A Paradox of Infinite Long Division // Un paradosso della divisione infinita

 Everything started from the aspects provided in the Post "An easy...".(I still haven't found a simple explanation.)

         I will extract from there the equation 

$26^3-24^3-12^3+1^3=(2^3+1)\cdot 225$

which expresses that  $2025=9\cdot 225$.

(under construction)

longdivision

\begin{array}{ccccccc}\;\;\;26^3&-24^3&\;&-12^3&+1&\vdash&2^3+1\\-26^3&\;&-13^3&&&&13^3-12^3-\left(\frac{13}{2}\right)^3+\left(\frac{13}{4}\right )^3-\left (\frac{13}{8}\right)^3+\left (\frac{13}{16}\right)^3-\left (\frac{13}{32}\right )^3+\cdots\\\hline \setminus&-24^3&-13^3&-12^3&+1\\\;&24^3&\;&12^3\\\hline \;&\setminus&-13^3&\setminus&+1\\\;&\;&13^3&+\left(\frac{13}{2}\right)^3\\\hline \;&\;&\setminus&\left(\frac{13}{2}\right)^3&\;&+1\\\;&\;&\;&-\left(\frac{13}{2}\right)^3&-\left(\frac{13}{4}\right)^3\\\hline \;&\;&\;&\setminus&-\left(\frac{13}{4}\right)^3&+1\\\;&\;&\;&\;&\left(\frac{13}{4}\right)^3+\left (\frac{13}{8}\right)^3\\\hline \;&\;&\;&\;&\left (\frac{13}{8}\right)^3&+1\\\;&\;&\;&\;&-\left(\frac{13}{8}\right)^3-\left(\frac{13}{16}\right)^3\\\hline \;&\;&\;&\;&-\left(\frac{13}{16}\right)^3&+1\\\;&\;&\;&\;&\left(\frac{13}{16}\right)^3+\left(\frac{13}{32}\right)^3\\\hline \;&\;&\;&\;&\left(\frac{13}{32}\right)^3&+1\\\:&\;&\;&\;&\dots&\dots\end{array}

An easy problem for... Leonhard EULER // Ein einfaches Problem für... Leonhard Euler

 I've messed up my blogs (petrell-man and/or mathematyka2), and none of them know how to type in LaTeX. And then I made a third one matemattica0, where a shy LaTeX form goes. 

              EULER is famous for elucidating Fermat's number  $F_5=2^{2^5}+1=641 \times 6700417$. He is the one who saw that  $641=2^7\times 5+1=2^4+5^4$.

That's why I thought that only he could help me with the problem:

              "1. (page 51)  We will call a positive integer  $n\;cool$  if it is a perfect square and there exists positive integers  $a,\;b,\;c,\;d$  different from each other, with  $a>b\;and\;c>d$ ,  such that  $n=a^3-b^3+c^3-d^3$.  For instance,  225  is cool because  $225=15^2\;\;and\;\;225=7^3-5^3+2^3-1^3$.  Show that :

a)  2025 is cool ;

b)  there are infinitely many cool numbers.

(Relu CIUPEA)" 

ANSWER CiP

a) $2025=45^2=11^3-3^3+9^3-2^3=16^3-15^3+11^3-3^3$

b)  $225\cdot k^6=(15k^3)^2=(7k^2)^3-(5k^2)^3+(2k^2)^3-(k^2)^3,\;k\in\mathbb{N}^*$ 

Solution CiP

                   a) was solved by ChatGPT. I chose the simplest of several equalities.

                    b) We multiply the example provided in the statement by  $k^6$.

$\blacksquare$

                   An infinite number of remarks can be made, but I still haven't managed to find a simple way, accessible to a junior, to achieve the result.

            At first I thought it was easy, starting from

$2025=9\cdot 225=(2^3+1^3)(7^3-5^3+2^3-1^3)=14^3-10^3+7^3-5^3+4^3-1^3.$

But there is no further path to the result in sight.

             Let's write 225 differently then. We know  $225=1^3+2^3+3^3+4^3+5^3$, and $3^3+4^3+5^3=6^3$, so  $225=1^3+2^3+6^3$  but even that doesn't get us where we want to go.

             ChatGPT gave us a few more representations that didn't quite fit our requirements :

$2025=15^3-11^3-3^3+2^3$

$2025=26^3-24^3-12^3+1^3$

$2025=29^3-26^3-17^3+5^3$

$2025=34^3-31^3-20^3+8^3$

When I asked him how he calculated it, he told me he used a code and the PYTHON language.

<end Rem>

"THE TREEPENNY OPERA" : Identities with sums of three binomials // "DIE DREIGROSCHEN OPERA" : Identitäten mit Summen von drei Binomien

 Inspiration from the title of a Bertolt BRECHT play.

                  We propose to establish algebraic identities of the form :

$x^2+(x+a)^2+(x+b)^2=(x+m)^2+(x+n)^2+(x+p)^2 \tag{1}$

where  $a,\;b,\;m,\;n,\;p\;$ are real numbers.

My greatest achievement is the formula in

 

                    PROPOSITION 1.  Let  $a,\;b\;$ be given numbers. It holds 

                      the algebraic identity :

$x^2+(x+a)^2+x+b)^2=\left (x+\frac{2(a+b)}{3}\right )^2+\left (x+\frac{2b-a}{3}\right )^2+\left ( x+\frac{2a-b}{3}\right )^2 \tag{2}$

          The proof can be done by direct calculation.

$\square$

For example

$x^2+(x+1)^2+(x+5)^2=(x+4)^2+(x+3)^2+(x-1)^2$

Some choices may be unfortunate, e.g.

$x^2+(x+1)^2+(x+2)^2=(x+2)^2+(x+1)^2+x^2$

$x^2+(x+2)^2+(x+4)^2=(x+4)^2+(x+2)^2+x^2$

$x^2+(x+3)^2+(x+6)^2=(x+6)^2+(x+3)^2+x^2$

while others give us equal terms

$x^2+(x+0)^2+(x+3)^2=(x+2)^2+(x+2)^2+(x-1)^2$

$x^2+(x+3)^2+(x+0)^2=(x+2)^2+(x-1)^2+(x+2)^2$

More seriously we have

$x^2+(x-1)^2+(x+4)^2=(x+2)^2+(x+3)^2+(x-2)^2$

but putting  $x-1=y$  we have  $(y+1)^2+y^2+(y+5)^2=(y+3)^2+(y+4)^2+(y-1)^2$  which is precisely the example from the beginning.

Another                                                     $x^2+(x-2)^2+(x+5)^2=(x+2)^2+(x+4)^2+(x-3)^2$

in which if we put  $x-3=y$ , write it backwards, somehow ordering the terms

$y^2+(y+5)^2+(y+7)^2=(y+1)^2+(y+3)^2+(y+8)^2$;

if instead we put  $x-2=z$  we have, ordering now

$z^2+(z+2)^2+(z+7)^2=(z-1)^2+(z+4)^2+(z+6)^2$

and if  $z-1=t$  and write it backwards, we have

$t^2+(t+5)^2+(t+7)^2=(t+1)^2+(t+3)^2+(t+8)^2$

which is the same as the identity in  $"y"$.

Finally

$x^2+(x-1)^2+(x+7)^2=(x+4)^2+(x+5)^2+(x-3)^2$

Ordering we have

$(x-1)^2+x^2+(x+7)^2=(x-3)^2+(x+4)^2+(x+5)^2\;;$

and putting  $x-1=y$  we obtain

$y^2+(y+1)^2+(y+8)^2=(y-2)^2+(y+3)^2+(y+6)^2\;;$

putting  $x-3=z$  in the backwards we obtain

$z^2+(z+7)^2+(z+8)^2=(z+2)^2+(z+3)^2+(z+10)^2$

so we get a bunch of other identities.

          We can write it somehow "in integers" on the (2)

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

where  $a$  and  $c$  are arbitrary numbers.

          Now let's see how we got to (2).

          Identifying in (1) the coefficients of  $x^1$  and  $x^0$  we have the relations

$\begin{cases} m+n+p=a+b\\m^2+n^2+p^2=a^2+b^2\end{cases} \tag{4}$

Eliminating the unknown  $n$  we obtain

  $m^2+(a+b-m-p)^2+p^2=a^2+b^2\Leftrightarrow2m^2+(a+b)^2-2m(a+b)-2p(a+b)+2mp+2p^2=a^2+b^2$

which divided by 2 and ordered by  $p $ gives us

$p^2+p(m-a-b)+m^2-m(a+b)+ab=0 \tag{5}$

The equation (5) in the unknown  $p$  has the discriminant

 $\Delta_p=(m-a-b)^2-4[m^2-m(a+b)+ab]=(a-b)^2+2m(a+b)-3m^2$. 

Assuming that it is a perfect square

$\Delta_p=(a-b)^2+2m(a+b)-3m^3=t^2 \tag{6}$

then the roots of the equation (5) will be

$p_{1,2}=\frac{a+b-m\pm t}{2} \tag{7}$

Writing (6) as an equation in the unknown  $m$

$3m^2-2m(a+b)+t^2-(a-b)^2=0 \tag{8}$

we set the condition that its half-discriminant is a perfect square

$\Delta^{'}_m=(a+b)^2-3[t^2-(a-b)^2]=s^2$

so

$s^2+3t^2=(a+b)^2+3(a-b)^2 \tag{9}$

The values ​​of m are given by

$m_{1,2}=\frac{a+b\pm s}{2} \tag{10}$

          Choosing in (9)  $s=a+b,\;t=a-b$  we obtain

           - from (10) : $m_{1,2}=\frac{a+b \pm (a+b)}{2}\in \left \{0,\;\frac{2(a+b)}{3} \right \}$; we take the second value

           - from (7) : $p_{1,2}=\frac{a+b-\frac{2(a+b)}{3} \pm (a-b)}{2} \in \left \{\frac{2b-a}{3},\;\frac{2a-b}{3}\right \}$.

Then  $n=a+b-m-p$  take the respectively values  $\frac{2a-b}{3},\;\frac{-a+2b}{3}$.

     If  $m=\frac{2(a+b)}{3},\;p_1=\frac{2b-a}{3},\;n=\frac{2a-b}{3}$  we obtain (2). And the values  $p_2=\frac{2a-b}{3},\;n_2=\frac{-a+2b}{3}$ give us the same formula (2) with  $a$  and  $b$  swapped.

          If we notice that in (9)  the right-hand side is  $4a^2-4ab+4b^2=(2a-b)^2+3b^2$  then we can still make the choice $s=2a-b,\;t=b$  but by the end we get the same formula (2) again. More than that, since we already know a solution for the quadratic equation (9) in the unknowns  $s\; and\; t$, we can find a general solution to it depending on two parameters  $\lambda,\;\mu$:

$s=\frac{(3\mu^2-\lambda^2)(a+b)-6\lambda \mu (a-b)}{\lambda^2+3\mu^2},\;\;\;t=\frac{(\lambda^2-3\mu^2)(a-b)-2\lambda \mu (a+b)}{\lambda^2+3\mu^2}$

We are not going in this direction anymore, but what has been said so far has shown that the following Proposition takes place:

                    PROPOSITION 2.  Let  $a,\;b,\;c\;$ be given numbers. We can

                          determine, in terms of  $a,\; b,\; c$ , two numbers $?(a,b,c)$,

                          $\;??(a,b,c)$  such that

$x^2+(x+a)^2+(x+b)^2=(x+c)^2+(x+?)^2+(x+??)^2 \tag{11}$

         Indeed, from (4) we obtain the conditions

$n+p=a+b-c,\;\;n^2+p^2=a^2+b^2-c^2$

which can express  $n$  and  $p$  in terms of  $a,\;b,\;c$ :

$n,p=\frac{a+b-c \pm \sqrt{a^2+b^2-3c^2-2ab+2bc+2ac}}{2}$

Conditions in which these expressions are rational, i.e. without radicals, we leave them to the study of others who are more skilled.

              Examples :        $x^2+(x+1)^2+(x+2)^2=(x+3)^2+(x-\imath \sqrt{2})^2+(x+\imath \sqrt{2})^2$

$x^2+(x+4)^2+(x+5)^2=(x+6)^2+(x+1)^2+(x+2)^2$

$x^2+(x+7)^2+(x+8)^2=(x+9)^2+(x+3-\sqrt{7})^2+(x+3+\sqrt{7})^2$

In a Triangle, a Median is equal to an Altitude : $m_B=h_A$ // In einem Dreieck hat die Mediane die gleiche Länge wie die Höhe

I didn't discover that an altitude and a median in a triangle can be equal.

However, if  $h_A$  is equal to  $m_A$  it also appears in the Problem below. Until then, let's digress a bit.

          Let's take their formulas. Altitude :  $h_A=\frac{2}{a}\cdot \sqrt{s(s-a)(s-b)(s-c)}$

where  $s=\frac{a+b+c}{2}$ (the notations are like here).

  Median :  $m_B=\frac{1}{2}\sqrt{2a^2+2c^2-b^2}$.

We prefer to write  $h_A=\frac{2}{a}\cdot \sqrt{\frac{a+b+c}{2}\cdot \frac{-a+b+c}{2}\cdot \frac{a-b+c}{2}\cdot \frac{a+b-c}{2}}$  or developed

$h_A=\frac{1}{2a}\cdot \sqrt{2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4} \tag{1}$

          The equality  $h_A^2=m_A^2$  is written : 

$\frac{1}{4a^2}(2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4)=\frac{1}{4}(2a^2+2c^2-b^2)\Leftrightarrow$

$\Leftrightarrow \frac{b^2}{2}+\frac{b^2c^2}{2a^2}+\not{\frac{c^2}{2}}-\frac{a^4}{4}-\frac{b^4}{4a^2}-\frac{c^4}{4a^2}=\frac{a^2}{2}+\not{\frac{c^2}{2}}-\frac{b^2}{4} \Leftrightarrow$

$\Leftrightarrow \frac{3b^2}{4}-\frac{b^4+c^4-2b^2c^2}{4a^2}=\frac{3a^2}{4}$  so in the end we get

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

I mention that we will not use this relationship today.

          I recently saw the Problem  E:17209 in a shop window :

In translation :

               "E:17209.  In triangle ABC, median  $BM$  and altitude  $AD$  are

             concurrent at point  $P$   and have the same length. Show that  $BP = 2PD$.

 Dragoș Ionuț MATEI, Constanța"

I am attaching a figure about this problem.

There are several cases of the figure, depending on the position of point D on line BC. In any of the cases we will prove, which immediately implies the conclusion of the problem :

ANSWER CiP

$\widehat{MBC}=30^{\circ}$

                    Solution CiP (Attention, the problem is proposed for Grade 6)

              The position of point D on line BC allows us to distinguish the cases :

$(i)\;\;D=B\;;\;\;(ii)\;\;D=C\;;\;\;(iii)\;\;B-D-C\;;\;\;(iv)\;\;B-C-D\;;\;\;(v)\;\;D-B-C$

          (i)  In this case triangle  $ABC$ has a right angle at vertex  $B$.

The hypothesis  $BM=AD$  shows that  $ABM$  is an equilateral triangle, so  $CAB$ is a 30 60 90  triangle. Hence  $\widehat{MBC}=30^{\circ}$. In the equation  $BP=2PD$  all segments are zero.

          (ii)  In this case  $ABC$  has a rightangle at vertex  $C$.

From  $AD=BM=2b$(say) it follow  $MC=BM/2$  so  $BMC$ is now a 30-60-90 triangle. Because $D=C,\;P=M$  then  $BM=2MC\Rightarrow BP=2PD$.

          (iii)  The figure looks like this

We will proof this case, but we will see that the proof also applies to the others.

          We draw, both in size and orientation  $DN\overset{=}{\parallel}MC$. Quadrilateral  $CMND$  is a parallelogram.

So is  $ADMN$; in this we have

$MN=AD=BM \tag{1}$

and since  $AD\perp BC$, we also have  $MN\perp CD$. But then  $CMDN$  is a rhombus, so   $CN=ND=CM$. Here too, the line  $DC$ is the perpendicular bisector of the segment  $[MN]$, and then

$BM=BN \tag{2}$

From (1) and (2) it follows that  $BM=MN=BN$, therefore the triangle  $BMN$  is equilateral. In it, $BC$  is the bisector of the  $\measuredangle{MBN}$, therefore  $\widehat{MBC}=30^{\circ}$.

     Triangle  $BPD$  is right-angled with  $\widehat{PBD}=30^{\circ}$, so  $BP=2PD$.

          (iv)  The figure looks like this

Same proof

          (iv)  The figure looks like this

Same proof

$\blacksquare\;\blacksquare$

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

 The page presents the problems from the fifth onwards proposed for the 7th grade (the first four are on the previous page)

As it says on the second cover: "In grades VII-XII, the first 8 problems are intended for deepening the subject and preparing for national exams, and the last 4 are addressed to those who wish to additionally prepare for competitions". 

So Problem S:E25.203 is considered easy. But still... ??

         In translation : 

    S:E25.203.  The positive real numbers  $a$  and  $b$  verify the relation  $a+b+ab=3.$

                a) Verify that, if  $a=2$  then  $a+b>2>\sqrt{a}+\sqrt{b}.$

                b)  Prove that  $a+b\geqslant \sqrt{a}+\sqrt{b}$,  for any  $a$  and  $b.$

ANSWER CiP

$\textbf{b)}\;\;\;  a+b\geqslant 2\geqslant \sqrt{a}+\sqrt{b} \tag{A}$.

 The  $"="$ sign occurs if and only if  $a=b=1$

                         Solution CiP

               a) From the given relation we obtain

$b=\frac{3-a}{1+a} \tag{1}$

and it is observed that the given numbers are within the range of values  $0<a,\;b<3$.

          $a=2\underset{(1)}{\Rightarrow}b=\frac{1}{3}$  and obviously  $a+b>2$.

     After that, we have  $\sqrt{a}+\sqrt{b}=\sqrt{2}+\sqrt{\frac{1}{3}}=\frac{3\sqrt{2}+\sqrt{3}}{3}.$  But

$(\sqrt{a}+\sqrt{b})^2=\frac{(3\sqrt{2}+\sqrt{3})^2}{9}=\frac{18+3+6\sqrt{6}}{9}=\frac{7+2\sqrt{6}}{3}<4$, because of  

$\sqrt{24}<\sqrt{25}\Rightarrow 2\sqrt{6}<5\Rightarrow 7+2\sqrt{6}<7+5 \Rightarrow \frac{7+2\sqrt{6}}{3}<4$.

Interestingly, we don't get a more precise result if we start with the "stronger" inequality  $\sqrt{288}<\sqrt{289}$. So

$2<\frac{289}{144}\Rightarrow \sqrt{2}<\sqrt{\frac{289}{144}}=\frac{17}{12}\Rightarrow 4\sqrt{2}<\frac{17}{3}\Rightarrow 6-4\sqrt{2}>6-\frac{17}{3}=\frac{1}{3}\Rightarrow \sqrt{4-4\sqrt{2}+(\sqrt{2})^2}>\sqrt{\frac{1}{3}}\Rightarrow$

$\Rightarrow \sqrt{(2-\sqrt{2})^2}>\sqrt{\frac{1}{3}}\Rightarrow 2-\sqrt{2}>\sqrt{\frac{1}{3}} \Rightarrow 2>\sqrt{2}+\sqrt{\frac{1}{3}}$.

               b)  Only after numerous numerical simulations was I led to the solution of this case, by introducing the intermediate term  $2$  between the two inequalities.

        In the first part we use

  $x+\frac{4}{x}\geqslant 4,\;for\;x>0 \tag{2}$

 ($\Leftrightarrow (\sqrt{x})^2-2\cdot \sqrt{x} \cdot \frac{2}{\sqrt{x}}+\left (\frac{2}{\sqrt{x}}\right )^2\geqslant 0\Leftrightarrow \left (\sqrt{x}-\frac{2}{\sqrt{x}}\right )^2\geqslant 0;\;"="\;iff\;\;x=2$)

We have  

$a+b\underset{(1)}{=}a+\frac{3-a}{1+a}=(a-1)+\left ( \frac{3-a}{1+a}+1 \right )=(a-1)+\frac{4}{1+a}=$

$=\left ((1+a)+\frac{4}{1+a}\right )-2\;\overset{(2)}{\underset{with\;x=1+a}{\geqslant}}\;4-2=2$

thus left side of (A). Equal sign occurs here iff  $1+a=2$  so  $a=1$  and hence  $b\overset{(1)}{=}\frac{3-1}{1+1}=1$.

For right side of (A),  $(\sqrt{ab}-1)^2\geqslant 0\Rightarrow ab-2\sqrt{ab}+1\geqslant 0 \;\overset{ab=3-a-b}{\Rightarrow}$

$\Rightarrow 3-a-b-2\sqrt{ab}+1\geqslant 0\Rightarrow a+b+2\sqrt{ab}\leqslant 4\Rightarrow (\sqrt{a}+\sqrt{b})^2\leqslant 4$

so  $\sqrt{a}+\sqrt{b}\leqslant 2$. The  $"="$  sign occurs if and only if  $a\cdot b=1$  which together with  $a+b=3-ab=2$  give  $a=b=1$.

$\blacksquare$

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.