Let , be the roots of the equation x^2-√2x+2=0.

Question

Let $\alpha,\beta$ be the roots of the equation $x^2-\sqrt2x+2=0$. Then $\alpha^{14}+\beta^{14}$ is equal to

Accepted Answer

$$x=\frac{\sqrt2\pm\sqrt{2-8}}2=\frac{\sqrt2\pm i\sqrt6}2=\sqrt2\cdot\frac{1\pm i\sqrt3}2=\sqrt2\,e^{\pm i\pi/3}=-\sqrt2\,\omega^2,\ -\sqrt2\,\omega$$

where $\omega$ is a primitive cube root of unity. So $\alpha=-\sqrt2\,\omega^2,\ \beta=-\sqrt2\,\omega$.

$$\alpha^{14}+\beta^{14}=(\sqrt2)^{14}\big(\omega^{28}+\omega^{14}\big)=2^7\big(\omega^{28}+\omega^{14}\big).$$

Since $28\equiv1\pmod3$ and $14\equiv2\pmod3$: $\omega^{28}=\omega,\ \omega^{14}=\omega^2$.

$$2^7(\omega+\omega^2)=128(-1)=-128.$$

Solutions · 1