Gheorghe ANDREI - COMPLEX NUMBERS (part II) - Some Problems -- 5 / Gheorghe ANDREI - NUMERE COMPLEXE (partea II) - Câteva Probleme -- 5 / Gheorghe ANDREI – LICZBY ZŁOŻONE (część II) – Niektóre problemy – 5

 I wrote about the book here. 

It's not for nothing that I wrote somewhere that THIS BOOK IS FULL OF SCRATCHES. (Maybe that's why the author withdrew Part I from sale)

                    We are now debating Problem 62***, page 157 (solved on pages 516-519).

       

                     " 62***.  Calculate  $P_n\; Q_n\;and\; R_n$, where  $n\in \mathbb{N},\;n\geqslant 3,\;a\in\mathbb{R}^*\;:$

 $$a)\;\;\;P_n=\prod_{k=0}^{n-1} \cos \left (a+\frac{k\pi}{n}\right )\;;$$

$$b)\;\;Q_n=\prod_{k=0}^{n-1}\sin \left (a+\frac{k\pi}{n}\right )\;;$$

$$c)\;\;R_n=\prod_{k=0}^{n-1}\tan \left (a+\frac{k\pi}{n}\right )."$$

ANSWER

$a)\;\;P_{2n}=\frac{(-1)^n\sin 2na}{2^{2n-1}}\;\;;\;\;P_{2n+1}=\frac{(-1)^n\cos (2n+1)a}{2^{2n}}$

$b)\;\;Q_n=\frac{\sin na}{2^{n-1}}$

$c)\;\;R_{2n}=-1\;\;;\;\;R_{2n+1}=(-1)^n\tan (2n+1)a$

Solution CiP

               We will start with b). A solution that, at some point, requires a square root extraction is found in TURTOIU Fanică - Probleme de Trigonometrie, Ed. Tehnică, București, 1979 (pages 86-87, Problem 2.38 - received by the kindness of Liviu PODGORNEI). This is a personal attempt.

          Let  $z_k=\cos\left (a+\frac{k\pi}{n}\right )+\imath \sin \left (a+\frac{k\pi}{n}\right )\;,\;\;0\leqslant k\leqslant n-1.$  We have

$|z_k|=1=z_k\cdot \bar z_k$  so  $\bar z_k=\cos \left (a+\frac{k\pi}{n}\right )-\imath \sin \left (a+\frac{k\pi}{n}\right )=\frac{1}{z_k}.$  In addition

$z_k^{2n}=\cos 2n \left (a+\frac{k\pi}{n}\right )+\imath \sin 2n\left (a+\frac{k\pi}{n}\right )=\cos 2na+\imath \sin 2na \tag{1}$

Then

$\sin \left (a+\frac{k\pi}{n}\right )=\frac{z_k-\bar z_k}{2\imath}=\frac{z_k-\frac{1}{z_k}}{2\imath}=\frac{z_k^2-1}{2\imath z_k} \tag{2}$

          From  (1)  it follows that the equation

$z^{2n}=\cos 2na+\imath \sin 2na \tag{3}$

has the roots  $\pm z_k\;,\;\;k\in \{0,\;1,\dots ,\;n-1 \}.$  Denoting

$f(z)=z^{2n}-(\cos na +\imath \sin na) \tag{4}$

we have  $f(z)=(z-z_0)(z-z_1)\dots (z-z_{n-1})(z+z_0)(z+z_1)\dots (z+z_{n-1})\;$,  or

$$f(z)=\prod_{k=0}^{n-1}(z^2-z_k^2) \tag{5}$$

From here we go further  $$\prod_{k=0}^{n-1}(z_k^2-1)=(-1)^n\prod_{k=0}^{n-1}(1-z_k^2)=(-1)^nf(1)\overset{(4)}{=}(-1)^n(1-\cos 2na-\sin 2na)=$$

$$=(-1)^n(2\sin^2na-2\imath \sin na \cos na)=(-1)^n(-2\imath^2\sin^2 na-2\imath \sin na \cos na)$$

so

$$\prod_{k=0}^{n-1}(z_k^2-1)=-2\imath (-1)^n\sin na(\cos na+\imath \sin na) \tag{6}$$

          On the other hand

$$\prod_{k=0}^{n-1}z_k=\prod_{k=0}^{n-1}\left [(\cos a+\imath \sin a)\left (\cos \frac{k\pi}{n}+\imath \sin\frac{k\pi}{n}\right )\right]=$$

$$=(\cos a+\imath \sin a)^n\prod_{k=0}^{n-1}\left (\cos \frac{k\pi}{n}+\imath \sin \frac{k\pi}{n}\right )=$$

$$=(\cos na+\imath \sin na)\left [\cos \left (\sum_{k=0}^{n-1}k\right )\frac{\pi}{n}+\imath \sin \left (\sum_{k=0}^{n-1}k\right )\frac{\pi}{n}\right ]=$$

$=(\cos na+\imath \sin na)\left [\cos (n-1)\frac{\pi}{2}+\imath \sin (n-1)\frac{\pi}{2}\right ]$

because  $\sum_{k=0}^{n-1}k=\frac{(n-1)n}{2}.$  Analyzing the cases  $n=4m,\;4m+1,\;4m+2,\;4m+3$  it is immediately seen that  $\cos (n-1)\frac{\pi}{2}+\imath \sin (n-1)\frac{\pi}{2}=-\imath^{n+1}$ , and then the previous calculation gives us

$$\prod_{k=0}^{n-1}z_k=-\imath^{n+1}(\cos na+\imath\sin na) \tag{7}$$

          Now we can finish

$$Q_n=\prod_{k=0}^{n-1}\sin \left (a+\frac{k\pi}{n}\right )\overset{(2)}{=}\frac{\prod_{k=0}^{n-1}(z_k^2-1)}{2^n\imath^n\prod_{k=0}^{n-1}z_k}\overset{(6)}{\underset{(7)}{=}}\frac{-2\imath(-1)^n\sin na(\cos na+\imath\sin na)}{2^n\imath^n(-\imath^{n+1})(\cos na+\imath \sin na)}=$$

$=\frac{\imath \cdot (-1)^n\sin na}{2^{n-1}\imath^{2n}\cdot \imath}=\frac{(-1)^n \sin na}{2^{n-1}(-1)^n}=\frac{\sin na}{2^{n-1}}.$  We got the answer of  b).

     Remark CiP   On page 518 the following stupid answer is given :

$$Q_n=\begin{cases}\frac{(-1)^n\cdot \sin na}{2^{n-1}}\;\;\;if\;\;n-even\\\frac{(-1)^{n+1}\cdot \sin na}{2^{n-1}}\;\;\;if\;\;n-odd\end{cases} \tag{Q}$$

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

                    a)  With the same notations as in point b), first we have

$\cos\left (a+\frac{k\pi}{n}\right )=\frac{z_k+\bar z_k}{2}=\frac{z_k^2+1}{2z_k} \tag{8}$

          Then, starting from  (5)  we see that  

$$f(\imath)=\prod_{k=0}^{n-1}(-1-z_k^2)=(-1)^n\prod_{k=0}^{n-1}(z_k^2+1)$$

so

$$\prod_{k=0}^{n-1}(z_k^2+1)=(-1)^nf(\imath) \tag{9}$$

          Then we can write

$$P_n=\prod_{k=0}^{n-1}\cos \left (a+\frac{k\pi}{n}\right )\overset{(8)}{=}\frac{\prod_{k=0}^{n-1}(z_k^2+1)}{2^n\prod_{k=0}^{n-1}z_k}\overset{(7)}{\underset{(9)}{=}}\frac{(-1)^nf(\imath)}{2^n(-\imath^{n+1})(\cos na+\imath \sin na)} \tag{10}$$

But  $f(\imath)\underset{(4)}{=}(-1)^n-\cos 2na-\imath \sin 2na$  and we will treat the  $n-even$  and  $n-odd$  cases separately.

          $\fbox{$n$-even}$

          $f(\imath)=1-\cos na-\imath \sin na=2\sin^2 na-2\imath \sin na \cos na=$

$=-2\imath^2\sin^2 na-2\imath \sin na \cos na=-2\imath \sin na(\cos na+\imath \sin na)$

In  (10)  also appears  $-\imath^{n+1}=-\imath^n\cdot \imath=(-\imath)(-1)^{\frac{n}{2}}$  and then we continue like this

$P_n=\frac{-2\imath \sin na(\cos na+\imath \sin na)}{2^n(-\imath)(-1)^{\frac{n}{2}}(\cos na+\imath \sin na)}=\frac{(-1)^{\frac{n}{2}}\sin na}{2^{n-1}}.$

          $\fbox{$n$-odd}$

          $(-1)^nf(\imath)=-(-1-\cos 2na-\imath \sin 2na)=2\cos^2 na+2\imath \sin na \cos na=$

$=2\cos na(\cos na+\imath \sin na)$

and because  $-\imath^{n+1}=-(-1)^{\frac{n+1}{2}}=-(-1)^{\frac{n-1}{2}+1}=-(-1)^{\frac{n-1}{2}}(-1)=(-1)^{\frac{n-1}{2}}$

we have  $P_n\underset{(10)}{=}\frac{2\cos na(\cos na+\imath \sin na)}{2^n(-1)^{\frac{n-1}{2}}(\cos na+\imath \sin na)}=\frac{(-1)^{\frac{n-1}{2}}\cos na}{2^{n-1}}.$  We got the answer of  a).

      Remark CiP   Although in the solution I used COPILOT, who first told me incorrectly that the minus sign always appears in the formula, then came back and corrected the mistake, we must admit that in the book, on page 517, the correct answer is given. Marius DRĂGAN and Sorin RĂDULESCU are cited as the authors of the solution.

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

                    c)  $R_{2n}=\frac{Q_{2n}}{P_{2n}}=\frac{\frac{\sin 2na}{2^{2n-1}}}{\frac{(-1)^n\sin 2na}{2^{2n-1}}}=(-1)^n\;;$

                         $R_{2n+1}=\frac{Q_{2n+1}}{P_{2n+1}}=\frac{\frac{\sin (2n+1)a}{2^{2n}}}{\frac{(-1)^n\cos (2n+1)a}{2^{2n}}}=(-1)^n\tan(2n+1)a.$

$\blacksquare\;\blacksquare\;\blacksquare$

An article about SOME TRIGONOMETRIC IDENTITIES // hg

 See the full article in GMB 7/2005, pages 289-293.

First identity :

(is formula (5) , page 290, in the cited article)

$$\prod_{k=0}^{n-1}\cos \frac{x+2k\pi}{n}=\frac{1}{2^{n-1}}\left [\cos\frac{n\pi}{2}+(-1)^{n+1}\cdot \cos x\right ] \tag{1}$$

           For  $x=\pi$  we obtain :

$\cos \frac{\pi}{n}\cdot \cos \frac{3\pi}{n} \dots  \cos\frac{(2n-1)\pi}{n}=\frac{1}{2^{n-1}}\left [\cos \frac{n\pi}{2}+(-1)^{n+1}\cdot \cos \pi \right]=\frac{1}{2^{n-1}}\left [\cos \frac{n\pi}{2}+(-1)^n\right ]$

i.e. Exercise 9a).

          For  $x=0$  we obtain :

$\cos \frac{2\pi}{n}\cdot \cos \frac{4\pi}{n}\dots \cos \frac{2(n-1)\pi}{n}=\frac{1}{2^{n-1}} \left [\cos \frac{n\pi}{2}+(-1)^{n+1}\cdot \cos 0\right ]=\frac{1}{2^{n-1}}\left [ \cos \frac{n\pi}{2}-(-1)^n\right ]$

i.e. Exercise 9b).

In particular we have the equalities :

$\cos \frac{\pi}{5}\cdot \cos \frac{3\pi}{5}\cdot \cos \frac{5 \pi}{5}\cdot \cos \frac{7\pi}{5}\cdot \cos \frac{9\pi}{5}=\frac{1}{2^4}\left [ \cos\frac{5\pi}{2}-1\right ]=-\frac{1}{16}\;;$

$\cos \frac{2\pi}{5}\cdot \cos \frac{4\pi}{5}\cdot \cos \frac{6\pi}{5}\cdot \cos\frac{8\pi}{5}=\frac{1}{2^4}\left [\cos \frac{5\pi}{2}+1\right ]=\frac{1}{16}.$

(in construction)

OLIMPIADA NAȚIONALĂ de MATEMATICĂ - cioburi // NATIONAL MATHEMATICS OLYMPIAD - some shards // NATIONALE MATHEMATIK-OLYMPIADE - Scherben

 (O încercare de regăsire a profesorului meu din Liceu, Vasile BIVOLARU. A plecat in GERMANIA, sotia, Ecaterina, a murit acolo  -  mi-a spus fiul lor, Cristian, pe Facebook)

          Scanning, little by little, my collection of  "Gazeta Matematica" magazines (aka GMB), I reached issue 6 from 1978 where the Topics from the 1977 Final Stage are published, the only one I participated in, in 10th grade. 

(in consrtuction)

C : 2816 - A concrete trigonometric equality // კონკრეტული ტრიგონომეტრიული თანასწორობა

The problem was published in GMB 12 / 2004; I don't think it's here. It was solved in GMB 6 / 2005.

    "C:2816.  Prove the equality  $\frac{\sin 110^{\circ}}{\sin 30^{\circ}}-\frac{\sin 60^{\circ}}{\sin 80^{\circ}}=1."$

{authors :  Romanța GHIȚĂ and Ioan GHIȚĂ, Blaj

               Solution CiP - taken from the cited source

          The given equality is successively equivalent to :

$\sin 110^{\circ}\cdot \sin 80^{\circ}=\sin 30^{\circ}\cdot \sin 60^{\circ}+\sin 30^{\circ}\cdot \sin 80^{\circ}\;\Leftrightarrow$

$\Leftrightarrow\;\;2\sin 70^{\circ}\cdot \sin 80^{\circ}=\sin 60^{\circ}+\sin 80^{\circ}\;\Leftrightarrow$

$\Leftrightarrow\;\;\cos 10^{\circ}-\cos 150^{\circ}=\sin 60^{\circ}+\sin 80^{\circ}$

which is obvious, because  $\cos 10^{\circ}=\sin 80^{\circ}\;\;and\;\; -\cos 150^{\circ}=\sin 60^{\circ}=\frac{\sqrt{3}}{2}.$

$\blacksquare$

          REMARK CiP   It would be interesting to study some trigonometric equations starting from the present equality, as I did in the post of September/7/ 2025  A forgotten trigonometric equation.

A problem with Symmetric Polynomials // Um problema com polinómios simétricos

 It is problem 3 016, 2/1977 page 59 from the "Revista Matematica a Elevilor din Timisoara" (Mathematical Magazine of Students from Timisoara).

         "3 016.  If  $a,\;b,\;c$  are nonzero real numbers, such that 

 $a+b+c=0$  and  $a^3+b^3+c^3=a^5+b^5+c^5$ , 

then  $a^2+b^2+c^2=\frac{6}{5}.$

{author : } Titu ANDREESCU, student, Timisoara"

Solution CiP

                    We will use the following algebraic identities :

$(a+b+c)^2-a^2-b^2-c^2=2(ab+bc+ca) \tag{1}$

$(a+b+c)^3-a^3-b^3-c^3=3(a+b)(b+c)(c+a) \tag{2}$

$(a+b+c)^5-a^5-b^5-c^5=$

$=5(a+b)(b+c)(c+a)(a^2+b^2+c^2+ab+bc+ca) \tag{3}$

          (1)  is obtained from the identity :

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$.

          For  (2)  let  $P(a,b,c)=(a+b+c)^3-a^3-b^3-c^3$. If  $b=-a$  then  $Pa,-a,c)=c^3-a^3+a^3-c^3=0$  so  $P$  has the factor  $a+b$. From symmetry it will also have the factors  $b+c,\;c+a$  so, being of degree 3, it results 

$P(a,b.c)=(a+b)(b+c)(c+a)\cdot Q(a,b,c)\;,\;\;deg(Q)=0$

hence  $Q(a,b,c)$  is a constant. As  $P(1,1,0)=2^3-1-1=6$, from  $6=(1+1)(1+0)(0+1)\cdot Q(1,1,0)$  it foloow   $Q=3.$

          For  (3)  let  $P(a,b,c)=(a+b+c)^5-a^5-b^5-c^5$.  $P(a,-a,c)=c^5-a^5+a^5-c^5=0$  so

$P(a,b,c)=(a+b)(b+c)(c+a)\cdot Q(a,b,c)\;,\;\deg(Q)=2 \tag{4}$

So  $ Q$  has the form  $Q(a,b,c)=m\cdot (a^2+b^2+c^2)+n\cdot (ab+bc+ca)$

If we put in  (4)  first  $a=b=1,\;c=0$,  and second  $a=b=c=1$  we get 

$P(1,1,0)=2^5-1-1=30=2(m\cdot 2+n),\;$

$\;\;P(1,1,1)=3^5-1-1-1=240=8(m\cdot 3+n\cdot 3)$

hence  $m=n=5$.

               Let's move on to solving the problem. In the condition  $a+b+c=0$ , we have :

 - from  (1)  

$\sum a^2=-2\sum ab \tag{5}$

  - from  (2)   $0-\sum a^3\;\underset{a+b=-c\;etc}{=}\;\;\;3(-c)(-a)(-b)$ , so

$\sum a^3=3abc \tag{6}$

  - from  (3)  $0-\sum a^5=5(-c)(-a)(-b)(\sum a^2+\sum ab)\overset{(5)}{=}-5abc(\sum a^2-\frac{1}{2}\sum a^2$c , so

$\sum a^5=\frac{5}{2}abc\sum a^2 \tag{7}$

But we also have  $\sum a^3=\sum a^5$ , so from  (6)  and  (7)  it results 

 $3abc=\frac{5}{2}abc \sum a^2$

hence the answer.

$\blacksquare$

 

A RIGHT TRIANGLE and ... one more // Un TRIUNGHI DREPTUNGHIC și ... încă unul // Правоаголен триаголник и ... уште еден

 It's a fairly basic geometry problem (proposed for sixth grade). 

    "E :  6209*.  In triangle  $ABC$  with  $\sphericalangle A=90^{\circ}$ the bisector of angle  $C$ intersects the perpendicular bisector of side  $AB$  in  $D$. Find  $\sphericalangle  CBD$.

 Ion MACREA, professor Brașov"

                    Figure of the problem (CiP) : 

               REMARK CiP  The order of important lines in a triangle is known to be: altitude -  - - bisectors - median - perpendicular bisectors. Since CF is the median and CA the altitude, line CD is between them, so point D is on the extension of segment EF, beyond F and not elsewhere.

ANSWER CiP

$\sphericalangle CDB=90^{\circ}$

                    Solution CiP 

               We rely on the following characterization of right triangles :

                           In a right triangle, the median to the hypotenuse

                          is equal to one half of the hypotenuse.

              Conversely, if in a triangle the median drawn from a vertex

                      has length equal to one half of the side to which it is drawn,

                       then the angle at that vertex is a right angle.

A complete figure is below. Point  $F$  is the midpoint of the leg  $[AB],\;\;DE\perp AB$.

(Basically,  $EF$  is the perpendicular bisector of side  $[AB]$.) 

          $FE,\;AC \perp AB \Rightarrow FE \parallel AC$  and then, 

once

 E is the midpoint of the segment BC                         (1)

 and the second time we have the equalities of angles

$\sphericalangle EDC=\sphericalangle ECD \;\;\;\overset{CD\; is\; the}{\underset{bisector\; of \;\sphericalangle ACB}{=}}\;\;\;\sphericalangle ACD\tag{2}$

             $(2)\Rightarrow\;\;CDE$  is isosceles triangle, with base  $ CD$  so

$DE=EC\underset{(1)}{=}EB=\frac{BC}{2}$

but this says that, in triangle  $BCD$  

    the median  $DE$  drawn from the vertex  $D$  has length equal to one half of the side  $BC$  to which it is drawn,

so  the angle  $\sphericalangle BDC$  at the vertex  $D$  is a right angle.

$\blacksquare$

Two High School Quizzes I Still Remember (9th Grade / Year 1) // Doua Extemporale de pe vremea Liceului (clasa a 9-a/anul I) // Дзве школьныя віктарыны, якія я дагэтуль памятаю (9 клас / 1 год)

          I found two Quizzes in my 1st year high school ALGEBRA textbook.(On the first page is my signature and a personal stamp. I had several math teachers at the beginning of that school year. 

         I will talk about the first of them, Vasile BIVOLARU, who only stayed for three weeks, being incorporated for a 6-month military internship, another time. 

I spent the first two years of high school with Professor Gheorghe DUMITRU, the one who gave me my grades.

First Quiz, with the last teacher

Second quiz

I also found among the pages of the book an Essay on Political Science. (The writing in this essay does not appear to be mine.)

Comments remain open...

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

ANSWER CiP

$$a=1,\;\;\;b=1$$

                         Solution CiP

               Let be       $\frac{a^2+b}{a+b-1}=m\in\mathbb{N},\;\;\;\frac{b^2+a}{a+b}=n\in\mathbb{N}\;\;\;\;\;\;\;\;\;(1)$

From (1)  we obtain :

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

          Subtracting the equations (2) we obtain :

$(a+b-1)\cdot m-(a+b)\cdot n=(a-b)(a+b-1) \tag{3}$

          We view  (3)  as a linear Diophantine equation with the unknowns  $m\; and\; n$. A particular, almost obvious solution of  (3)  is  $m_0=a-b\;,\;\;n_0=0.$

 Then, all integer solutions of the equation  (3)  will be :

$m_k=a-b+k\cdot (a+b)\;,\;\;\;n_k=k\cdot (a+b-1)\;\;,\;\;\;k\in\mathbb{Z} \tag{4}$

However, we are only interested in natural number solutions  $(m_k,n_k)\in\mathbb{N}^*\times \mathbb{N}^*$  so  $k>0.$

          The second equation in  (2)  is written

$k\cdot (a+b-1)\cdot (a+b)=b^2+a$

 from which we obtain :

$k=\frac{b^2+a}{(a+b)(a+b-1)} \tag{5}$

                    This is where I got stuck, although the intention was to see from  (5)  what the possible values ​​for  $k$  are.

     Examining the "First Solution" from the official Olympiad materials (see page 15, Problem 3) I found that I was on the right track. Moreover, the expression in  (5)  appears there with the notation  $"p"$  (which in Greek is written  $\pi$). “Diophantus would denote this number using the letter π, hence my choice to include it in the title of the post.”(Idea first shared with COPILOT.) That's how I chose, a little bit cool, sly, the title of the Post there.

"Returning to our sheep," as the proverb would say, we see that :

$k< 1\;\Leftrightarrow\;b^2+a< a^2+2ab+b^2-a-b\;\Leftrightarrow$

$\Leftrightarrow\;a(a-2)+b(2a-1)>0$

for  $a\geqslant 2$. So in the case  $a\geqslant 2$  there are no solutions.

          For  $a=1,\;\;m=\frac{1+b}{b}=1+\frac{1}{b}\;,\;\;n=\frac{b^2+1}{b+1}=b-1+\frac{2}{b+1}$. and  $m\;,n$  are integers if and only if  $b=1$. 

          We got the answer.

$\blacksquare$

O CONTRADICTIE LA CARE NU TE ASTEPTAI // A CONTRADICTION 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,  $a,\;b \geqslant 2.$

          We denote  $s=a+b$;  in this case we have  $s\geqslant 4$.

$n=\frac{a+b+b^2-b}{a+b}=\frac{s+b(b-1)}{s}=1+\frac{b(b-1)}{s} \tag{2}$

$m=\frac{(s-b)^2+b}{s-1}=\frac{s^2-2sb+b^2+b}{s-1}= s-2b+1+\frac{b(b-1)+1}{s-1}\tag{3}$

(For (3) we process the numerator like this  $s^2-2sb+b^2+b=(s^2-s)\color{Red}{+s}\underset{=-2b(s-1)}{\underbrace{-2sb+2b}}-b+b^2\color{Red}{-1}+1=$

$=(s-1)(s+1-2b)+b^2-b+1$)

We will show that we can have neither  $m\in \mathbb{Z}$  nor  $n\in \mathbb{Z}$.

We assume the opposite :

$m\in\mathbb{Z}\;\;AND\;\; n\in\mathbb{Z} \tag{4}$

     Because  $b(b-1)\geqslant 2$,  we have  $n\in\mathbb{Z}\;\overset{(2)}{\Rightarrow} \;b(b-1)=k \cdot s,\;k\in\mathbb{N}.$  But  $s>b\Rightarrow k\cdot s=b(b-1)>k\cdot b$,  hence

$\color{Red}{\fbox{k<b-1}}\tag{5}$

On the other hand  $m\in\mathbb{Z}\overset{(3)}{\Rightarrow} s-1\mid ks+1=k(s-1)+(k+1)$,  so  $s-1 \mid k+1$,  and hence  $k+1\geqslant s-1$  from where  

$k\geqslant s-2=a+b-2=(a-1)+(b-1)$.

From the previous row it follows :

$\color{Red}{\fbox{$k\geqslant b-1$}}\tag{6}$

But  (5) and (6) are contradictory.

          

          And here, as in the first example, the contradiction appeared somewhere you didn't expect it. So (4) is false, so at least one of the numbers  $m,\;n$  is not an integer.

UN NUMAR $\pi$ CIUDAT CARE APARE IN REZOLVARE // A STRANGE NUMBER $\pi$ THAT APPEARS IN THE SOLUTION // ΕΝΑΣ ΠΑΡΑΞΕΝΟΣ ΑΡΙΘΜΟΣ $\pi$ ΠΟΥ ΕΜΦΑΝΙΖΕΤΑΙ ΣΤΗ ΛΥΣΗ

This is about the problem exposed in the Post here. The problem has two official solutions (see pages 15-16). I discussed the Second Solution in the Post here. The First Solution inspired me to continue the solution in the Post here(here I also explained why I will use the Greek letter  $\pi$  for the "p" symbol).

               First Solution

               

               Let be  $m=\frac{a^2+b}{a+b-1}=a+1-\frac{ab-1}{a+b-1}$

$n=\frac{b^2+a}{a+b}=b+1-\frac{ab+b}{a+b}$

If  $m\;,n$  are natural number, then the number

$p\;(\pi\; for\;me)=\frac{ab+b}{a+b}-\frac{ab-1}{a+b-1}=\frac{b^2+a}{(a+b)(a+b-1}>0 \tag{1}$

(!! Notice that the number in  (1)  coincides with the number  $k$  in the mentioned Post)

We have  $p<1\;\Leftrightarrow\;b^2+a<a^2+2ab+b^2-a-b\;\Leftrightarrow$

$\Leftrightarrow\;a(a-2)+b(2a-1)>0$

and the last inequality is true for  $a\geqslant 2$,  so in the case  $a\geqslant 2$  there are no

solutions. For  $a=1$  we obtain the solution  $(1,1)$.

$\blacksquare$

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>