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

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

 

           $\blacklozenge$fghn

                   

<end CiP comments>$\blacklozenge$

<de continuat>

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

$\blacksquare$