O Problemă cu cașu' rezolvată de CIOBANU // Problem 1300 - CMJ vol 56 issue 2

 The problem 1300 is compounded by my compatriots. 

The title of the post is intended to be an innocent joke.

          We have to show that: $e^{\frac{\pi \imath}{7}}$ is a solution of the equation

$$\imath \sqrt{7}\cdot z^6-z^5+z^4+z^3+z^2+z-1 =0\tag{E}$$

                 Solution CiP

               We have equality [(1) for $n=3$]:

$$\sin \frac{\pi}{7} \cdot \sin \frac {2\pi}{7} \cdot \sin \frac{3\pi}{7}=\frac{\sqrt{7}}{8} \tag {1}$$

     Let $z=e^{\frac{\pi \imath}{7}}$. According to the De Moivre's and Euler's formulas we have

$z^7=e^{\pi \imath}=-1, \;\;|z|=1,\;\bar z=\frac{1}{z},\;\;\sin \frac{k\pi}{7}=\frac{z^k-\bar z ^k}{2\imath}=\frac{z^{2k}-1}{2\imath \cdot z^k},\;k=1,\;2,\;3$  so

$$z^7+1=0\quad ; \quad z^6-z^5+z^4-z^3+z^2-z+1=0 \tag{C}$$

$$\sin \frac{\pi}{7}=\frac{z^2-1}{2\imath \cdot z},\; \quad \sin \frac{2\pi}{7}=\frac{z^4-1}{2\imath \cdot z^2},\; \quad \sin \frac{3\pi}{7}=\frac{z^6-1}{2\imath \cdot z^3}.\tag{S}$$

Substituting the values ​​(S) into (1) we obtain, after multiplying by $(2\imath)^3=-8\imath$

$$-\sqrt{7} \imath \cdot z^6=(z^2-1)(z^4-1)(z^6-1) \Leftrightarrow$$

$$\Leftrightarrow\;\;-\sqrt{7}\imath \cdot z^6=z^{12}-z^{10}-z^8+z^4+z^2-1 \tag{2}$$

          We will apply the Long Division Algorithm to the polynomial $P=X^{12}-X^{10}-X^8+X^4+X^2-1$  to both $Q=X^7+1$ , $Q_1=X^6-X^5+X^4-X^3+X^2-X+1$.

$\begin{array}{c|c}X^{12}&+0X^{11}&-X^{10}&+0X^9&-X^8&+0X^7&+0X^6&+0X^5&+X^4&+0X^3&+X^2&+0X&-1&Q=X^7+1\\\hline\\-X^{12}&&&&&&&-X^5 &&&&&&D=X^5-X^3-X\\\hline \\&&-X^{10}&&-X^8&&&-X^5&+X^4&&+X^2&&-1&\\\hline &&+X^{10}&&&&&&&+X^3&&&&&\\\hline &&&&-X^8&&&-X^5&+X^4&+X^3&+X^2&&-1\\ \hline\\&&&&&&&-X^5&+X^4&+X^3&+X^2&+X&-1&=R\\\end{array}$

In the first case we obtain

$$P=Q \cdot D+R \tag{3}$$

and in the second case

$\begin{array}{c|c}X^{12}&+0X^{11}&-X^{10}&+0X^9&-X^8&+0X^7&+0X^6&+0X^5&+X^4&+0X^3&+X^2&+0X&-1&Q_1=X^6-X^5+X^4-X^3+X^2-X+1\\\hline -X^{12}&+X^{11}&-X^{10}&+X^9&-X^8&+X^7&-X^6&&&&&&&D_1=X^6+X^5-X^4-X^3-X^2-X\\\hline &X^{11}&-2X^{10}&+X^9&-2X^8&+X^7&-X^6&+0X^5&+X^4&+0X^3&+X^2&+0X&-1\\\hline &-X^{11}&+X^{10}&-X^9&+X^8&-X^7&+X^6&-X^5\\\hline &&-X^{10}&+0X^9&-X^8&+0X^7&+0X^6&-X^5&+X^4&+0X^3&+X^2&+0X&-1\\\hline &&+X^{10}&-X^9&+X^8&-X^7&+X^6&-X^5&+X^4\\\hline &&&-X^9&+0X^8&-X^7&+X^6&-2X^5&+2X^4&+0X^3&+X^2&+0X&-1\\\hline &&&+X^9&-X^8&+X^7&-X^6&+X^5&-X^4&+X^3\\\hline &&&&-X^8&+0X^7&+0X^6&-X^5&+X^4&+X^3&+X^2&+0X&-1\\\hline &&&&+X^8&-X^7&+X^6&-X^5&+X^4&-X^3&+X^2\\\hline &&&&&-X^7&+X^6&-2X^5&+2X^4&+0X^3&+2X^2&+0X&-1\\\hline &&&&&+X^7&-X^6&+X^5&-X^4&+X^3&-X^2&+X\\ \hline &&&&&&&-X^5&+X^4&+X^3&+X^2&+X&-1&=R_1 \end{array}$

$$P=Q_1 \cdot D_1+R_1 \tag{4}$$

I got the same result $R=R_1=-X^5+X^4+X^3+X^2+X-1$. The equality (2) is written

$$-\sqrt{7} \imath \cdot z^6=P(z)=Q(z) \cdot D(z)+R(z)$$

but according to (C) we have $Q(z)=0=Q_1(z)$ and so 

$$-\imath \sqrt{7} \cdot z^6=-z^5+z^4+z^3+z^2+z-1$$

which is equivalent to the equation (E)

$\blacksquare$