Doua Extemporale de pe vremea Liceului (clasa a 9-a/anul I)

 

(in construction)

A 7TH GRADE PROBLEM FROM NMO // O PROBLEMA DE CLASA A 7-A DE LA ONM

                     We will solve the following problem (It is problem 3 from NMO 2026, proposed for grade 7 page 5):

               Determine the pairs  $(a,b)$ of nonzero natural numbers for which the

 numbers  $\frac{a^2+b}{a+b-1}$  and  $\frac{b^2+a}{a+b}$  are natural.

{author} Lucian DRAGOMIR, Oțelu Roșu

Vocabulary  :  NMO = National Mathematics Olympiad

(in construction)

O CONTRADICTIE LA CARE NU TE ASTEPTAI // A CONTRACTION YOU DIDN'T EXPECT // EGY ÖSSZEHÚZÓDÁS, AMIRE NEM SZÁMÍTOTTAD

           Seeing the "Second solution" (on page 16) to the problem in the post 

A 7TH GRADE PROBLEM FROM NMO // O PROBLEMA DE CLASA A 7-A DE LA ONM 

reminded me of the demonstration, which had long annoyed me, of the fact that  $\sqrt{2}$  is irrational. I consulted the textbook when I was in the first year of high school (today's 9th grade) :

BOGDANOF Z., GEORGESCU-BUZAU E., PANAITOPOL L.  

-  ALGEBRA : Manual pentru anul I licee,

Ed. Did. si Ped., Bucuresti, 1975

 Here's what it says on page 23 (starting with the 5th; minor retouching by CiP) : 

"There is no rational number whose square is 2.

(in the context of the statement that "the equation  $x^2=2$  has no solutions in the set of rational numbers"). The proof is done by contradiction :

Suppose there is a rational number  $\frac{m}{n}$  such that  $\frac{m^2}{n^2}=2\;\;\;\;\;(1)$

   We can assume that the fraction  $\frac{m}{n}$  is irreducible (otherwise we simplify with the greatest common divisor of the numbers m and n).

(1)$\;\;\Rightarrow m^2=2n^2\Rightarrow 2\mid m^2\Rightarrow 2\mid m$

so  $m=2q\;(q\in\mathbb{Z})$. Then  $(2q)^2=2n^2\Leftrightarrow 2q^2=n^2$, so  $2\mid n^2$  and then  $2\mid n$.

     Therefore, the fraction  $\frac{m}{n}$  is not irreducible (both the numerator  $m$ and the denominator  $n$ are divisible by 2). A contradiction was obtained with the hypothesis that the fraction $\frac{m}{n}$  is irreducible."

          This made me unhappy, because in fact the contradiction is NOT with the hypothesis, but with an assumption made during the demonstration. It is possible, I have heard, to logically formalize such a path.

(I am also unhappy because I fell in love with a woman ALINA, to whom I plan to dedicate a page on this blog, with our story (I still don't know for sure, at this time, how it will end). She is a big fan of a song.... I also promoted it on Twitter, I'm putting it here too : Three Seconds - Sia Ft Damian Marley

)

          Second Solution

          If  $a=1$  then

  $m=\frac{a^2+b}{a+b-1}=\frac{1+b}{b}=1+\frac{1}{b},\;\;n=\frac{b^2+a}{a+b}=\frac{b^2+b-b-1+2}{1+b}=b-1+\frac{2}{1+b}$

  and  $m,\;n$  are integers if and only if  $b=1$.

If  $b=1$  then  $m=\frac{a^2+1}{a}=a+\frac{1}{a},\;\;n=\frac{1+a}{a+1}=1$  and  $m$  is integer if and only if  $a=1$.

          Let, from now on,  $a,\;b \geqslant 2.$

(in construction)

proof by contradiction

UN NUMAR $\pi$ CIUDATCARE APARE IN REZOLVARE

(in construction)

Gheorghe ANDREI - COMPLEX NUMBERS (part II) - Some Problems -- 4 / Gheorghe ANDREI - NUMERE COMPLEXE (partea II) - Câteva Probleme -- 4 / Gheorghe ANDREI – LICZBY ZŁOŻONE (część II) – Niektóre problemy – 4

 I wrote about the book here.

                       We discuss some Identities with two or more complex variables. 

               I took a look at the classic works, of which I have the following:

                       @ EVGRAFOV M., BÉJANOV K., SIDOROV Y., FÉDORUK M, CHABOUNINE M.

RECUEIL DE PROBLÈMES SUR LA THÉORIE DES FONCTIONS ANALYTIQUES

MIR, Moscou, 1974

                     @@ Problems in the theory of functions of a complex variable , authors        L. Volkovysky, G. Lunts, I. Aramanovich (translated from the Russian by Victor Shiffer)  Mir ; Moscow; 1972 (I only have it in paper format)

                   @@@ ВОЛКОВЫСКИЙ Л. И., ЛУНЦ Г. Л., АРАМАНОВИЧ И. Г.

Сборник задач по теории функций комплексного переменного

ФИЗМАТЛИТ, Москва, 2004

                @@@@ VOLKOVYSKY I., LUNTS G., ARAMANOVICH I.

Problems in the Theory of Functions of a Complex Variable

MIR, Moscow, 1977 

              @@@@@ VOLKOVYSKII L[ev] I[zrailevich], LUNTS, G[rigorii] L[‘vovich],

             ARAMANOVICH I[saak] G[enrikhovich]

A Collection of Problems on Complex Analysis

DOVER PUBLICATIONS, INC. New York, 1991

    

.....................................................................................................................................

          From  @   we take problem 1.68 from page 19 and 1.59 from page 18 :

         1.68.  Show that for all complex values ​​of  $z$  and  $\zeta$  the following equalities are valid :

   1. $|z+\zeta|^2+|z-\zeta|^2=2|z|^2+2|\zeta|^2$

   2. $|z\bar \zeta+1|^2+|z-\zeta|^2=(1+|z|^2)(1+|\zeta|^2)$

   3. $|z\bar \zeta-1|^2-|z-\zeta|^2=(|z|^2-1)(|\zeta|^2-1)$

 

           1.59.  Show that for no matter what complex number  $z$,  the formula below is valid :

$|\sqrt{z^2-1}+z|+|\sqrt{z^2-1}-z|=|z-1|+|z+1|.$

            From  @@  we take problems 9, 10, 13  from page 12 :

             9.  Prove the identity :

$|z_1+z_2|^2+|z_1-z_2|^2=2(|z_1|^2+|z_2|^2.$

           10.  Prove the identity :

$|1-\bar z_1z_2|^2-|z_1-z_2|^2=(1-|z_1|^2)(1-|z_2|^2).$

           13.  Prove the foloowing identities :

  $$1)\;\;(n-2)\sum_{k=1}^n|a_k|^2+\bigg |\sum_{k=1}^na_k\bigg|^2=\sum_{1\leqslant k<s\leqslant n}|a_k+a_s|^2;$$

$$2)\;\;n\sum_{k=1}^n|a_k|^2-\bigg | \sum_{k=1}^na_k \bigg |^2=\sum_{1\leqslant k<s\leqslant n}|a_k-a_s|^2.$$

We will list from the basic work in the title the issues more or less related to this topic.

          Problem 35* (page 25)  Show that for any  $z\in \mathbb{C}$  the relation holds :

$|\sqrt{z^2-1}+z|+|\sqrt{z^2-1}-z|=|z-1|+|z+1|.$

          Problem 53 (page 27)  Show that :

$2|\sqrt{z^2-1}+z|=|z+1|+|z-1|+\sqrt{(|z+1|^2+|z-1|^2)-4},\;\;\;\forall z\in \mathbb{C}.$

          Problem 55 (page 27)   If  $z\in \mathbb{C}$  then

$|\sqrt{z^2+2z}+z+1|+|\sqrt{z^2+2z}-(z-1)|+|z-2|=$

$=|\sqrt{z^2-2z}+z-1|+|\sqrt{z^2-2z}-(z-1)|+|z+2|.$

          Problem  1 (page 28-29)   To check the equalities :

a)  $|z_1+z_2|^2+|z_1-z_2|^2=2(|z_1|^2+|z_2|^2);$

b)  $|z_1|^2+|z_2|^2-z_1 \bar z_2-\bar z_1 z_2=|z_1-z_2|^2;$

c)  $|z_1\bar z_2+1|^2+|z_1-z_2|^2=(|z_1|^2+1)(|z_2|^2+1);$

d)  $|z_1 \bar z_2-1|^2-|z_1-z_2|^2=(|z_1|^2-1)(|z_2|^2-1);$

e)  $|z_1+2z_2|^2-2|z_1+z_2|^2=2|z_2|^2-|z_1|^2.$

          Problem  2 (page 29)  If  $z_1,\;z_2\in \mathbb{C}$,  then :

a)  $|z_1 \bar z_2+1|^2+|z_2 \bar z_1+1|^2+2|z_1-z_2|^2=2(1+|z_1|^2)(1+|z_2|^2);$

b)  $z_1 \bar z_2-1|^2+|z_2 \bar z_1-1|^2-2|z_1-z_2|^2=2(1-|z_1|^2)(1-|z_2|^2);$

c)  $|1-\bar z_1z_2|^2-|z_1-z_2|^2=(1+|z_1z_2|)^2-(|z_1|+|z_2|)^2.$

          Problem  4 (page 29)  BERGSTROM identity : 

 for  $u,\;v \in \mathbb{C}\;\;and \;\;a,\;b\in \mathbb{R}^*,\;a+b \neq 0$

$\frac{|u|^2}{a}+\frac{|v|^2}{b}-\frac{|u+v|^2}{a+b}=\frac{|bu-av|^2}{ab(a+b)}.$

ANSWER CiP / Solutions CiP

  

                    @1.68 1) or @@9  or  #1a) (p. 28)        $|z_1+z_2|^2+|z_1-z_2|^2=$

$=(z_1+z_2)\cdot \overline{(z_1+z_2)}+(z_1-z_2)\cdot \overline{(z_1-z_2)}=(z_1+z_2)(\bar z_1+\bar z_2)+(z_1-z_2)(\bar z_1-\bar z_2)=$

$=(z_1 \bar z_1+z_2 \bar z_1+z_2\bar z_1+z_2 \bar z_2)+(z_1 \bar z_1-z_1\bar z_2-z_2 \bar z_1+z_2 \bar z_2)=$

$=2z_1 \bar z_1+2z_2 \bar z_2=2(|z_1|^2+|z_2|^2)$

(to be continue)

Gheorghe ANDREI - COMPLEX NUMBERS (part II) - Some Problems -- 3 / Gheorghe ANDREI - NUMERE COMPLEXE (partea II) - Câteva Probleme -- 3 / Gheorghe ANDREI - KOMPLEXNÉ ČÍSLA (časť II) - Niektoré problémy -- 3

 I wrote about the book here.

               We discuss problems no. ##3, 4, page 21 (solved on page 177) and #38, page 25 (solved on page 185).

                Problem #3  "Let  $\varepsilon$  be a complex cube root of unity. Prove that

$|z-1|^2+|z-\varepsilon|^2+|z_\varepsilon^2|^2=3(|z|^2+1)"$

                Problem #4   "Let  $\varepsilon$  be a complex cube root of unity. Prove that

$|z+u|^2+|z+\varepsilon u|^2+|z+\varepsilon^2 u|^2=3(|z|^2+|u|^2)$

                         whatever  $z,\;u\in\mathbb{C}$  are."

                                  Problem #38  "Prove that :

                                           a)   $(z-1)^2+(z-\varepsilon)^2+(z-\bar\varepsilon)^2=3z^2$

                where  $\varepsilon=\cos\frac{2\pi}{3}+\imath \sin\frac{2\pi}{3};$

                                          $\textbf{b)}\;\;\;(z-1)^2+(z-\varepsilon)^2+(z-\varepsilon ^2)^2+\dots+(z-\varepsilon ^{n-1})^2=nz^2$

where  $\varepsilon=\cos \frac{2\pi}{n}+\imath \sin \frac{2\pi}{n};$

 $$\textbf{c)}\;\;\; \sum_{k=0}^{n-1}|z-\varepsilon ^k|^2=n(|z|^2+1|)."$$

                        Solution of #4(CiP - Same as the solution on page 177) 

      The cube roots of unity are  $\left \{1,\;-\frac{1}{2}\pm \imath \frac{\sqrt{3}}{2}\right \}$. Let  $\varepsilon \neq 1$  be one of them. We have

$\varepsilon ^3=1,\;\;\;1+\varepsilon+\varepsilon ^2=0,\;\;\;\bar \varepsilon =\varepsilon ^2 \;\;\;\varepsilon \cdot \bar \varepsilon =1\tag{U3}$

          Using that  $|z|^2=z \cdot \bar z,\;\;(\forall) z\in \mathbb{C}$  we have

$|z+u|^2+|z+\varepsilon u|^2+|z+\varepsilon ^2 u|^2=$

$=(z+u)\cdot \overline{(z+u)}+(z+\varepsilon u)\cdot \overline{(z+\varepsilon u)}+(z+\varepsilon ^2 u)\cdot \overline{(z+\varepsilon ^2 u)}=$

$=(z+u)(\bar z +\bar u)+(z+\varepsilon u)(\bar z+\bar \varepsilon \bar u)+(z+\varepsilon ^2 u)(\bar z+\overline{\varepsilon ^2}\bar u)\overset{(U_3)}{=}$

$=(z\bar z+z\bar u+\bar z u+u\bar u)+(z \bar z+\varepsilon ^2 z \bar u+\varepsilon \bar z u+u \bar u)+(z\bar z+\varepsilon z \bar u+\varepsilon ^2 \bar z u+\varepsilon ^3 u \bar u)=$

$=3|z|^2+z\bar u (1+\varepsilon^2+\varepsilon)+\bar z u (1+\varepsilon +\varepsilon^2)+3|u|^2\underset{(U_3)}{=}3|z|^2+3|u|^2.$

$\blacksquare$

                        Solution of #3(CiP)

               If  $u=-1$,  the statement of Problem  #3 is obtained, or the calculation can be redone in this particular case.

$\blacksquare$

                REMARK CiP   On the cited page 177, the author of the book mentions another possible solution, using the identity(according to page 41, Exercise #10a)) :

$|z_1+z_2|^2+|z_2+z_3|^2+|z_3+z_1|^2=|z_1|^2+|z_2|^2+|z_3|^2+|z_1+z_2+z_3|^2$

Damn, I couldn't do this.

<end REM>

                        Solution of #38(CiP)

               a)  We have

  $(z-1)^2+(z-\varepsilon)^2+(z-\bar \varepsilon)^2 \overset{(U_3)}{=}(z-1)^2+(z-\varepsilon)^2+(z-\varepsilon^2)^2=$

$=(z^2-2z+1)+(z^2-2\varepsilon z+\varepsilon^2)+(z^2-2\varepsilon^2z+\varepsilon^4)=$

$=3z^2-2(1+\varepsilon+\varepsilon^2)z+(1+\varepsilon^2+\varepsilon)\underset{(U_3)}{=}3z^2$

$\blacksquare$

               b)   Let  $\varepsilon=\cos\frac{2\pi}{n}+\imath \sin \frac{2\pi}{n}$; we have  $\varepsilon ^n=1$,  and since  $1-\varepsilon ^n=(1-\varepsilon)(1+\varepsilon+\varepsilon^2+\dots+\varepsilon ^{n-1})=0$  it follows

$$\sum_{k=0}^{n-1}\varepsilon ^k=0. \tag{1}$$

But we also have

$$\sum_{k=0}^{n-1}(\varepsilon ^k)^2=0;\tag{2}$$

this is verified by distinguishing between odd and even cases for  $n$ :

$$n=2m+1\Rightarrow\sum_{k=0}^{2m}\varepsilon ^{2k}=\sum_{k=0}^m \varepsilon ^{2k}+\sum_{k=m+1}^{2m}\varepsilon^{2k}\;\;\;\;\;\overset{\varepsilon^{2m+1}=1}{\underset{\varepsilon^{2m+2}=\varepsilon^1,\;\varepsilon^{2m+4}=\varepsilon^3,\dots, \varepsilon^{4m}=\varepsilon^{2m-1}}{=}}\;\;\sum_{l=0}^{2m}\varepsilon^l\overset{(1)}{=}0;$$

$$n=2m\Rightarrow\sum_{k=0}^{2m-1}\varepsilon^{2k}=\sum_{k=0}^{m-1}\varepsilon^{2k}+\sum_{k=m}^{2m-1}\varepsilon^{2k}\;\;\;\;\;\overset{\varepsilon^{2m}=1}{\underset{\varepsilon^{2m}=1,\;\varepsilon^{2m+2}=\varepsilon^2,\dots, \varepsilon^{4m-2}=\varepsilon^{2m-2}}{=}}\;\;=2\cdot \sum_{l=0}^{m-1}\varepsilon^{2l}=0$$

the latter based on identity  $1-\varepsilon^{2m}=(1-\varepsilon^2)(1+\varepsilon^2+\varepsilon^4+\dots+\varepsilon^{2m-2})=0$.

          So we calculate 

$$\sum_{k=0}^{n-1}(z-\varepsilon^k)^2=\sum_{k=0}^{n-1}(z^2-2z\varepsilon^k+(\varepsilon^k)^2=nz^2-2z\cdot \sum_{k=0}^{n-1}\varepsilon^k+\sum_{k=0}^{n-1}(\varepsilon ^k)^2\;\;\underset{(1)\;(2)}{=}nz^2$$

$\blacksquare$

                    c)  Let us note that, along with (1), we also have its conjugate relation :

$$\sum_{k=0}^{n-1}(\bar \varepsilon)^k=0 \tag{3}$$

We obtain by an easy calculation :

$$\sum_{k=0}^{n-1}|z-\varepsilon^k|^2=\sum_{k=0}^{n-1}(z-\varepsilon^k)\cdot \overline{(z-\varepsilon^k)}=\sum_{k=0}^{n-1}(z-\varepsilon^k)(\bar z-\bar \varepsilon ^k)=$$

$$=\sum_{k=0}^{n-1}(z\cdot \bar z-\bar z \cdot \varepsilon^k-z\cdot \bar \varepsilon^k+\varepsilon^k\cdot \bar \varepsilon^k)=$$

$$\overset{z\cdot \bar z=|z|^2}{\underset{\varepsilon \cdot \bar \varepsilon=|\varepsilon|^2=1}{=}}\;\;\;\sum_{k=0}^{n-1}(|z|^2+1)-\bar z\cdot \sum_{k=0}^{n-1}\varepsilon^k-z\cdot \sum_{k=0}^{n-1}\bar \varepsilon^k\overset{(1)}{\underset{(3)}{=}}n(|z|^2+1).$$

                             REMARK CiP   These formulas take place starting from  $n=2$ :

$|z-1|^2+|z+1|^2=2(|z|^2+1);$

$n=3$ is the subject of exercise  #3.

$\blacksquare$

Gheorghe ANDREI - COMPLEX NUMBERS (part II) - Some Problems -- 2 / Gheorghe ANDREI - NUMERE COMPLEXE (partea II) - Câteva Probleme -- 2 // Gheorghe ANDREI - اعداد مختلط (قسمت دوم) - برخی از مسائل - 2

 You can find the book in my Electronic Library. The author, Gh. ANDREI gets lost among the crowd of Titu ANDREESCU...(current version today ; you can still access the library by scanning the QR code from the beginning)

                                                                                       < begin preview>

ANDREESCU Titu, DOSPINESCU Gabriel, MUSHKAROV Oleg

NUMBER THEORY: CONCEPTS AND PROBLEMS

XYZ Press, Plano_TX, 2017

ANDREI Gheorghe

NUMERE COMPLEXE : partea a II-a

Ed. GIL, Zalău, 2004/2005-ebook(cadou de Craciun 25 Dec 2025 de la fiul meu

                                                                                                                     CIOBANU Victor)

ANDREIAN-CAZACU Cabiria, DELEANU Aristide, JURCHESCU Martin

TOPOLOGIE. CATEGORII. SUPRAFEȚE RIEMANNIENE

Ed. ACADEMIEI [R.S.R.], București, 1966 

< end preview>

 This post is number 2, because we believe that the post here would be number 1.

 

            Problem #2 (page 21, solved on page 177 ; in translation)

"Let  $z\in \mathbb{C}\setminus \{\pm 1\}$, with the property  $\imath \frac{z-1}{z+1}\in\mathbb{R}$. Determine  $|z|$."

ANSWER CiP

$$|z|=1$$

                    Solution CiP

                Let  $\mathbb{R}\ni a=\imath \cdot \frac{z-1}{z+1}.$  We have (because $\imath ^2=-1\Rightarrow \frac{1}{\imath}=-\imath$)

$\frac{z-1}{z+1}=\frac{a}{\imath}=-a\imath$

so

 $z=\frac{1-a\imath}{1+a\imath} \tag{1}$

hence, according to Property #20, page 13,

$|z|=\frac{|1-a\imath|}{|1+a\imath|}=\frac{\sqrt{1+a^2}}{\sqrt{1+a^2}}=1$

$\blacksquare$

                 REMARKS CiP

               $1^r$.  Continuing in (1), we have  $z=\frac{(1-a\imath)^2}{(1+a\imath)(1-a\imath)}$  so

$z= \frac{1-a^2}{1+a^2}-\frac{2a\imath}{1+a^2}\;\overset{b=-a}{=}\;\frac{1-b^2}{1+b^2}+\frac{2b\imath}{1+b^2}\tag{2}$

               $2^r$.  The reciprocal statement is also true :

$|z|=1\Rightarrow (\exists)a\in\mathbb{R}\;\;s.t.\;\;z=\frac{1-a^2}{1+a^2}-\frac{2a}{1+a^2}\imath\Leftrightarrow (\exists)b\in\mathbb{R}\;\;s.t.\;\;z=\frac{1-b^2}{1+b^2}+\frac{2b}{1+b^2}\imath \tag{3}$

Indeed, according to page 16, Trigonometric form of complex numbers, #1, we have

$z=\cos \theta+\imath \sin \theta$,  and  $\cos \theta=\frac{1-b^2}{1+b^2},\;\;\sin \theta =\frac{2b}{1+b^2},\;\;\;b=\tan \theta /2$

hence (3).

$\square$<end REM>