A Curious Inequality with... Radicals // ... रेडिकलहरूसँगको एउटा जिज्ञासु असमानता

 Nu mai stiu

Dupa cum xice pe coperta

(to be continue)

Neculai STANCIU (Buzău) in the MATHEMATICAL JOURNAL // Neculai STANCIU (Buzău) în GAZETA MATEMATICĂ

 In the recent issue of the Exercise Supplement (aka SGM) he proposed the Problem S:E25.199. The problem statement is :

                    "Consider  $\Delta ABC$  a triangle such that  $\measuredangle A=2\measuredangle C$.

                      Prove that  $\frac{AB}{BC}=\frac{1}{2\cos C}.$

Neculai STANCIU, Buzău"

Solution CiP

We construct  $AD$ - the bisector of angle  $\widehat{BAC}$. We have 

$$\measuredangle CAD=\measuredangle BAD=\frac{\measuredangle BAC}{2}=\measuredangle C$$

          We now construct  $BE\parallel AD,\;E\in AC$. We have for the angles formed with the secant $AB$  that

$\measuredangle BAD=\measuredangle ABE=\measuredangle C \tag{1}$

and the exterior angle  $\widehat{BAC}$ of triangle  $\Delta ABE$ shows us that

$\measuredangle BAC=\measuredangle ABE+\measuredangle AEB\Leftrightarrow 2\measuredangle C\overset{(1)}{=}\measuredangle C+\measuredangle AEB\Rightarrow \measuredangle AEB=\measuredangle C$

From all the angles equal to  $C$  in the figure, we see that we have the isosceles triangles  $BAE$  and  $CBE$, so

$ AB=AE\;,\;BC=BE\tag{2}$

Constructing the height  $AF$  in the isosceles triangle  $ABE,\;AF\perp BE$, it will also be the median, so  $BF=EF=BE/2$. Then, in triangle  $ABF$

$\cos C\underset{(1)}{=}\cos \widehat{ABF}=\frac{BF}{AB}=\frac{BE/2}{AB}\underset{(2)}{=}\frac{BC/2}{AB}=\frac{BC}{2\cdot AB}$,  which is equivalent to the statement  $\frac{AB}{BC}=\frac{1}{2\cos C}.$

$\blacksquare$

          Remark CiP  The solution does not use any advanced trigonometry formulas, only definitions, otherwise from the Law of Sines in triangle  $ABC$  we could easily obtain

$\frac{AB}{\sin C}=\frac{BC}{\sin A}\Rightarrow \frac{AB}{BC}=\frac{\sin C}{\sin A}=\frac{\sin C}{\sin 2C}=\frac{\sin C}{2\sin C \cdot \cos C}=\frac{1}{2\cos C}$.

Starting from a Trigonometric Identity without variables // Ξεκινώντας από μια Τριγωνομετρική Ταυτότητα χωρίς μεταβλητές

           We have on my Page, at No. 6, the Identity

$\frac{\tan10^{\circ}}{\tan20^{\circ}\cdot \tan40^{\circ}}=\frac{1}{\sqrt{3}}$

It is equivalent to  $\frac{\tan10^{\circ}}{\tan20^{\circ}\cdot \tan40^{\circ}}=\tan30^{\circ}$  or else

$$\tan20^{\circ}\cdot \tan30^{\circ}\cdot \tan40^{\circ}=\tan10^{\circ} \tag{1}$$

               Let's consider the equation

$$ \tan x\cdot \tan\frac{3x}{2}\cdot \tan 2x=\tan \frac{x}{2}\tag{2}$$

(1) shows that  $x=20^{\circ}$  is a solution for (2).

The problem of completely solving this equation remains open for now.

A forgotten trigonometric equation // Egy elfeledett trigonometrikus egyenlet

 We will solve the following trigonometric equation here :

$$\tan(30^{\circ}-x)\cdot \tan(30^{\circ}+x)=\frac{\tan x}{\sqrt{3}} \tag{1}$$

I first mentioned it in the Post here. There we also showed that the equation  (1)  is equivalent to the equation

$$t^3-3\sqrt{3}\cdot t^2-3\cdot t+\sqrt{3}=0,\;\;t=\tan x \tag{2}$$

ANSWER CiP

$$x_k^{\circ}=20^{\circ}+k\cdot 60^{\circ},\;\;k\in\mathbb{Z} \tag{3}$$

                    Solution CiP

              The calculation has already been done, but we are presenting it for convenience.

$(1)\overset{t=\tan x}{\Leftrightarrow} t^3-3\sqrt{3}\cdot t^2-3\cdot t+\sqrt{3}=0\Leftrightarrow \sqrt{3}=\frac{3t-t^3}{1-3t^2}\Leftrightarrow$

$\Leftrightarrow \sqrt{3}=\frac{3\cdot \tan x-\tan^3x}{1-3\cdot \tan^2x}\Leftrightarrow \tan 60^{\circ}=\tan 3x\Leftrightarrow 3x=60^{\circ}+k\cdot 180^{\circ},\;\;k\in \mathbb{Z}$

From here we get the answer  (3)

$\blacksquare$

          Remark CiP   It all started with solving the first equation from No. 6 on Page here. Now, after we have solved the equation  (1) (see also the Post from September 3, 2025), we can write the identities :

$$\frac{\tan10^{\circ}\cdot \tan50^{\circ}}{\tan20^{\circ}}=\frac{1}{\sqrt{3}}\tag{4}$$

$$\frac{\tan10^{\circ}\cdot \tan70^{\circ}}{\tan40^{\circ}}=\frac{1}{\sqrt{3}}\tag{5}$$

$$\frac{\tan50^{\circ}\cdot \tan70^{\circ}}{\tan80^{\circ}}=\frac{1}{\sqrt{3}}\tag{6}$$

          But  (4), considering that  $\frac{1}{\tan20^{\circ}}=\cot20^{\circ}$, is exactly the first identity from No. 6. Seeking to express all quantities as tangents of angles not exceeding  $45^{\circ}$, we further have  $\frac{\tan10^{\circ}\cdot \tan50^{\circ}}{\tan20^{\circ}}=\frac{\tan10^{\circ} \cdot \color{Red}{\cot40^{\circ}}}{\tan20^{\circ}}=\frac{\tan10^{\circ}\cdot \color{Red}{\frac{1}{\tan40^{\circ}}}}{\tan20^{\circ}}=\frac{\tan10^{\circ}}{\tan20^{\circ}\cdot \tan40^{\circ}}=\frac{1}{\sqrt{3}}$

which is the second entity from the aforementioned No. 6.

     Doing the same thing with  (5) : 

$\frac{1}{\sqrt{3}}=\frac{\tan10^{\circ}\cdot \color{Red}{\cot20^{\circ}}}{\tan40^{\circ}}=\frac{\tan10^{\circ}\cdot \color{Red}{\frac{1}{\tan20^{\circ}}}}{\tan40^{\circ}}=\frac{\tan10^{\circ}}{\tan20^{\circ}\cdot \tan{40^{\circ}}}$

and with (6) :

$\frac{1}{\sqrt{3}}=\frac{\color{Red}{\cot40^{\circ}\cdot \cot20^{\circ}}}{\color{Red}{\cot10^{\circ}}}=\frac{\frac{1}{\tan40^{\circ}}\cdot \frac{1}{\tan20^{\circ}}}{\frac{1}{\tan10^{\circ}}}=\frac{\tan10^{\circ}}{\tan20^{\circ}\cdot \tan40^{\circ}}$

So the second of the identities in No. 6 is the most important.

              Let's demonstrate it directly then. We will see that the pattern is the same.

$\frac{\tan10^{\circ}}{\tan20^{\circ}\cdot \tan40^{\circ}}=\frac{\frac{\sin10^{\circ}}{\cos10^{\circ}}}{\frac{\sin20^{\circ}}{\cos20^{\circ}}\cdot \frac{\sin40^{\circ}}{\cos40^{\circ}}}=\frac{8\cdot \sin10^{\circ}\cdot \cos20^{\circ}\cdot \cos40^{\circ}}{8\cdot \cos10^{\circ}\cdot \sin20^{\circ}\cdot \sin40^{\circ}}\tag{7}$

The numerator in (7) is  $4\cdot \frac{\sin20^{\circ}}{\cos10^{\circ}}\cdot \cos20^{\circ}\cdot \cos40^{\circ}=\frac{1}{\cos10^{\circ}}\cdot 2\cdot \sin40^{\circ}\cdot \cos40^{\circ}=\frac{1}{\cos10^{\circ}}\cdot \sin80^{\circ}=1$

The denominator in (7) is  $4\cdot \cos10^{\circ}\cdot(\cos20^{\circ}-\cos60^{\circ})=4\cos10^{\circ}\cdot \cos20^{\circ}-4\cos10^{\circ}\cdot \frac{1}{2}=$

$=2(\cos10^{\circ}+\cos30^{\circ})-2\cos10^{\circ}=2\cos10^{\circ}+2\cdot \frac{\sqrt{3}}{2}-2\cos10^{\circ}=\sqrt{3}$

So the final value in (7) is  $\frac{1}{\sqrt{3}}$.

<end Rem>

The Expressions for Roots of Cubic Polynomial // تعبيرات جذور كثيرة الحدود التكعيبية، مرة أخرى بمساعدة اللاهوت

                We have discussed this issue before. We will discuss the similarities and differences with the current situation at the end.

          Here we consider the problem : 

                   Let  $\tau$  be one of the roots of the equation

$t^3-3\sqrt{3}\cdot t^2-3\cdot t+\sqrt{3}=0 \tag{1}$

                   The other two roots are

$\tau_2=\frac{\sqrt{3}}{2}\cdot \tau^2-5\cdot \tau+\frac{\sqrt{3}}{2}\;,\;\;\tau_3=-\frac{\sqrt{3}}{2}\cdot \tau^2+4\cdot \tau+\frac{5\sqrt{3}}{2} \tag{2}$ 

SOLUTION  CiP

               From the fact that  $\tau$  checks  (1)  we obtain some useful formulas in future calculations

$\tau^3=3\sqrt{3}\cdot \tau^2+3\cdot \tau -\sqrt{3} \tag{3}$

$\tau^4=30\cdot \tau^2+8\sqrt{3} \cdot \tau-9 \tag{4}$

$\frac{1}{\tau}=-\frac{\sqrt{3}}{3}\cdot \tau^2+3\cdot \tau+\sqrt{3} \tag{5}$

          Indeed, $\tau^3-3\sqrt{3}\cdot \tau^2-3\cdot \tau +\sqrt{3}=0 \Rightarrow \tau^3=3\sqrt{3} \cdot \tau^2+3\cdot \tau-\sqrt{3}$,  i.e. (3). Then  $\tau^4=\tau \cdot \tau^3\overset{(3)}{=}$

$=\tau (3\sqrt{3}\cdot \tau^2+3\cdot \tau-\sqrt{3})=3\sqrt{3}\cdot \tau^3+3\cdot \tau^2-\sqrt{3}\cdot \tau\overset{(3)}{=}3\sqrt{3}(3\sqrt{3}\cdot \tau^2+3\cdot \tau-\sqrt{3})+3\cdot \tau^2-\sqrt{3}\cdot \tau=30\cdot \tau^2+8\sqrt{3}\cdot \tau-9$ i.e. (4). Finally, $\sqrt{3}=-\tau^3+3\sqrt{3}\cdot \tau^2+3\cdot \tau=\tau \cdot (-\tau^2+3\sqrt{3}\cdot \tau+3)\Rightarrow\frac{1}{\tau}=\frac{-\tau^2+3\sqrt{3}\cdot \tau+3}{\sqrt{3}}=-\frac{1}{\sqrt{3}}\cdot \tau^2+3\cdot \tau+\sqrt{3}$, i.e. (5).

          From the first formula of Vieta  $\tau_1+\tau_2+\tau_3=3\sqrt{3}$  we get

$$\tau_2+\tau_3=3\sqrt{3}-\tau \tag{6}$$

and from the third  $\tau_1\cdot \tau_2\cdot \tau_3=-\sqrt{3}$  we deduce

$$\tau_2\cdot \tau_3=-\frac{\sqrt{3}}{\tau} \tag{7}$$

(6) and (7) show that  $\tau_{2,3}$  are the roots of the quadratic equation

$t^2-(3\sqrt{3}-\tau) \cdot t-\frac{\sqrt{3}}{\tau}=0 \tag{8}$

          Let's calculate its discriminant:  $\Delta_2=(3\sqrt{3}-\tau)^2+\frac{4\sqrt{3}}{\tau}=$

$\overset{\color{Red}{!!!}}{=}27-6\sqrt{3}\cdot \tau+\tau^2+\frac{4\sqrt{3}\cdot \tau}{\color{Red}{\tau^2}}=\frac{\tau^4-6\sqrt{3}\cdot \tau^3+27\cdot \tau^2+4\sqrt{3}\cdot \tau}{\color{Red}{\tau^2}}=$

$\overset{(4)}{\underset{(3)}{=}}\frac{(30\cdot \tau^2+8\sqrt{3}\cdot \tau-9)-6\sqrt{3}(3\sqrt{3}\cdot \tau^2+3\cdot \tau-\sqrt{3})+27 \tau^2+4\sqrt{3}\cdot \tau}{\color{Red}{\tau^2}}=\frac{3\cdot \tau^2-6\sqrt{3}\cdot \tau+9}{\color{Red}{\tau^2}}=\frac{(\sqrt{3}\cdot \tau-3)^{\color{Red}2}}{\color{Red}{\tau^2}}$

<  Hallelujah, I got a perfect square  !!!  > 

Let's do a little more calculation to make writing easier :  $\pm\sqrt{\Delta_2}=\frac{\sqrt{3}\cdot \tau-3}{\tau}=$

$=(\sqrt{3}\cdot \tau-3)\cdot \frac{1}{\tau}\overset{(5)}{=}(\sqrt{3}\cdot \tau-3)\left (-\frac{\sqrt{3}}{3}\cdot \tau^2+3\cdot \tau+\sqrt{3}\right )=$

$=-\tau^3+\sqrt{3}\cdot \tau^2+3\sqrt{3}\cdot \tau^2-9\cdot \tau+3\cdot \tau-3\sqrt{3}=$

$\overset{(3)}{=}-(3\sqrt{3}\cdot \tau^2+3\cdot \tau-\sqrt{3})+4\sqrt{3}\cdot \tau^2-6\cdot \tau-3\sqrt{3}=\color{Blue}{\sqrt{3}\cdot \tau^2-9\cdot \tau-2\sqrt{3}}$

 !!! Remember this expression !!!

With these, the equation  (8)  has the roots

$\tau_{2,3}=\frac{3\sqrt{3}-\tau \pm \sqrt{\Delta_2}}{2}=\frac{3\sqrt{3}-\tau \pm (\sqrt{3}\cdot \tau^2-9\cdot \tau-2\sqrt{3})}{2} \tag{9}$

from where we obtain the expressions  (2).

$\blacksquare$

               REMARKS CiP

                    1.  In the calculation of  $\Delta_2$  that follows immediately after the equation  (8), I made a trick so that at least the denominator would be a perfect square. Then I was lucky enough !!  Compare with the Post  << Not only God does not help, but also Allah does not "put you in trouble">>.

If in (7) we had used (5), obtaining  $\tau_2\cdot \tau_3=\tau^2-3\sqrt{3}\cdot \tau-3$, then for the equation (8) we would have had  $\Delta_2=(3\sqrt{3}-\tau)^2-4(\tau^2-3\sqrt{3}\cdot \tau-3)=39+6\sqrt{3}\cdot \tau -3\cdot \tau^2.$  Seeing this expression of  $\Delta_2$  we wondered, as we did in the Post How? can we recognize a perfect square??, "how the hell" can we notice - compare with (9) - that

$$\color{Blue}{39+6\sqrt{3}\cdot \tau-3\cdot \tau^2=(\sqrt{3}\cdot \tau^2-9\cdot \tau-2\sqrt{3})^2}$$

Problem NOT solved yet.

                    2.  The theoretical part about when and how such a statement is possible was exposed in  the  Post "The Rational Expressions for Roots of Cubic Polynomial". There are some differences, but only apparent.

          First of all, that the polynomial  $f=X^3-3\sqrt{3}\cdot X^2-3\cdot X+\sqrt{3}$  is not in  $\mathbb{Q}[X]$  but in  $\mathbb{Q}(\sqrt{3})[X].$

          Then we need to make sure that the polynomial $f$  is irreducible. Maybe this is not that important, and I don't have an immediate justification at hand. 

          Third, its discriminant must be a perfect square (see Proposition in the Post). From  $f$  we obtain the depressed cubic by substitution  $t=u+\sqrt{3}$, obtainig the polynomial  $u^3-12\cdot u-8\sqrt{3}$. Its discriminant, the same as for  $f$, is  $-4p^3-27q^2=4\cdot12^3-27\cdot(64\cdot3)=1\;728$  so equal to $3\cdot 576=(24\sqrt{3})^2$, hence perfect square in  $\mathbb{Q}(\sqrt{3})$.

<end REM's>

Phương trình đại số có nghiệm lượng giác //Algebraic Equation with Trigonometric Solution

 We are showing here what we have left from a post from yesterday. I was helped for this in the AOPS Forum.

               The roots of the equation

$$t^3-3\sqrt{3}\cdot t^2-3\cdot t+\sqrt{3}=0 \tag{E}  $$

             are  $\tan20^{\circ},\;\;\tan80^{\circ},\;\;\tan140^{\circ}$

Adapted Solution by CiP

$(E)\Leftrightarrow \sqrt{3}\cdot (1-3t^2)=3t-t^3 \Leftrightarrow \sqrt{3}=\frac{3t-t^3}{1-3t^2}\;\underset{t=\tan x}{=}\frac{3\cdot \tan x-\tan^3x}{1-3\cdot \tan^2x}=\tan (3x)$

so, taking the equality of the extreme terms,  $3\cdot x=60^{\circ}+k\cdot 180^{\circ},\;\;k\in \mathbb{Z}$, hence  $x_k=20^{\circ}+k\cdot 60^{\circ}$. Returning to the variable  $t$,  we have the roots

$t_1=\tan( 20^{\circ}+3k\cdot 60^{\circ})=\tan (20^{\circ}+k\cdot 180^{\circ})=\tan20^{\circ},$

$t_2=\tan(20^{\circ}+(3k+1)\cdot 60^{\circ})=\tan(80^{\circ}+k\cdot 180^{\circ})=\tan80^{\circ}$

$t_3=\tan(20^{\circ}+(3k+2)\cdot 60^{\circ})=\tan(140^{\circ}+k\cdot 180^{\circ})=\tan140^{\circ}$.

$\blacksquare$

          Remark Cip  Vieta's formulas for the roots of the equation (E) lead to the equalities

$\tan20^{\circ}-\tan40^{\circ}+\tan80^{\circ}=3\sqrt{3}$

$\tan20^{\circ}\cdot \tan 40^{\circ}+\tan 40^{\circ}\cdot \tan 80^{\circ}-\tan 20^{\circ}\cdot \tan 80^{\circ}=3$

$\tan20^{\circ}\cdot \tan 40^{\circ}\cdot \tan80^{\circ}=\sqrt{3}$

<end Rem>

A trigonometric identity without variables

 We will prove the trigonometric identity

$$\tan10^{\circ}\cdot \tan50^{\circ}\cdot \cot20^{\circ}=\frac{1}{\sqrt{3}} \tag{1}$$

The problem appears at Solutions in GMB 3/2005, number O:1063, page 132. You can find more about this magazine here and here.

Official Solution

      Using that we have

 $\sin50^{\circ}=\cos(90^{\circ}-50^{\circ})=\cos40^{\circ},\;\;\cos10^{\circ}=\sin(90^{\circ}-10^{\circ})=\sin80^{\circ}$

and  $\cos50^{\circ}=\sin(90^{\circ}-50^{\circ})=\sin40^{\circ}$  we obtain

$\tan10^{\circ}\cdot \tan50^{\circ}\cdot \cot20^{\circ}=\frac{\sin10^{\circ}}{\cos10^{\circ}} \cdot \frac{\sin50^{\circ}}{\cos50^{\circ}} \cdot \frac{\cos 20^{\circ}}{\sin20^{\circ}}=\frac{\sin10^{\circ}\cdot \cos40^{\circ}\cdot \cos20^{\circ}}{\sin80^{\circ}\cdot \sin40^{\circ}\cdot \sin20^{\circ} } \tag{2}$

          Up in (2) we have, from  $\sin20^{\circ}=2\cdot \sin10^{\circ}\cdot \cos10^{\circ}$  if we replace  $2\cdot \sin10^{\circ}$ :

$8\cdot \sin10^{\circ}\cdot \cos20^{\circ}\cdot \cos40^{\circ}=\frac{4\cdot \sin20^{\circ}\cdot \cos20^{\circ}\cos40^{\circ}}{ \cos10^{\circ}}=\frac{2\cdot \sin40^{\circ}\cdot \cos40^{\circ}}{\cos10^{\circ}}=\frac{\sin80^{\circ}}{\cos10^{\circ}}=1.$

          Down in (2) we have, using  $2\sin u\sin v=\cos(v-u)-\cos(v+u)$, replacing  $\sin80^{\circ}=\cos10^{\circ}$, and using  $2\cos u \cos v=\cos(v-u)+\cos(v+u)$

$8\cdot \sin20^{\circ}\cdot \sin40^{\circ}\cdot \sin80^{\circ}=4(\cos20^{\circ}-\cos60^{\circ})\cdot \cos10^{\circ}=$

$=4\cdot \cos20^{\circ}\cdot \cos10^{\circ}-4\cdot \frac{1}{2}\cdot \cos10^{\circ}=2(\cos10^{\circ}+\cos30^{\circ})-2\cdot \cos10^{\circ}=$

$=2\cdot \cos10^{\circ}+2\cdot \frac{\sqrt{3}}{2}-2\cdot \cos10^{\circ}=\sqrt{3}$

These values ​​entered in  (2) give the answer.

$\blacksquare$

               Just as "After the battle, everyone is a general", we will also provide a solution.

                    Solution CiP

                Lemma CiP      $t_1=\tan20^{\circ}$  is  one of the roots of the equation

$$t^3-3\sqrt{3}\cdot t^2-3\cdot t+\sqrt{3}=0 \tag{3}$$

               Proof of Lemma

          In the triple angle formula  $\tan 3\theta=\frac{3\tan \theta-\tan^3 \theta}{1-3\tan^2 \theta}$  we substitute

$\theta=20^{\circ},\;\;t=\tan \theta$  and because  $\tan3\theta=\tan 60^{\circ}=\sqrt{3}$, we have

$\frac{3t-t^3}{1-3t^2}=\sqrt{3}\Leftrightarrow\;3t-t^3=\sqrt{3}-3\sqrt{3}\cdot t^2\Leftrightarrow\;t^3-3\sqrt{3}\cdot t^2-3\cdot t+\sqrt{3}=0$

that is (3).

qed Lemma$\square$

          To prove (1), we successively transcribe it (1)$\;\Leftrightarrow$

$\Leftrightarrow\;\frac{\tan(30^{\circ}-20^{\circ})\cdot \tan(30^{\circ}+20^{\circ})}{\tan20^{\circ}}=\frac{1}{\sqrt{3}}\Leftrightarrow\;\tan(30^{\circ}-20^{\circ})\cdot \tan(30^{\circ}+20^{\circ})=\frac{\tan20^{\circ}}{\sqrt{3}}$

and we have the equivalent problem: 

              Verify that the equation 

$$\tan(30^{\circ}-x)\cdot \tan (30^{\circ}+x)=\frac{\tan x}{\sqrt{3}} \tag{4}$$

               is satisfied by  $x=20^{\circ}$.

The left side of  (4)  is written :

$$\frac{\sin(30^{\circ}-x) \cdot \sin(30^{\circ}+x)}{\cos(30^{\circ}-x)\cdot \cos(30^{\circ}+x)}\;\;\overset{2\sin u\sin v=\cos((v-u)-\cos(v+u)}{\underset{2\cos u\cos v=\cos(v-u)+\cos(v+u)}{=}}$$

$=\frac{\cos2x-\cos60^{\circ}}{\cos2x+\cos60^{\circ}}\;\;\underset{\tan x=t}{=}\;\;\frac{\frac{1-t^2}{1+t^2}-\frac{1}{2}}{\frac{1-t^2}{1+t^2}+\frac{1}{2}}=\frac{2-2t^2-1-t^2}{2-2t^2+1+t^2}=\frac{1-3t^2}{3-t^2}$

so we need to show that the equation  $\frac{1-3t^2}{3-t^2}=\frac{t}{\sqrt{3}}$  is verified by  $t=\tan20^{\circ}$. But the equation with the unknown  $t$ is written equivalently

$\sqrt{3}-3\sqrt{3}\cdot t^2=3\cdot t-t^3\;\Leftrightarrow\;t^3-3\sqrt{3}\cdot t^2-3\cdot t+\sqrt{3}=0$

which the Lemma shows has the solution  $t=\tan20^{\circ}$.

$\blacksquare$

          Remark CiP  By this we have NOT completely solved the equation  (4) nor what are the other two roots of the equation  (3).

<end Rem>

BUBBLE's algebra // Algebra lui BULĂ

           Bubble is a comic character. In Romanian, his name also has a trivial connotation.

          Everyone knows that multiplication is distributive over addition...

$$a\cdot (b+c)=a\cdot b+a\cdot c \tag{$\cdot$ / +}$$

Here we imply a certain "priority" of multiplication over addition, so as not to write excessively pedantically   $a\cdot (b+c)=(a\cdot b)+(a\cdot c)$.

          What if addition were distributive to multiplication ?

A very narrow-minded math teacher would compare a student who thinks like this to Bubble. This would be written

$$a+b\cdot c=(a+b)\cdot (a+c) \tag{+/$\cdot$}$$

or, excessively pedantically, like this    $a+(b\cdot c)=(a+b)\cdot (a+c).$

          Let's not forget that it is possible for two operations to be mutually distributive of each other e.g. in the case of Boolean algebras.

          I thought about a similar situation when I was developing initial assessment tests. I gave a problem like this to 8th graders :

                 Let  $a,\;b,\;c$  be real numbers that satisfy 

$a+b+c=1 \tag{1}$

                  Prove that  $a+b\cdot c=(a+b)(a+c) \tag{2}$

         There are many ways to demonstrate this, from a complete lack of reaction to the statement, to the most complicated calculations. Which? would be the shortest, and therefore most elegant, demonstration?? I would say that this would be :

$a+b\cdot c=a\cdot 1+bc \underset{(1)}{=}a(a+b+c)+bc=a^2+ab+ac+bc=a(a+b)+c(a+b)=(a+b)(a+c)$

 

          The problem in the image is related to the same context.

(I don't think there's any need for translation of E : 6134*.) "I. SAFTA, Pitești" is a well-known name in the field of problem-solving.

                                                       Solution CiP

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

$=\frac{a^2+2ab+b^2}{2}+\frac{b^2+2bc+c^2}{2}+\frac{c^2+2ca+a^2}{2}=$

$=\underline{\frac{(a+b)^2}{2}+\frac{(b+c)^2}{2}+\frac{(c+a)^2}{2}}=$

$\overset{a+b=1-c}{\underset{and \;the\; others}{=}}\frac{(1-c)^2}{2}+\frac{(1-a)^2}{2}+\frac{(1-b)^2}{2}=\underline{\left (\frac{1-c}{\sqrt{2}}\right )^2+\left ( \frac{1-a}{\sqrt{2}}\right )^2+\left( \frac{1-b}{\sqrt{2}}\right )^2}$

$\blacksquare$

Regarding the Problem 16 667

 Here, page 47, the solution was published. I haven't found the Statement yet.

     We select the property in the photo.

 ANSWER CiP

We will show that one of the numbers  $a_k$  is equal to 1, 

and the others are zero. Then  $b=k$.

                              Solution CiP

               We will treat the case  $n=4$  in detail, so for

$a_1,\;\;a_2,\;\;a_3,\;\;a_4\geqslant 0 \tag{1}$

we have equations

\begin{cases}a_1+a_2+a_3+a_4=1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(2)\\1\cdot a_1+2\cdot a_2+3\cdot a_3+4\cdot a_4=b\;\;\;\;\;\;\;\;\;\;\;\;(3)\\1^2\cdot a_1+2^2\cdot a_2+3^2\cdot a_3+4^2\cdot a_4=b^2\;\;\;\;\;(4)\end{cases}

Subtracting (2) from (3) we have  $b-1=a_2+2\cdot a_3+3\cdot a_4\geqslant 0$,  so  $b\geqslant 1.$

          $\textbf{If}\;\;\bf{b\geqslant 4}$  then 

$b^2-4\cdot b\;\underset{(4)-4\times (3)}{=}(a_1+4a_2+9a_3+16a_4)-4\cdot (a_1+2a_2+3a_3+4a_4)$,  so

$0\leqslant b^2-4b=-3a_1-4a_2-3a_3\overset{(1)}{\leqslant} 0 \tag{5}$

Therefore, in (5) we have everywhere the sign  $"="$. In particular

$b=4,\;\;a_1=a_2=a_3=0,\;\;and\;\;a_4\overset{(2)}{=}1 \tag{6}$

        $\textbf{Let}\;\;\bf{b\in[i,i+1)}$  for some  $i\in\{1,\;2,\;3\}$. We calculate first

$b^2-b\cdot i\;\overset{(4)}{\underset{i\times (3)}{=}}(1a_1+4a_2+9a_3+16a_4)-(i\cdot a_1+2i\cdot a_2+3i\cdot a_3+4i\cdot a_4)$  so

$b^2-bi=(1-i)a_1+2(2-i)a_2+3(3-i)a_3+4(4-i)a_4 \tag{7}$

On the other hand

$b-i\cdot 1\overset{(3)}{\underset{i\times (2)}{=}}(a_1+2a_2+3a_3+4a_4)-(i\cdot a_1+i\cdot a_2+i\cdot a_3+i\cdot a_4)=$

$=(1-i)a_1+(2-i)a_2+(3-i)a_3+(4-i)a_4$

which, multiplied by  $b$  gives us

$b^2-bi=b(1-i)a_1+b(2-i)a_2+b(3-i)a_3+b(4-i)a_4 \tag{8}$

Equating (7) and (8)

$(1-i)a_1+2(2-i)a_2+3(3-i)a_3+4(4-i)a_4=b(1-i)a_1+b(2-i)a_2+b(3-i)a_3+b(4-i)a_4$

we obtain

$(b-1)(i-1)a_1+(b-2)(i-2)a_2+(b-3)(i-3)a_3+(b-4)(i-4)a_4=0 \tag{9}$

               For  $i=1$, (so  $1\leqslant b<2$), the first term cancels out, and all coefficients of   $a_2,\;a_3,\;a_4$  are  $>0$.

Therefore  $a_2=a_3=a_4=0$,  so (cf. (2))  $a_1=1$  and then  $b\overset{(3)}{=}1$.

                For  $i=2$, (so  $2\leqslant b<3$), then the second term cancels out, and all coefficients of  $a_1,\;a_3,\;a_4$  are  $>0$.

Therefore  $a_1=a_3=a_4=0$,  so (cf.(2))  $a_2=1$  and then  $b\overset{(3)}{=}2$.

                For  $i=3$, (so  $3\leqslant b <4$), then the third term cancels out, and all coefficients of  $a_1,\;a_2,\;a_4$  are  $>0$.

Therefore  $a_1=a_2=a_4=0$, so (cf.(2))  $a_3=1$  and then  $b\overset{(3)}{=}3$.

The last three conclusions, together with (6), validate the answer in this particular case.

               The general case extends what is shown in case  $n=4$. So let be the numbers

$a_1,\;a_2,\;\dots,a_n\;\geqslant 0 \tag{11}$

that verify the relations

$$a_1+a_2+\dots+a_n=\sum_{k=1}^na_k=1 \tag{12}$$

$$a_1+2\cdot a_2+\dots+n\cdot a_n=\sum_{k=1}^nk\cdot a_k=b \tag{13}$$

$$a_1+4\cdot a_2+\dots +n^2\cdot a_n=\sum_{k=1}^nk^2\cdot a_k=b^2 \tag{14}$$

We have

$$b-1\overset{(13),(12)}{=}\sum_{k=1}^nk\cdot a_k-\sum_{k=1}^na_k=\sum_{k=1}^n(k-1)\cdot a_k=\sum_{k=2}^n(k-1)a_k\geqslant 0$$

so  $b\geqslant 1$.

          If  $b\geqslant n$  then

$$0\leqslant b^2-n\cdot b\overset{(14),(13)}{=}\sum_{k=1}^nk^2\cdot a_k-n \sum_{k=1}^nk\cdot a_k=\sum_{k=1}^n(k^2-nk)a_k=\sum_{k=1}^{n-1}k(k-n)a_k\leqslant 0$$

and according to (11) it result  $a_1=\dots=a_{n-1}=0,\;b=n,\;a_n\overset{(12)}{=}1$.

          If  $b\in[i,i+1)$  for some  $i\in\{1,\dots,n-1\}$, then, on the one hand

$$b^2-i\cdot b\;\overset{(14),(13)}{=}\sum_{k=1}^nk^2a_k-i\cdot \sum_{k=1}^nka_k=\sum_{k=1}^nk(k-i)a_k \tag{15}$$

On the other hand

$$b-i\;\overset{(13),(12)}{=}\sum_{k=1}^nka_k-i\cdot \sum_{k=1}^na_k=\sum_{k=1}^n(k-i)a_k$$

which, multiplied by  $b$  gives us

$$b^2-ib=\sum_{k=1}^nb(k-i)a_k \tag{16}$$

Equating (15) and (16)  we obtain

$$\sum_{k=1}^n(b-k)(i-k)a_k=0$$

or, excluding the term  $a_i$ wich has coefficient zero

$(b-1)(i-1)a_1+\dots+(b-i+1)a_{i-1}+(b-i-1)(-1)a_{i+1}+\dots (b-n)(i-n)a_n=0$.

Let's analyze the coefficients of the numbers  $a_k$  in the equation above :

          for  $k<i\;\Rightarrow\;(b-k)(i-k)\;\overset{k<i\leqslant b}{>}0;$

          for  $k>i\;\Rightarrow\;(b-k)(i-k)=(k-b)(k-i)\overset{k\geqslant i+1>b}{>}0$

Then it turns out that  $a_1=\dots=a_{i-1}=0=a_{i+1}=\dots =a_n$,  so  $a_i\overset{(12)}{=}1$  and  $b\underset{(13)}{=}i.$

$\blacksquare$

PLM inseamna o trivialitate.... PLV este prescurtarea oficiala

 Am postat in Aug/01/2024 Decizia de Recalculare a pensiei.

Un comentator anonim, cam sugubat, s-a interesat de Decizia de trecere la Pensia pentru Limita de Varsta. Prescurtat PLV. I-am raspuns ... la fel de trivial. Scuze cititorilor mai sensibili...

Aici sunt cele 3 pagini din Decizie...

Talonul de pensie pe luna AUGUST 2025, insemnat cu compararile din "Marea Recalculare"

Am incasat si niste diferente pntru lunile MAI-JUL, cum se vede mai jos

Voi posta si Talonul pe SEPtembrie, care asa va fi de aici incolo....cat voi trai.

Edit Sep 8, 2025 : Iata Talonul ....final:

A Problem in "GHEBA"

 The author was a celebrity in the second half of the 20th century for his Mathematical Problem Collections for middle school students.

          Ironically, his name has come up in a controversy. See here and here.

          I solved it from his 1973 collection. The 1975 edition, slightly modified, is here. A list of exercises is selected here and here.

          

         I liked the following problem which seems like Elementary Arithmetic :

               "Un obiect se vinde cu 39 lei, castigandu-se atat la suta cat a costat

                 obiectul. Care a fost costul obiectului ?"

In translation : 

"An object is sold for 39 lei, earning a percentage of the cost of the object.

             What was the cost of the object ?"    

              The answer is 30 lei.  That is, an object that costs 30 lei was sold for 30% more. That is, a commercial addition of  $30\cdot \frac{30}{100}=9$ lei. So the selling price is

30 lei  +  9 lei = 39 lei.

          In my personal edition the problem appears on page 136, Problem #5. In the 1975 edition the problem appears on page 201, Problem #5. Among more modern editions, the problem appears on page 192, Problem  #22.

               How can this problem be solved arithmetically ?

I don't know the answer, but I solved it using algebra. In fact, the problem is included in the Chapter on 2nd Grade Equations.

          Solution CiP

          Let  $x$  be the initial price of the object. The object is being sold for  $x\text{%}$  more. The commercial markup is therefore  $x\cdot \frac{x}{100}=\frac{x^2}{100}$. The selling price of the object will be  $x+\frac{x^2}{100}$. 

     So we have the quadratic equation

$x+\frac{x^2}{100}=39$

     In real numbers the equation has two solutions  $x_1=30,\;\;x_2=-130$. Of these, only the first has significance for our problem.

$\blacksquare$

A Problem That Has a Chance of Becoming a Theorem // Problema, kuri gali tapti teorema

          I heard the expression in the title from a somewhat megalomaniac fellow mathematician.

          It does, and we will try to prove the following formula :

                    Let be the nonzero numbers  $a_1,\;\dots a_m$. The following equation holds :

$a_1\cdot \left ( \frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+\dots +\frac{1}{a_m}\right )+(a_2-a_1)\cdot \left ( \frac{1}{a_2}+\frac{1}{a_3}+\dots +\frac{1}{a_m}\right )+$

$+(a_3-a_2)\cdot \left (\frac{1}{a_3}+\dots +\frac{1}{a_m}\right )+\dots+(a_{m-1}-a_{m-2})\cdot \left ( \frac{1}{a_{m-1}}+\frac{1}{a_m}\right )+$

$+(a_m-a_{m-1})\cdot \frac{1}{a_m}=m \tag{1}$

I seem to see that a "proof without words" is imminent.

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

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

 

           $\blacklozenge$The hard truth about the sequence $(x_n)$ is, this sequence being the protocol of a slow and sad decomposition of a rational number  $x_1=\frac{a}{b}$ to the form - pay attention to colors -

$\frac{a}{b}=\color{Blue}s+\frac{1}{k_1}+\dots+\frac{1}{k_{\color{Red}m}} \tag{E_1}$

where

$s=[x_1],\;\;k_1<k_2<\dots<k_m \tag{E_2}$

$k_i,\;i=1,\dots m,$  are positive integers, and

$x_n=\frac{1}{k_i}+\dots +\frac{1}{k_m}\;\;for\;\;k_{i-1}<n\leqslant k_i  \tag{E_3}$

          Moreover, the sum of the nonzero terms of the sequence is equal to (!!colors!!)

$x_1+x_2+\dots+x_{k_m}=\color{Blue}s+\color{Red}m \tag{S}$

     Returning to the previous examples  

          In Example 1,  $x_7=0,\;\;x_6=x_5=x_4=x_3=\frac{1}{6},\;\;x_2=x_1=\frac{2}{3}=\frac{1}{2}+\frac{1}{6}.$ So  $\color{Blue}{s=0},\;\color{Red}{m=2},\;k_1=2,\;k_2=6.$

           In Example 2,  $x_5=0,\;\;x_4=x_3=\frac{1}{4},\;\;x_2=x_1=\frac{3}{4}=\frac{1}{2}+\frac{1}{4}.$ So $\color{Blue}{s=0},\;\;\color{Red}{m=2},\;k_1=2,\;k_2=4.$

           In Example 3,  $x_{16}=0,\;\;x_{15}=x_{14}=\dots=x_4=\frac{1}{15},\;\;x_3=x_2=\frac{2}{5}=\frac{1}{3}+\frac{1}{15}.$ So  $\color{Blue}{s=3},\;\;\color{Red}{m=2},\;k_1=3,\;k_2=15.$

           In Example 4,  $x_7=0,\;\;x_6=x_5=\dots=x_2=\frac{1}{6}$. So  $\color{Blue}{s=1},\;\;\color{Red}{m=1},\;k_1=6.$

           In Example 5,  $x_{232}=0,\;\;x_{231}=x_{230}=\dots=x_{12}=\frac{1}{231},\;\;x_{11}=x_{10}=\dots=x_4=\frac{2}{21}=\frac{1}{11}+\frac{1}{231},\;\;$

            $x_3=x_2=\frac{3}{7}=\frac{1}{3}+\frac{1}{11}+\frac{1}{231}.$ So  $\color{Blue}{s=0},\;\;\color{Red}{m=3},\;k_1=3,\;k_2=11,\;k_3=231.$

            All relations (E_1), (E_3), (S) are verified.

          $\blacklozenge$A final example, treated more expeditiously:

              Let  $x_1=\frac{7}{8}$. The calculations show that

$\color{Blue}{s=0},\;\;k_1=2,\;\;k_2=3,\;\;k_3=24,\;\;\color{Red}{m=3}$

$x_{25}=0,\;\;x_{24}=\dots=x_4=\frac{1}{24},\;\;x_3=\frac{3}{8}=\frac{1}{3}+\frac{1}{24},\;\;x_2=x_1=\frac{7}{8}=\frac{1}{2}+\frac{1}{3}+\frac{1}{24}$

and

$x_1+\dots +x_{24}=2\cdot \frac{7}{8}+\frac{1}{3}+21\cdot \frac{1}{24}=3$                  

<end CiP comments>$\blacklozenge$

                    Proving that the sequence contains

only a finite number of nonzero terms

Let's prove (SJ_1) that is, if   $x_1$ is a rational number then the sequence contains a term (and from here on all terms) equal to zero.

          If  $x_1\in\mathbb{Z}$, then  $\{x_1\}=0$  and statement (SJ_1) is true.

          Otherwise,  $x_2\;\overset{def}{=}\;\frac{\{1\cdot x_1\}}{1}=\{x_1\} \in (0;1)$  and we will clarify this case in the following. 

     So, let

$x_2=\frac{a}{b},\;\;a<b \tag{1}$

 and  $a\; and\; b$  are positive coprime integers.

We choose  $k$-the smallest natural number such that 

$x_2 \geqslant \frac{1}{k} \tag{2}$

Obviously  $k\geqslant 2$.

               If  $k=2$, that is  $\frac{a}{b}=x_2\geqslant 2$ (so $2a-b\geqslant 0$ we'll use that later)

we have  $2x_2-1\geqslant 0$ and, as  $x_2<1$, we also have  $2x_2-1<1$. Using the periodicity of the function  $\{\cdot\}\;(i.e.\;\{\alpha\}=\{\alpha-1\})$  we find

$x_3=\frac{\{2x_2\}}{2}=\frac{\{2x_2-1\}}{2}\underset{0\leqslant 2x_2-1<1}{===}\frac{2x_2-1}{2}=x_2-\frac{1}{2} \tag{3}$

With (1) this is written

$x_3=\frac{2a-b}{2b},\;\;\;0\leqslant 2a-b \;\overset{\color{Red}!a<b}{<}\;a \tag{4}$

     So, if  $x_3$  is not eqal zero, then the numerator of  $x_3$  in irreducible form, is  $\leqslant 2a-b$  hence (cf. (4)) less than the numerator of  $x_2$.

<end case $k=2$>

              If  $k>2$  we will prove the following facts :

(i)    $2\leqslant i \leqslant k\;\;\Rightarrow\;\;x_i=x_2$

(ii)   $x_{k+1}=x_2-\frac{1}{k} \tag{5}$

          We show (i) one by one, up close and close.

         Obviously (i) holds for  $i=2$. 

We assume the statement is true for some  $i$, with  $i\geqslant 2\;\;and\;i+1\leqslant k$. We have  $i\leqslant k-1$  and, considering how we defined the number  $k$, it results (cf. (2))  $x_2<\frac{1}{i}.$  But then

$x_{i+1}=\frac{\{i \cdot x_i\}}{i}\underset{\overbrace{x_i=x_2}}{==}\frac{\{i \cdot x_2\}}{i}\overset{\underbrace{ix_2<1}}{==}\frac{i\cdot x_2}{i}=x_2$

 so  (i)  is also true for  $i+1$.(It's like a kind of Induction.)

             For  (ii) , since from the way we chose the number $k,\;\;k-1$ does not verify  (2) - that is  $x_2<\frac{1}{k-1}$-, we have the inequalities

$$\frac{1}{k}\overset{(2)}{\leqslant} x_2\underset{(1)}{=}\frac{a}{b}<\frac{1}{k-1}\overset{k>2}{<}\frac{2}{k} \tag{6}$$

Then  $b\leqslant k\cdot a<2b$, so  $0\leqslant ka-b<b\;\left ( \Leftrightarrow 0\leqslant \frac{ka-b}{b}<1\right )$,  and

  $x_{k+1}=\frac{\{k\cdot x_k\}}{k}=\frac{\{k\cdot x_2\}}{k}=\frac{\left\{\frac{ka}{b}\right\}}{k}=\frac{\left \{\frac{ka}{b}-1\right \}}{k}=\frac{\left \{\frac{ka-b}{b}\right \}}{k}=\frac{\frac{ka-b}{b}}{k}=\frac{ka-b}{kb}=x_2-\frac{1}{k} \tag {7}$

     From  $(k-1)a<b$ (cf.(6)) we have  $ka-b<a$  and from  $x_{k+1}=\frac{ka-b}{kb}$  we see that the numerator of  $x_{k+1}$  in irreducible form, is  $\leqslant ka-b$  hence less than the numerator  $a$  of  $x_2$.

<end case $k>2$>

(Notice that  (3)  is embedded in  (7) for  $k=2$.)

          "As the numerators cannot decrease infinitely, at some moment the next term will be  0." (We have reproduced the words of the authors of the Official Solution.)

<<end of Proof (SJ_1)>> $\blacksquare$

Before moving on, let's try to find out the 

Structure of  the Sequence  $(x_n)_{n\geqslant 1}$  

and its Connection with the Egyptian Fractions

Let  $k_m$  be the index of the last nonzero term. From  $x_{k_m+1}=0$, applying the analogous of (5)

$x_{k_m+1}=x_{k_m}-\frac{1}{k_m} \tag{8}$ 

it result

$x_{k_m}=\frac{1}{k_m} \tag{9'}$

(The same thing can be obtained from  $x_{k_m+1}=\frac{\{k_m\cdot x_{k_m}\}}{k_m}=0\;\;\overset{\underbrace{x_n<\frac{1}{n-1}}}{\Rightarrow}\;\;k_m\cdot x_{k_m}=1\;\;etc$  )

     Let  $k_{m-1}$  be the largest index  $<k_m$  for wich  $x_{k_{m-1}}\neq x_{k_m}.$  We have

$\frac{1}{k_m}=x_{k_m}=x_{k_m-1}=\dots =x_{k_{m-1}+1} \tag{10'}$

abd from  (8)  written for  $k_{m-1}\;:$

$x_{k_{m-1}+1}=x_{k_{m-1}}-\frac{1}{k_{m-1}}\overset{(10')}{\Leftrightarrow}\frac{1}{k_m}=x_{k_{m-1}}-\frac{1}{k_{m-1}}$

it result

$x_{k_{m-1}}=\frac{1}{k_{m-1}}+\frac{1}{k_m} \tag{9''}$

     If  $k_{m-2}$  is the largest index $<k_{m-1}$  for wich  $x_{k_{m-2}}\neq x_{k_{m-1}}$, that is

$\frac{1}{k_{m-1}}+\frac{1}{k_m}=x_{k_{m-1}}=x_{k_{m-1}+1}=\dots=x_{k_{m-2}+1} \tag{10''}$

then we get

$\frac{1}{k_{m-1}}+\frac{1}{k_m}\;\overset{(10'')}{=}$$x_{k_{m-2}+1}\overset{(8)}{\underset{for\;k_{m-2}}{=}}\;x_{k_{m-2}}-\frac{1}{k_{m-2}}$

and it result

$x_{k_{m-2}}=\frac{1}{k_{m-2}}+\frac{1}{k_{m-1}}+\frac{1}{k_m} \tag{9'''}$

(9'), (9''), (9''')  as well as  (10'), (10'')  confirm the equations  (E_3)  from CiP_Comments above.

          Further, in the words of the Authors of the Official Solutions :

 "Restoring the sequence backwards in this way, we arrive at the desired formula."

It's about the formula  (E_1), but the statement doesn't seem very convincing.

          Let's say it's still good. But I hope it doesn't create confusion that I used the fraction  $\frac{a}{b}$  in (1), in a different context. 

          I will now present one last example, to show that we can start the algorithm with any real number. The reader is advised to treat Examples 1-5 and the final one in my comments at the beginning in the same way. Let's choose a favorite number of Archimedes:

$x_1=-\frac{22}{7}$

$\color{Blue}s=[x_1]=\left [-4+\frac{6}{7}\right ]=\color{Blue}{-4}$

$x_2=\{x_1\}=\frac{6}{7}\;;$

we have  $\frac{6}{7}\geqslant \frac{1}{2}$  so we choose in  (2)  $k_1=2$. The next term will therefore be different from  $x_2$. Let's find it :

$x_3=\frac{\{2x_2\}}{2}=\frac{\left \{\frac{12}{7}\right \}}{2}=\frac{\left \{\frac{5}{7}\right \}}{2}=\frac{\frac{5}{7}}{2}=\frac{5}{14}\;;$

the calculation is also in agreement with  (3). We have  $\frac{5}{14}\geqslant 3$ so we choose in  (2)  $k_2=3$. Because  $k_2-k_1=3-2=1$, this value appears only once in the string. Let's move on.

$x_4=\frac{\{3x_3\}}{3}=\frac{\left\{\frac{15}{14}\right \}}{3}=\frac{\frac{1}{14}}{3}=\frac{1}{42}$

and obvously  $x_4\geqslant \frac{1}{42}$  so  $k_3=42$. All of the following  $k_3-k_2=42-3=39$  terms have the same value :

$x_4=x_5=\dots=x_{42}=\frac{1}{42}$

and finally  $x_{43}=0.$  So  $\color{Red}{m=3}$.

          The calculation of the sum of nonzero terms has, in the official solution, the aspect

$x_1+x_2+\dots +x_{k_m}=s+k_1 \cdot \frac{1}{k_1}+k_2\cdot \frac{1}{k_2}+\dots +k_m\cdot \frac{1}{k_m}=s+m$

This did not appear obviously in the calculations of the previous examples, although the result was correct. That this is indeed the case follows if we write the terms in the form :

$x_4=\color{Violet}{\frac{1}{42}}\;\;\;\;\}\;k_3-k_2=39\;values$

$x_3=\color{Teal}{\frac{1}{3}}+\color{Violet}{\frac{1}{42}}\;\;\;\}\;k_2-k_1=1\;value$

$x_2=\color{Red}{\frac{1}{2}}+\color{Teal}{\frac{1}{3}}+\color{Violet}{\frac{1}{42}}\;\;\}\;k_1-1=1\;value$

$x_1=\color{Blue}{-4}+\color{Red}{\frac{1}{2}}+\color{Teal}{\frac{1}{3}}+\color{Violet}{\frac{1}{42}}\;\;\}\;1\;value$

We see that each value $\frac{1}{k_i}$  appears  $k_i$ times. So  this sum is  $\color{Blue}s+\color{Red}m=-4+3=-1$.

$\blacksquare\;\blacksquare$

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 $\blacksquare$