\section*{Problem 3 - 081223}

Provide twelve real numbers \(a_{1}, \ldots, a_{6}, b_{1}, \ldots, b_{6}\) such that for every real number \(x\) the equation holds:

\[
x^{12}+1=\left(x^{2}+a_{1} x+b_{1}\right)\left(x^{2}+a_{2} x+b_{2}\right)\left(x^{2}+a_{3} x+b_{3}\right) \cdot\left(x^{2}+a_{4} x+b_{4}\right)\left(x^{2}+a_{5} x+b_{5}\right)\left(x^{2}+a_{6} x+b_{6}\right)
\]
}

Choose

\[
\begin{gathered}
b_{1}=\cdots=b_{6}=1, \quad a_{1}=\sqrt{2+\sqrt{3}}, \quad a_{2}=\sqrt{2}, \quad a_{3}=\sqrt{2-\sqrt{3}} \\
a_{4}=-a_{3}, \quad a_{5}=-a_{2} \text { and } a_{6}=-a_{1}
\end{gathered}
\]

Then we have

\[
\left(x^{2}+a_{i} x+b_{i}\right)\left(x^{2}+a_{7-i} x+b_{7-i}\right)=\left(x^{2}+1+a_{i} x\right)\left(x^{2}+1-a_{i} x\right)=x^{4}+\left(2-a_{i}^{2}\right) x^{2}+1
\]

so

\[
\begin{gathered}
T:=\left(x^{2}+a_{1} x+b_{1}\right)\left(x^{2}+a_{2} x+b_{2}\right)\left(x^{2}+a_{3} x+b_{3}\right)\left(x^{2}+a_{4} x+b_{4}\right)\left(x^{2}+a_{5} x+b_{5}\right)\left(x^{2}+a_{6} x+b_{6}\right) \\
=\left(x^{4}+(2-(2+\sqrt{3})) x^{2}+1\right) \cdot\left(x^{4}+(2-2) x^{2}+1\right) \cdot\left(x^{4}+(2-(2-\sqrt{3})) x^{2}+1\right) \\
=\left(x^{4}+1\right) \cdot\left(x^{4}+1-\sqrt{3} x^{2}\right) \cdot\left(x^{4}+1+\sqrt{3} x^{2}\right) \\
=\left(x^{4}+1\right) \cdot\left(x^{8}+2 x^{4}+1-3 x^{4}\right)=\left(x^{4}+1\right) \cdot\left(x^{8}-x^{4}+1\right) \\
=x^{12}+x^{8}-x^{8}-x^{4}+x^{4}+1=x^{12}+1
\end{gathered}
\]

Remark:

It is \(x^{24}-1=\left(x^{12}-1\right)\left(x^{12}+1\right)\). Therefore, the complex roots of the polynomial \(x^{12}+1\) are precisely those of \(x^{24}-1\) that are not roots of \(x^{12}-1\).

The complex roots of \(x^{n}-1\) are \(\zeta_{n}^{k}=\cos \left(k \cdot \frac{2 \pi}{n}\right)+i \cdot \sin \left(k \cdot \frac{2 \pi}{n}\right)\), where \(i\) runs through the integers from 0 to \(n-1\).

Thus, we obtain the complex roots of the polynomial \(x^{12}+1\) as

\[
x_{k}=\zeta_{24}^{2 k-1}=\cos \left((2 k-1) \cdot \frac{2 \pi}{24}\right)+i \cdot \sin \left((2 k-1) \cdot \frac{2 \pi}{24}\right)
\]

(The 24th roots of unity with even \(k\) are also 12th roots of unity, and thus roots of \(x^{12}-1\), which should be excluded here.)

We obtain the factorization of the polynomial \(x^{12}+1\) into its linear factors as

\[
x^{12}+1=\left(x-x_{1}\right) \cdot\left(x-x_{2}\right) \cdot \ldots \ldots\left(x-x_{12}\right)
\]

Since \(\cos (2 \pi-\phi)=\cos (-\phi)=\cos (\phi)\) and \(\sin (2 \pi-\phi)=\sin (-\phi)=-\sin (\phi)\), we can now combine the first and last of these complex linear factors, the second and second-to-last, and so on: For \(k\) in \(\{7 ; 8 ; \ldots ; 12\}\), we have

\[
\begin{gathered}
\left(x-x_{k}\right)\left(x-x_{13-k}\right)=\left(x-\cos \left((2 k-1) \cdot \frac{2 \pi}{24}\right)-i \cdot \sin \left((2 k-1) \cdot \frac{2 \pi}{24}\right)\right) \\
=\left(x-\cos \left((2 k-1) \cdot \frac{2 \pi}{24}\right)+i \cdot \sin \left((2 k-1) \cdot \frac{2 \pi}{24}\right)\right)= \\
=x^{2}-2 x \cos \left((2 k-1) \cdot \frac{2 \pi}{24}\right)+\cos ^{2}\left((2 k-1) \cdot \frac{2 \pi}{24}\right)+\sin ^{2}\left((2 k-1) \cdot \frac{2 \pi}{24}\right) \\
=x^{2}-2 x \cos \left((2 k-1) \cdot \frac{2 \pi}{24}\right)+1
\end{gathered}
\]

which is a purely real quadratic polynomial. Setting \(a_{k}:=-2 \cos \left((2 k-1) \cdot \frac{2 \pi}{24}\right)\), we obtain exactly the values mentioned above.