If $\mathrm{\alpha }, \mathrm{\beta }$ are roots of the equation ${\mathrm{x}}^2-4\mathrm{x}+8=0$ then for any $\mathrm{n}\in \mathrm{N}, {\mathrm{\alpha }}^{2\mathrm{n}}+{\mathrm{\beta }}^{2\mathrm{n}}=$

$\begin{array}{rcl}{\mathrm{x}}^2-4\mathrm{x}+8& =& 0\end{array}$

$\begin{array}{rcl}\mathrm{x}& =& \frac{4\pm \sqrt{16-32}}{2}=\frac{4\pm 4\mathrm{i}}{2}=2(1\pm \mathrm{i})\end{array}$

$\begin{array}{rcl}\mathrm{\alpha }& =& 2(1+\mathrm{i}) \mathrm{\beta }=2(1-\mathrm{i})\end{array}$

$\begin{array}{rcl}{\mathrm{\alpha }}^{2\mathrm{n}}+{\mathrm{\beta }}^{2\mathrm{n}}& =& 2^{2\mathrm{n}} \left({\left(1+\mathrm{i}\right)}^{2\mathrm{n}}+{(1-\mathrm{i})}^{2\mathrm{n}}\right)\\ & =& 2^{2\mathrm{n}}·2^{\mathrm{n}}\left({\left(\frac{1}{\sqrt{2}}+\frac{\mathrm{i}}{\sqrt{2}}\right)}^{2\mathrm{n}}+{\left(\frac{1}{\sqrt{2}}-\frac{\mathrm{i}}{\sqrt{2}}\right)}^{2\mathrm{n}}\right)\\ & =& 2^{3\mathrm{n}} \left({\left(\cos \frac{\mathrm{\pi }}{4}+\mathrm{i} \sin \frac{\mathrm{\pi }}{4}\right)}^{2\mathrm{n}}+{\left(\cos \frac{\mathrm{\pi }}{4}-\mathrm{i} \sin \frac{\mathrm{\pi }}{4}\right)}^{2\mathrm{n}}\right)\\ & =& 2^{3\mathrm{n}}\left(\cos \frac{2\mathrm{n\pi }}{4}+\mathrm{i} \sin \frac{2\mathrm{n\pi }}{4}+\cos \frac{2\mathrm{n\pi }}{4}-\mathrm{i} \sin \frac{2\mathrm{n\pi }}{4}\right)\\ & =& 2^{3\mathrm{n}}\left(2\cos \frac{\mathrm{n\pi }}{2}\right)\\ & =& 2^{3\mathrm{n}+1} \cos \frac{\mathrm{n\pi }}{2} \end{array}$