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>

PROBLEM E : 12 955 ABOUT the LACK of PERFECT SQUARES

 In GMB 5/2005 magazine, page 230 :

          E:12955  Prove that for any  $x\in\mathbb{N}$, the number  $x^2+13x+40$  

                            is not a perfect square.

([Authors:] Ioana CRĂCIUN și Gh. CRĂCIUN, Plopeni)

To avoid complicated Diophantine equation techniques, we looked for the interleaving of the number  $x^2+13x+40$  between two consecutive squares.

ANSWER CiP

$(x+6)^2<x^2+13x+40<(x+7)^2 \tag{1}$

                           Solution CiP

$0<x+4<x+13\;\;\;\;|+(x^2+12x+36)$

$\Rightarrow x^2+12x+36<x^2+13x+40<x^2+14x+49\Rightarrow (x+6)^2<x^2+13x+40<(x+7)^2$

hence (1).

$\blacksquare$

PROBLEM C : 2870 from PROBLEMS for the ANNUAL COMPETITION of "GAZETA MATEMATICĂ"

          In GMB 5/2005 magazine, pages 234 (Romanian version) and 236 (English version).

          The problem says : 

     C:2870.  Find the minimum value of :  $E(x)=\frac{x^2-2x-1}{x^2-2x+3}$,  for  $x\in\mathbb{R}$.

[Author :] Vasile PREDAN, Curtea de Argeș

ANSWER CiP

$-1=E(1)\leqslant E(x)<1$

                     Solution CiP  We will imitate the solution from the Post here.

           The number  $\lambda$  is a value of  $E(x)\;\;\Leftrightarrow$

$\Leftrightarrow (\exists)x\in\mathbb{R}\;\;\mathbf{s.t.}\;\;E(x)=\lambda$

$\Leftrightarrow (\exists)x\in\mathbb{R}\;\;\mathbf{s.t.}\;\;x^2-2x-1=\lambda \cdot (x^2-2x+3)$

$\Leftrightarrow (\exists)x\in\mathbb{R}\;\;\mathbf{s.t.}\;\;(1-\lambda)x^2+2(\lambda-1)x-(3\lambda+1)=0 \tag{1}$

$\Leftrightarrow$  the quadratic equation (1) has solution(s).

[For  $\lambda=1$  equation (1) isn't quadratic both then in hasn't solution.].

     Using the half discriminant ($\Delta'=(b')^2-ac$) formula, the roots of equation (1) are

$x_{1,2}=\frac{1-\lambda\pm\sqrt{2-2\lambda^2}}{1-\lambda} \tag{2}$

The requirement that the roots be real numbers, $|\Delta'\geqslant 0$  implies  $2-2\lambda^2\geqslant 0\;\Leftrightarrow$

$\Leftrightarrow\;\lambda^2\leqslant 1\;\Leftrightarrow\;-1\leqslant \lambda<1.$ From (2) we obtain the expressions of the roots as follows  $x_{1,2}=1\pm \sqrt{\frac{2(1+\lambda)}{1-\lambda}}.$

$\blacksquare$

       Writing differently,  $E(x)=1-\frac{4}{(x-1)^2+2}\;\;\Rightarrow\;\; -1=1-\frac{4}{0+2}\leqslant 1-\frac{4}{(x-1)^2+2}<1$,  with the same answer.

A RATIONAL VALUE of COSINE // مقدار گویای کسینوس

               In the magazine GAZETA MATEMATICĂ No 4, 1978, on covers 3-4 the following Problem appears (as a solution to the one proposed in GMB 11/1977, page 455 Problem#3) :

          Problem 3 page 455 / No. 11/1977  : Prove that if   $\alpha \in \mathbb{R}$  and 

      $\cos(\alpha \pi)=\frac{1}{3}$  then  $\alpha$  is irrational. (The angle  $\alpha \pi$ is considered in radians)

(We will return to this issue at the end)

               In the past, in my youth, I was more diligent. In the magazine GMB 10/2014 (so 35 years after the one mentioned in the preamble), in the -recently established- Column PROBLEMS for NATIONAL EXAMS, in the 12th grade, page 478, we see problems 27 and 28. (We made a system of cards with them, by grade, etc.; that's why I said "diligence")

This statement

Solving

As you can see, in Problem 2c), the question arises whether at the value  $q\pi$  for which  $\cos q\pi=\frac{3}{5}$, we have  $q\in\mathbb{Q}$  or not. When resolving this point, I remembered the problem in the Preamble. The answer is NO.

So, $\cos q\pi=\frac{3}{5}\;\Rightarrow\; q\notin \mathbb{Q}$.

         We assume the opposite (see line (1) in the third picture) : 

$\exists q\in \mathbb{Q}\;\;s.t.\;\;\cos q\pi=\frac{3}{5}\tag{1}$

Let  $q=\frac{m}{n},\;m\in\mathbb{Z},\;n\in \mathbb{N}$, with their greatest common divisor  $(m,n)=1$. Then

 the set of values ​​of the string  $\{\cos kq\pi;\;q\in\mathbb{Z}\}$ is finite :                           (2)

$\pm\cos 0q\pi,\;\pm\cos q\pi,\;\pm\cos 2q\pi,\dots,\;\pm\cos (n-1)q\pi \tag{3}$

     Indeed, we will show that, whatever  $k\in\mathbb{Z},\;\cos kq\pi$  has one of the values  (3). Writing, with Euclid's division lemma,  $k=n\cdot t+l,\;\;t\in\mathbb{Z},\;l\in\{o,\;1,\;\dots,\;n-1\}$,  we have

$\cos kq\pi=\cos k(\frac{m}{n})\pi=\cos (nt+l)\frac{m}{n}\pi=\cos (mt+l\frac{m}{n})\pi=\cos (mt\pi+lq\pi)=\pm\cos lqt$

qed$\square$

       On the other hand, we will show that if  $k$ is a power of 2, then  $\cos k\pi$  takes on an infinity of values. More precisely, we show, by mathematical induction, the equality :

$\cos 2^kq\pi=\frac{a_k}{5^{2^k}},\;\;\;a_k\in\mathbb{Z},\;\;\;5\not{\mid}a_k \tag{4}$

For $k=0,\;\;\cos 2^0q\pi=\cos q\pi=\frac{3}{5}=\frac{3}{5^{2^0}}$, and we have  $a_0=3,\;5\not{\mid }a_0.$ Then  $\cos 2q\pi=2\cos ^2q\pi-1=2(\frac{3}{5})^2-1=-\frac{7}{25}=\frac{-7}{5^{2^1}}$, so  $a_1=-7\;and\;5\not{\mid}a_1$.

Assuming (4) true for a fixed  $k$, we have

$\cos 2^{k+1}q\pi=\cos 2(2^kq\pi)=2\cos^2(2^kq\pi)-1=2\cdot (\frac{a_k}{5^{2^k}})^2-1=\frac{2a_k^2-5^{2^k\cdot 2}}{5^{2^k\cdot 2}}=\frac{2a_k^2-5^{2^{k+1}}}{5^{2^{k+1}}}$

and taking  $a_{k+1}=2a_k^2-5^{2^{k+1}}$  we have  $5\not{\mid}a_{k+1}$  because  $5\not{\mid}a_k$. So (4) is thrue for  $k+1$. Moreover, the values ​​given by (4) are all distinct, because if we had  $\frac{a_k}{5^{2^k}}=\frac{a_l}{5^{2^l}},\;k<l$, it would result  $a_l=5^{2^l-2^k}\cdot a_k$  so  $5\mid a_l$ - false.

$\blacksquare$

               The same argument is used in the original problem in the Preamble, where, from  $\cos \alpha \pi=\frac{1}{3}$,  results :

$\cos 2^k\alpha \pi=\frac{r}{3^{2^k}},\;\;r\in\mathbb{Z},\;\;\;3\not{\mid}r$

Problem 2954 from REVISTA MATEMATICĂ a ELEVILOR din TIMIȘOARA

 You can see issue 1-1977 from which I took the issue (page 62), here.

A larger collection is here.

             2 954.  Determine the real numbers  $x$  for which the number

$\frac{x}{x^2-5x+7}$  is an integer.

(Matematika v Șkole)

ANSWER CiP

$x\in \{0,\;2,\;\frac{7}{3},\;3,\;\frac{7}{2}\}$

                    Solution CiP

          If  $\frac{x}{x^2-5x+7}=k\in\mathbb{Z}$  then  $x$  is the root of the quadratic equation

$kx^2-(5k+1)x+7k=0 \tag{1}$

The equation (1) has the discriminant  $\Delta_x=(5k+1)^2-28k^2=1+10k-3k^2$.

     But  $1+10k-3k^2=\frac{28}{3}-\left (\frac{25}{3}-10k+3k^2\right )=\frac{28}{3}-3\cdot \left(\frac{25}{9}-\frac{10}{3}k+k^2\right)=$

$=\frac{28}{3}-3\cdot \left (\frac{5}{3}-k\right)^2\leqslant \frac{28}{3}$,  so  

$\Delta_x\leqslant \frac{28}{3}\tag{2}$

     For the equation (1) to have rational roots, it must be  $\sqrt{\Delta_x}\in\mathbb{Q}$,  in fact  $\sqrt{\Delta_x}\in\mathbb{N}$. This happens if  (cf. (2))

$\sqrt{\Delta_x}\in\{0,\;1,\;2,\;3\} \tag{3}$

     Solving the equations  $1+10k-3k^2=0\;or\;1\;or\;2\;or\;3$  we obtain the integer values

$k\in\{0,\;2,\;3\}$

for which, we obtain from the equation (1) the values ​​of  $x$  in the answer.

$\blacksquare$

          Correction

          My answer is incomplete, as the commenter noted. 

          For  $x=3\pm\sqrt{2}$, the number  $\frac{x}{x^2-5x+7}$  takes the value  $1$, also an integer.

I wrongly assumed that  $x$  is a rational number. Let's analyze the sign of  $\Delta_x$.

          Setting the condition  $\Delta_x \geqslant 0\;\Rightarrow\; 1+10k-3k^2\geqslant 0$  which gives us the admissible values

$k\in\left ( \frac{5-\sqrt{28}}{3}\;,\;\frac{5+\sqrt{28}}{3}\right )\cap \mathbb{Z}=\{0,\;\color{Red}1,\;2,\;3\}$

Considering that the roots of the equation (1) are  $x_{1,2}=\frac{5k+1\pm\sqrt{\Delta_x}}{2k}$, for  $k=1$  we find  $x=3\pm\sqrt{2}$, so the correct answer is

$x\in \{0,\;3-\sqrt{2},\;2,\;\frac{7}{3},\;3,\;\frac{7}{2},\;3+\sqrt{2}\}$

$\color{Red}{!!!}\square \color{Red}{!!!}$

PROBLEM E : 6185 from the magazine GAZETA MATEMATICĂ

          From GMB 4/1978 page 164.

In translation :

                         "E : 6185*.  Knowing that the numbers  4830,  448230,  44482230

         are products of two consecutive natural numbers, find these consecutive

                      numbers. Generalization.

[Author : ] I. I. Mihailov, Ivanovo, U.R.S.S."

ANSWER CiP

$4\;830\;=\;69\;\times\;70$

$448\;230\;=\;669\;\times\;670$

$44\;482\;230\;=\;6\;669\;\times\;6\;670$

[CiP addition] $4\;444\;822\;230\;=\;66\;669\;\times\;66\;670$

                                  Generalization :

$\underset{n+1}{\underbrace{4\;\dots\;4}}\;8\;\underset{n}{\underbrace{2\;\dots\;2}}\;30\;=\;\underset{n+1}{\underbrace{6\;\dots\;6}}\;9\times \underset{n}{\underbrace{6\;\dots\;6}}\;70 \tag{G}$

                         Solution CiP

           If a number  $A$  is a product of consecutive numbers, i.e.  $A=k(k+1)$, then we have

$A=k(k+1) \Rightarrow k^2<A<(k+1)^2\Rightarrow k<\sqrt{A}<k+1\Rightarrow k=[\sqrt{A}]$

([X]=the integer part of the number X) We have :

$\sqrt{4830}=69,4...$  and it is verified by calculation that  $4830=69\times 70$

$\sqrt{448230}=669,4...$  and it is verified by calculation that  $448230=669\times 670$

$\sqrt{44482230}=6669,4...$  and it is verified by calculation that  $44482230=6669\times 6670$

The next example in the answer, verified by calculation, is my creation. I also formulated the generalization, which we will verify below.

          To carry out the calculations we need to replace  $1\;\dots\;1$  with the more precise expression

$\underset{n}{\underbrace{1\;\dots\;1}}=\frac{1}{9}\cdot (10^n-1) \tag{1}$

Indeed  $\underset{n}{\underbrace{1\;\dots\;1}}=\frac{1}{9}\cdot \underset{n}{\underbrace{9\;\dots\;9}}=\frac{1}{9}\cdot (1\underset{n}{\underbrace{0\;\dots\;0}}-1)=\frac{1}{9}\cdot(10^n-1).$

          Analyzing the position of the digits of the number  $4\dots482\dots230$, we obtain that the value of the number on the left of the relation  (G) is :

$\underset{n+1}{\underbrace{4\;\dots\;4}}\underset{n+3}{\underbrace{\;8\;\overset{n+2}{\overbrace{\underset{n}{\underbrace{2\;\dots\;2}}\;\underset{2}{\underbrace{30}}}}}}=\underset{n+1}{\underbrace{4\dots4}}\cdot 10^{n+3}+8\cdot 10^{n+2}+\underset{n}{\underbrace{2\dots2}}\cdot 100+30=$

$\overset{(1)}{=}4\cdot \frac{1}{9}\cdot (10^{n+1}-1)\cdot 10^{n+3}+8\cdot 10^{n+2}+2\cdot \frac{1}{9}\cdot(10^n-1)\cdot 100+30=$

$=\frac{4}{9}\cdot 10^{2n+4}-\frac{4}{9}\cdot 10^{n+3}+8\cdot 10^{n+2}+\frac{2}{9}\cdot 10^{n+2}-\frac{2}{9}\cdot 100+30=$

$=\frac{4}{9}\cdot 10^{2n+4}+\frac{34}{9}\cdot 10^{n+2}+\frac{70}{9} \tag{2}$

     On the right side of (G) we have :

$\underset{n+1}{\underbrace{6\dots6}}9\times \underset{n}{\underbrace{6\dots6}}70=\left [\frac{6}{9}\cdot (10^{n+1}-1)\cdot 10+9\right ]\times \left[\frac{6}{9}\cdot (10^n-1)\cdot 100+70\right ]=$

$=\left [\frac{2}{3}\cdot 10^{n+2}+\frac{7}{3}\right]\cdot \left [\frac{2}{3}\cdot 10^{n+2}+\frac{10}{3}\right ]=\frac{4}{9}\cdot 10^{2n+4}+\frac{2}{3}\cdot 10^{n+2}\cdot \left (\frac{7}{3}+\frac{10}{3}\right )+\frac{70}{9}=$

$=\frac{4}{9}\cdot 10^{2n+4}+\frac{34}{9}\cdot 10^{n+2}+\frac{70}{9} \tag{3}$

The results in (1) and (2) prove the equality in (G).

$\blacksquare$

A PROBLEM that JUMPED like a KANGAROO

           The author of the problem is George STOICA from Canada.

Although proposed for 12th grade (the last grade before college), it only requires elementary knowledge of quadratic equations and some common sense knowledge of natural numbers.

          Here we denote  $\mathbb{N}=\{0,\;1,\;2,\dots \}$  be the set of natural numbers,  $\mathbb{Q}$  the set of rational numbers. We state the following, almost obvious, statement as a

              Lemma.   Let  $m\in\mathbb{N}$  such that  $\sqrt{m}\in\mathbb{Q}$.    Then  $\sqrt{m}\in\mathbb{N}$.

                  Proof CiP  $\sqrt{m}$ admits the representation  $\sqrt{m}=\frac{q}{r},\;\;q,\;r\in\mathbb{N},\;r\neq0$, where the greatest common divisor of the numbers  $q$ and  $r$,  $(q,r)=1$. By the Fundamental Theorem of Arithmetic, we have the (unique) writings  

$m=p_1^{e_1}\cdot p_2^{e_2}\cdot \dots \cdot p_k^{e_k},\;\;q=p_1^{f_1}\cdot p_2^{f_2}\cdot \dots \cdot p_k^{f_k},\;\;r=p_1^{g_1}\cdot p_2^{g_2}\cdot \dots \cdot p_k^{g_k}\;\;\;\;k\geqslant 1 \tag{1}$

where  $p_1<p_2<\dots <p_k$ are primes and  $e_{1-k},\;f_{1-k},\;g_{1-k}$  are nonnegative integers. In (1), some  $e_i,\;f_i,\;g_i$  can be zero and the condition  $(q,r)=1$  requires that, for any index  $i$, one of  $f_i\; or\; g_i$  be zero.

From the equality  $r^2\cdot m=q^2$  and the uniqueness of the representation as a product of powers of prime numbers, it follows

$2\cdot g_i+e_i=2\cdot f_i$

so all the exponents  $e_i$  are even numbers and then  $\sqrt{m}=p_1^{e_1/2}\cdot p_2^{e_2/2}\cdot \dots \cdot p_k^{e_k/2} \in\mathbb{N}$.

$\square$Lemma

             The issue originally appeared in the GMB Supplement 3/2025, as S:L25.144 on page 15. Then he jumped (like a kangaroo !!) in GMB 6-7-8/2025 , as 29 181 on page 374. (Nice jump, otherwise the problem would not have benefited from ever publishing a solution.) The problem statement is :

                    "Let  $P$  be a polynomial with integer coefficients and  $a,\;b\in\mathbb{Z}$  such 

               that  $P(a)\cdot P(b)=-(a-b)^2$. Show that  $P(a)=-P(b)=\pm(a-b)$."

ANSWER CiP

An EXAMPLE of such a polynomial is  $P(X)=2\cdot X-a-b$

                        Solution (as it appears in GMB 1/2026, page 38, adapted by CiP)

               If  $a=b$, that is,  $P(a)^2=0$, we have  $P(a)=0$  so the conclusion is obvious.  Next we assume  $a\neq b$.

                From the algebraic identity

  $\frac{a^k-b^k}{a-b}=a^{k-1}+a^{k-2}b+\dots+ab^{k-2}+b^{k-1},\;\;\forall k\geqslant 1$

it follows that for any polynomial  $P$ , the number  $\alpha=\frac{P(a)-P(b)}{a-b}\in \mathbb{Z}$.

          For rational numbers 

$x_1=\frac{P(a}{a-b},\;\;\;x_2=-\frac{P(b}{a-b} \tag{2}$

we have  $x_1+x_2=\frac{P(a)-P(b)}{a-b}=\alpha,\;\;x_1\cdot x_2=-\frac{P(a)\cdot P(b)}{(a-b)^2}=1$ so the numbers (2) are the roots of the equation with integer coefficients

$x^2-\alpha \cdot x+1=0. \tag{3}$

Then the discriminant of the equation (3) is a rational number. But  $\Delta=\alpha^2-4$  and then

$\mathbb{Q}\ni \sqrt{\alpha^2-4}=\beta\;\;\overset{\textbf{Lemma}}{\Rightarrow}\;\; \beta \in \mathbb{N}$

We have   $\alpha^2-\beta^2=4=(\alpha+\beta)(\alpha-\beta)$, so,  $\alpha+\beta\;and\;\alpha-\beta$  having the same parity, it results

$\alpha+\beta=\pm2=\alpha-\beta$

and from here  $\beta=0,\;\alpha=\pm2$.

          So from the equation (3) we get  $x_1=x_2=1\;or\;x_1=x_2=-1$. Therefore

$1\overset{(2)}{=}\frac{P(a)}{a-b}=-\frac{P(b)}{a-b}$

or

$-1\underset{(2)}{=}\frac{P(a)}{a-b}=-\frac{P(b)}{a-b}.$

It results from here  $P(a)=-P(b)=a-b\;\;or\;\;P(a)=-P(b)=-(a-b)$.

$\blacksquare$

مسئله ۶۷۶۷ از سری B مجله ریاضی، رومانی // Problem 17 384 from a "Gazeta Matematică seria B" , Romania

 به پست وبلاگ دیگرم هم سر بزنید

I've written about this number before here. Today I'm posting another problem.

                  " 17 384.  Show that if  $a+b+c=abc,\;(a,\;b,\;c \neq 0)$, then 

$a\left (\frac{1}{b}+\frac{1}{c} \right )+b\left (\frac{1}{c}+\frac{1}{a} \right )+c\left ( \frac{1}{a}+\frac{1}{b}\right )+3=ab+bc+ca.$

 Dan SECLEMAN, student, Craiova"

Solution CiP

We will do a calculation in which at some point we make the substitutions

$a+b=abc-c,\;\;\;b+c=abc-a,\;\;\;a+c=abc-b \tag{S}$

 From left to right, we march :   $a(\frac{1}{b}+\frac{1}{c})+b(\frac{1}{c}+\frac{1}{a})+c(\frac{1}{a}+\frac{1}{b})+3=$

$=\underline{\underline{\frac{a}{b}}}+\frac{a}{c}+\frac{b}{c}+\underline{\frac{b}{a}}+\underline{\frac{c}{a}}+\underline{\underline{\frac{c}{b}}}+3=\frac{1}{a}\underline{(b+c)}+\frac{1}{b}\underline{(\underline{a+c})}+\frac{1}{c}(a+b)+3\overset{(S)}{=}$

$\underset{(S)}{=}\frac{1}{a}(abc-a)+\frac{1}{b}(abc-b)+\frac{1}{c}(abc-c)+3=(bc-1)+(ac-1)+(ab-1)+3=$

$=ab+bc+ca.$ QED

A solution “pulled out of a hat”. An explanation of the trick // Uma solução "tirada da cartola". Uma explicação do truque

 In the last issue of the Exercise Supplement of the Mathematics Gazette, most of the problems seemed trivial. Maybe this S:L 25. 411 is too, but look at the solution I found.

  "S:L25.411.  Consider the sequence  $(x_n)_{n\in\mathbb{N}}$, defined by  $x_{n+1}=\frac{4x_n+1}{2x_n+3}$, for any 

           $n\in\mathbb{N}$  and  $x_0=0$. Show that  $$\lim_{n \to \infty}x_n=1.$$

[Author] Ludovica LAZĂR, Năsăud"

ANSWER  CiP

$x_n=\frac{5^n-2^n}{5^n+2^{n+1}}=\frac{1-\left (\frac{2}{5}\right )^n}{1+2\cdot \left (\frac{2}{5}\right )^n}\;\underset{n \to \infty}{\to} \;1$

   

                           Solution CiP

               Let's consider the sequence

$y_n=\frac{x_n-1}{2\cdot x_n+1}. \tag{1}$

The reverse connection is

$x_n=\frac{1+y_n}{1-2\cdot y_n}. \tag{2}$

The initial conditions are

$x_0=0,\;\;\;y_0=-1. \tag{3}$

          Let's calculate now

$y_{n+1}\;\overset{(1)}{\underset{n\to n+1}{=}}\;\frac{x_{n+1}-1}{2\cdot x_{n+1}+1}\overset{def}{=}\frac{\frac{4x_n+1}{2x_n+3}-1}{2\cdot \frac{4x_n+1}{2x_n+3}+1}=\frac{4x_n+1-2x_n-3}{8x_n+2+2x_n+3}=\frac{2x_n-2}{10x_n+5}=\frac{2}{5}\cdot \frac{x_n-1}{2x_n+1}\overset{(1)}{=}\frac{2}{5}\cdot y_n.$

The equality of the extreme terms tells us that the sequence  $(y_n)_{n\in\mathbb{N}}$  is a geometric progression. So

$y_n=y_0\cdot \left (\frac{2}{5}\right )^n\overset{(3)}{=}-\left (\frac{2}{5}\right )^n,\;\;n\in\mathbb{N} \tag{4}$

          Substituting the value from  (4)  into  (2)  we obtain the formula for  $x_n$  mentioned in the Answer, as well as the value of the limit.

$\blacksquare$

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

               Recurrence relations of the form 

$x_{n+1}=\frac{a\cdot x_n+b}{c\cdot x_n+d}\;,\;\;a\cdot d-b\cdot c\neq 0 \tag{H}$

have an unquantified prevalence in Problem Collection books. 

In the book BĂTINEȚU D.M. "Şiruri..." (ALBATROS Publishing House, Bucharest, 1979 - click on the link to download) I came across problem 1.184, page 174-176, together with the solution. With the mention "Bacalaureat, France, 1975".

THAT'S WHERE I TAKEN THE IDEA OF SOLUTION

Luckily, the statement suggested the solution method. In translation

"1.184.  Consider the sequences  $(u_n)_{n\in \mathbb{N}},\;(v_n)_{n\in \mathbb{N}}$  where  $u_0=1$  and

$u_n=\frac{2\cdot u_{n-1}-1}{2\cdot u_{n-1}+5}\;,(\forall)n\in\mathbb{N}^*\;\;;\;\;v_n=\frac{2\cdot u_n+1}{u_n+1}\;,(\forall) n\in\mathbb{N}. \tag{5}$

  $a^{\circ}$)  Show that the sequence  $(v_n)_{n\in\mathbb{N}}$  is a geometric progression,

                         calculate its ratio and  $\lim_{n\to \infty}v_n$.

                                  $b^{\circ}$)  Express  $u_n$  in terms of  $n$  and establish its nature.

(Bacalaureat, Franța, 1975)"

       [The answers are :  $v_n=\frac{3}{2}\cdot \left (\frac{3}{4} \right )^n,\;u_n=\frac{1-v_n}{v_n-2}=\frac{2\cdot 4^n-3^{n+1}}{3^{n+1}-4^{n+1}}\underset{n\to \infty}{\to} -\frac{1}{2}$]

               Fortunately for Romanian-speaking readers, there is the book (I think I'm missing the electronic version) AVADANEI C., AVADANEI N., BORȘ C., CIUREA C.  "De la matematica elementară spre matematica superioară" (Ed. ACADEMIEI [R.S.] România, București, 1987)". Here, Chapter I is dedicated to a systematic study of sequences defined by homographic recurrence relations, pages 7 - 68.

See also here  !!

[In this book I found Problem 1.184 above, with the mention that it appeared under number 15 896 in GMB no. 9, 1978. Unfortunately the citation is erroneous, it is probably a number from 1976, I have not identified it exactly yet.]

Magic Trick (1) Explanation

For 

$x_{n+1}=\frac{4\cdot x_n+1}{2\cdot x_n+3} \tag{6}$

let's look for a transformation of type 

$y_n=\frac{a\cdot x_n+b}{c\cdot x_n+d} \tag{7}$

such that the latter is a geometric progression. 

A simple calculation gives us  $y_{n+1}\underset{(7)}{=}\frac{a\cdot x_{n+1}+b}{c\cdot x_{n+1}+d}\overset{(6)}{=}\frac{a\cdot \frac{4x_n+1}{2x_n+3}+b}{c\cdot \frac{4x_n+1}{2x_n+3}+d}=\frac{(4a+2b)\cdot x_n+a+3b}{(4c+2d)\cdot x_n+c+3d}.$

          Let's try to find two numbers  $\lambda\;and\;\mu$  such that the previous result is equal to  $\left (\frac{\lambda}{\mu}\right )\cdot y_n$.  We want to have the equality

$\frac{(4a+2b)x_n+a+3b}{(4c+2d)x_n+c+3d}=\frac{\lambda}{\mu}\cdot \frac{ax_n+b}{cx_n+d} \tag{8}$

Then, let's try, maybe we have a chance to succeed, to fulfill the equations

$(4a+2b)\cdot x_n+a+3b=\lambda \cdot (a\cdot x_n+b) \tag {9'}$

$(4c+2d)\cdot x_n+c+3d=\mu \cdot (c\cdot x_n+d) \tag {9"}$

Oh, and if we're not asking too much, it wouldn't hurt to have the equations

$\begin{cases}4a+2b=\lambda \cdot a\\\;a+3b=\lambda \cdot b\\4c+2d=\mu\cdot c\\\;c+3d=\mu\cdot d\end{cases} \tag{10}$

     Let's choose from (10) the first two equations :

$\begin{cases}(4-\lambda)\cdot a+\;\;\;\;\;\;\;\;\;\;\;2b=0\\\;\;\;\;\;\;\;\;\;\;\;\;\;a+(3-\lambda)\cdot b=0.\end{cases} \tag{11}$

For the system of equations with unknowns  $a\; and \;b$  to admit non-trivial solutions, it is necessary and sufficient that its matrix be singular, that is

$\left |\begin{matrix}4-\lambda&2\\1&3-\lambda\end{matrix}\right |=0 \tag{12}$

$\Leftrightarrow \lambda^2-7\cdot\lambda +10=0 \tag{13}$

Equation (13) has the roots  $\lambda_1=2,\;\;\lambda_2=5$. 

Choosing  $\lambda_1=5$ the system (11) is written  $\begin{cases}2\cdot a+2\cdot b=0\\a+b=0\end{cases}$  hence  $b=-a$.

          The last two equations in (10) lead to the same system (11) (of course with different notations), and we have the same equation (12)2. Now we choose  $\mu=5$  and we have the system  $\begin{cases}-c+2\cdot d=0\\c-2\cdot d=0\end{cases}$  hence  $c=2\cdot d$.

          So the possible transformations in (7) are  $y_{n+1}=\frac{a\cdot x_n-a}{2d\cdot x_n+d}=\frac{a}{d}\cdot \frac{x_n-1}{2\cdot x_n+1}$. With  $a=1,\;d=1$  we find what we chose in (1). 

                    REMARK  CiP

                   We see in  (H)  a homographic (or fractionally-linear) transformation

$x\mapsto \frac{a\cdot x+b}{c\cdot x+d},\;\;ad-bc \neq 0$

This can be represented by a matrix

$A=\left (\begin{matrix}a&b\\c&d\end{matrix}\right )$

The connection is not just a convention: there is a deep structural correspondence.

          These constitute Möbius transformation. They form a group, often noted  $\mathbb{PSL}(2)$.

            The composition of transformations corresponds to the multiplication of matrices.

             Since matrices obtained from a particular one and multiplied by an arbitrary nonzero scalar give us the same homographic transformation, it is natural that they act on a projective vector  $\textbf{x}=\left (\begin{matrix}x\\1\end{matrix} \right )$.

             It is also not difficult to see that  (12) is the characteristic equation associated with the matrix  $\left (\begin{matrix}4&2\\1&3\end{matrix}\right )$. Why it is not exactly the matrix associated with the transformation (6), i.e.  $\left( \begin{matrix}4&1\\2&3\end{matrix}\right)$  but its transpose, I leave to your discretion.