Prove: $(\frac{1+i\sqrt{7}}{2})^4+(\frac{1-i\sqrt{7}}{2})^4=1$ $$\left(\frac{1+i\sqrt{7}}{2}\right)^4+\left(\frac{1-i\sqrt{7}}{2}\right)^4=1$$
I tried moving the left exponent to the RHS to then make difference of squares exp. $(x^2)^2$. Didn't get the same on both sides though. Any help?

Let $(a_n)_n$ the sequence verifying $\begin{cases} a_0=2\\ a_1=1\\a_{n+2}=a_{n+1}-2a_n\end{cases}$
The characteristic equation of this linear recurrence relation is $x^2=x-2$ 
Whose roots are $\dfrac{1\pm i\sqrt{7}}2$.
Thus $a_n=\alpha r^n+\beta {\bar r}^n$ and given the initial conditions then $\alpha=\beta=1$.
So $a_n=\left(\dfrac{1+i\sqrt{7}}2\right)^n+\left(\dfrac{1-i\sqrt{7}}2\right)^n$
We are asked to calculate $a_4$ ?


*

*$a_2=a_1-2a_0=1-4=-3$

*$a_3=a_2-2a_1=-3-2=-5$

*$a_4=a_3-2a_2=-5+6=1$


Use the integer sequence to calculate any other power you desire...
