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

$\blacksquare$

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).

АXA! Немам ПОИМ!! // AHA! I have NO IDEA!!

 It's a parody of the title of Martin ERICKSON's  book "Aha! Solutions" 

          The Problems Given at the Dam for JBMO in North Macedonia fell into my hands. I said I'd try my hand at Problem 1. 

            "1.  Let  $n>1$ be a natural number and  $m>2$  a divisor of  $2n$. Prove that

               the number  $n^2$ can be written as the sum of  $m$ nonzero perfect squares."

          I have no idea how to solve this, so for now I'm just showing a few

          EXAMPLES CiP

           a) $n=2$ :  $2<m \mid 4\;\;\Rightarrow\;\;m=4$.  We have the equation

$2^2=1+1+1+1$

          b) $n=3$ :  $2<m \mid 6\;\;\Rightarrow\;\;m=3$ or $m=6$. We have the equations

$3^2=1+4+4$

$3^2=1+1+1+1+1+4$

          c) $n=4$ :  $2<m \mid 8\;\;\Rightarrow\;\;m=4$ or $m=8$. We have the equations

$4^2=4+4+4+4$

$4^2=1+1+1+1+1+1+1+9$

Let's hope that inspiration will help me solve the problem.

Edited Friday 06 June  The following Lemma, which I will now formulate only as a Conjecture, would be helpful:

               Lemma CiP (Conjecture) Whatever the prime number  $p>2$, the

                                                  number  $p^2$ can be written as the sum of  $p$ squares.

          So we have an equation like this

$$p^2=a_1^2+a_2^2+\dots +a_p^2=\sum_{i=1}^pa_i^2 \tag{P}$$

          Examples:

$3^2=\underset{3-terms}{\underbrace{1+4+4}}$

$5^2=\underset{5-terms}{\underbrace{4+4+4+9}}$

$7^2=\underset{7-terms}{\underbrace{1+1+4+9+9+9+16}}$

For the following example we proceed by trial and error:

$11^2=\underset{4-terms}{\underbrace{5+16+36+64}}\;\overset{we\;replace\; 16\; with\; a}{\underset{sum\;of\;several\;terms}{=}}\;\underset{7-terms}{\underbrace{5+4+4+4+4+36+64}}=$

and now if we replace the term  $5$ with the sum  $1+1+1+1+1$  we are lucky to obtain the desired result, so

$11^2=\underset{11-terms}{\underbrace{1+1+1+1+1+4+4+4+4+36+64}}$

Until we find a proof for the Lemma, let us observe that if for two prime numbers $p>2$ and $q>2$ we have decompositions  (P) and

$$q^2=\sum_{j=1}^qb_j^2$$

then because

$$\left ( \sum_{i=1}^pa_i^2\right )\cdot \left (\sum_{j=1}^qb_j^2 \right )=\sum_{i,j=1}^{i=p,j=q}a_i^2b_j^2$$

we obtain a sum of  $p\cdot q$  squares 

$$p^2\cdot q^2=\sum_{i,j=1}^{i=p,j=q}(a_i\cdot b_j)^2 \tag{PQ}$$

<end Edit 06 June>

Weekend Edition (There are no days off in Mathematics. When you are struggling with a problem, you are constantly thinking about it.) 

          But what if the property in yesterday's Lemma holds for any number  $n>2$ ? 

          Also by trying, as for number  $121$ , I obtained the following equations:

$4^2=\underset{4-terms}{\underbrace{4+4+4+4}}$

$6^2=\underset{6-terms}{\underbrace{1+1+1+4+4+25}}$

$8^2=\underset{8-terms}{\underbrace{1+1+1+1+1+1+9+49}}$

$9^2=\underset{9-terms}{\underbrace{1+1+1+1+1+4+4+4+64}}$

Ugh!, no pattern is visible. Worse, if we start from equation for  $3^2=1+4+4$ and square it (or rather multiply it by itself),

$3^2 \cdot 3^2=(1+4+4)\cdot (1+4+4)=\dots $ (the $3\times 3=9$ terms obtained by multiplication are all squares)

 we find an equation for  $9^2$ that differs from what I obtained:

$9^2=\underset{9-terms}{\underbrace{1+4+4+4+4+16+16+16+16}}$

So, we don't have unique writings for the representations we're looking for. Complicated stuff...

<end Weekend Edition>

          I couldn't be patient anymore and I consulted the solution to the problem. As expected, the problem should be quite simple. I'm saved !

          You can also consult the solution from where I got the problem statement.

          As a consolation for how much I've been struggling these days, the statement that I considered above as a Lemma is true. In fact, the following are true:

    (a)   For any number  $n\geqslant 3$, the number  $n^2$ can be written as a sum

            of  $n$ nonzero perfect squares.

    (b)   For any number  $n\geqslant 2$, the number  $n^2$ can be written as a sum

             of  $2\cdot n$ nonzero perfect squares.

 To my shame, these statements are almost trivial. If you only showed them, you would get 2 points on the solution scale. For just one, you would get nothing.

        Proof of (a) : We have that

  $n^2=(n^2-4\cdot n +4)+4\cdot n-4=(n-2)^2+4\cdot (n-1)$  so

$$n^2=\underset{1-term}{\underbrace{(n-2)^2}}+\underset{n-1\;-terms}{\underbrace{4+4+\dots+4}}$$

     Examples (that differ from those I found) : 

$6^2=4^2+4+4+4+4+4$

$7^2=5^2+4+4+4+4+4+4$

$8^2=6^2+4+4+4+4+4+4+4$

$9^2=7^2+4+4+4+4+4+4+4+4$

$10^2=8^2+4+4+4+4+4+4+4+4+4$

$11^2=9^2+4+4+4+4+4+4+4+4+4+4$

       Proof of (b) :  We have that

$n^2=(n^2-2\cdot n+1)+2n-1=(n-1)^2+(2n-1)\cdot 1$   so

$$n^2=\underset{1-term}{\underbrace{(n-1)^2}}+\underset{2\cdot n-1\;-terms}{\underbrace{1+1+\dots +1}}$$

Note that (a) is the particular case with  $m=n$ of the Problem, and (b) is the particular case  $m=2n$ of it.

                    Official Solution (adapted by CiP)

Let  $k=\frac{2n}{m}$; it is, from the divisibility condition, a natural number, $k\geqslant 1$. From

$n^2=(n^2-2 n k+k^2)+2nk-k^2\;\;\overset{2n=k\cdot m}{=}\;(n-k)^2+km\cdot k-k^2=(n-k)^2+k^2 \cdot (m-1)$

and because  $n-k=\frac{2n}{2}-k=\frac{k\cdot m}{2}-k=k\cdot \left (\frac{m}{2}-1\right )>0$

we have

$$n^2=(n-k)^2+\underset{m-1\;-terms}{\underbrace{k^2+\dots +k^2}}$$

QED $\blacksquare$

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...