

$p\cdot q= (p\cdot c)\cdot b$  we obtain  $q=c\cdot b$. But  $q$  is also prime so we can have  $c=1$ (when  $q=b,\;p=a$  which contradicts  $a>p$)  or $b=1$ , in which case  $q=c$  and $a=pq$.

      If  $p\mid b$,  then $b=p\cdot d$  for a certain  $d\in\mathbb{N}$. Then, from  $p\cdot q=a(p\cdot d)$  we obtain  $q=a\cdot d$, but $q$ being prime and  $a>1$  it follow  $d=1,\;q=a$  wich contradicts  $a>q$.

With these,  (a)  is proven.

(b) If  $k\in \mathbb{N}$ is such that  $p^2+3pq+q^2=k^2$  then we get

$$pq=k^2-p^2-2pq-q^2=k^2-(p+q)^2=(k-p-q)\cdot (k+p+q)$$

From the above it can be seen that the number  $a:=k+p+q>p,\;q$  divides the product  $pq$. According to (a) we must have $a=pq$  that is, equivalent to

$k+p+q=pq\;\Rightarrow\;k=pq-p-q\;\;\Rightarrow\;\;p^2+3pq+q^2=(pq-p-q)^2\;\:\Leftrightarrow$

$\Leftrightarrow\;pq=p^2q^2-2p^2q-2pq^2\;\;\Leftrightarrow\;\;1=pq-2p-2q\;\Leftrightarrow\;5=pq-2p-2q+4\;\Leftrightarrow$

$$\Leftrightarrow\;\;\;5=(p-2)(q-2).$$ 

Then it follows that  $p-2=1$  or  $p-2=5$. We get the answer.

          Verification:  $3^2+3\cdot 3\cdot 7+7^2=9+63+49=121=11^2$.

$\blacksquare$

An UNSOLVED completely problem from Sierpiński

 It is Problem #4, page 16 mentioned in the cited work. Edited from the manuscript.

In translation : 

                           "Prove that there are infinitely many natural numbers  $n$  for which

                          the number  $4n^2+1$ is divisible by both  $5$  and  $13$."

               The author's[WS] solution is on page 38. (The text is written in green.) In translation:

                         "ANSWER : All numbers in the arithmetic progression

 $65k+56,\;\;k=0,\;1,\;2\dots$

[I noticed at the bottom of the page in the first photo that there is no indication of how to find the answer. Then comes its Solution :]

          Indeed, if  $n=65k+56$ , $k\geqslant 0$ is an integer, then  $n\equiv 1\;(mod\;5)$ and $n\equiv 4\;(mod\;13)$ from where  $4n^2+1\equiv 0\;(mod\;5)$ and  $4n^2+1\equiv 0\;(mod\;13)$, so that 

$5\mid 4n^2+1$  and  $13\mid 4n^2+1.$

$\color {Green}{\blacksquare}$"

                I solved the problem without knowing this answer. [Text written in blue; in translation:]

ANSWER CiP

The numbers that have the property in the statement are exactly those of the form

$65k+4,\;\;65k+9,\;\;65k+56,\;\;65k+61$  where  $k=0,\;1,\;2,\dots\;.$

                   Solution CiP

               Since  $5$  and  $13$  are coprime, we have

$5\mid A\;\;and\;\;13 \mid A\;\;\;\Leftrightarrow\;\;\;5\cdot 13 \mid A$

Let  $n=65k+r$ ;  then  $4n^2+1=4(65k+r)^2+1=4(65^2k^2+2\cdot 65\cdot r+r^2)+1=$

$=65\cdot(4\cdot 65k^2+8r)+4r^2+1=65k_1+4r^2+1.$

Trying to choose a number  $r$  so that  $4r^2+1=65$  we get  $r^2=16$. For example  $r=4$, so we have an infinity of numbers, of the form  $n=65k+4$, with the property in the statement. The problem would be solved [, but not completely.]

$\color {Blue}{\blacksquare}$

       I noted in red, further on, that we can also have  $r=-4$, obtaining another infinity of convenient numbers. 

         - And all the convenient values ​​found in my answer were obtained by checking all the possibilities

$65k,\;65k\pm1,\;65k\pm2,\;\dots,\;65k\pm32$

     Finally, I also noted, [written in black], that : The numbers in ANSWER CiP form the sequence

 A203464 in OEIS ("The On-Line Encyclopedia of Integer Sequences).