1973 IMO Shortlist Problems/Cuba 2

Problem

Decompose the polynomial

$\displaystyle x^{2m} - 2 |a|^{m} x^{m} \cos ( m\theta ) + a^{2m}$

into $\displaystyle m$ polynomials of second degree with real coefficients.

Solution

We use the quadratic formula to find that the polynomial is zero when

$\displaystyle x^{m} = |a|^{m}(\cos ( m\theta ) \pm i \sin( m \theta ) )$ ,

which clearly occurs when $\displaystyle x = |a| \left[ \cos \pm \left( \theta+\frac{ 2k\pi }{ m }\right)+i\sin \pm \left( \theta+\frac{ 2k\pi }{ m }\right) \right]$ , for all integral $\displaystyle k$ .

Hence

$\begin{matrix} x^{2m}-2|a|^{m}x^{m}\cos ( m \theta )+a^{2m}& = & \displaystyle \prod_{k=0}^{m-1}\left[ x-|a| \mbox{ cis}\left( \theta+\frac{ 2k\pi }{ m }\right) \right] \left[ x-|a| \mbox{ cis} \left( -\left( \theta+\frac{ 2k\pi }{ m } \right) \right) \right] \\ & = & \displaystyle \prod_{k=0}^{m-1}\left( x^{2}-2 |a| \cos \left( \theta+\frac{2k\pi}{m}\right) x+|a|^{2}\right) \end{matrix}$

Q.E.D.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

Resources

1973 IMO Shortlist Problems
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=603067 Discussion on AoPS/MathLinks]

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1973 IMO Shortlist Problems/Cuba 2

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