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?

Yet another way:   let $a=\frac{1+i\sqrt{7}}{2}$ and $b=\frac{1-i\sqrt{7}}{2}$ then $a+b=1, ab=2$ so $a,b$ are the roots of the quadratic $x^2-x+2=0\,$.
Therefore $a^2=a-2\,$, then successively:


*

*$a^3 = a\cdot a^2 = a \cdot(a-2) = a^2-2a = (a-2)-2a=-a-2$

*$a^4= a \cdot a^3 = a(-a-2) = -a^2-2a=-(a-2)-2a=-3a+2$
The same holds for $\,b\,$ so $\;a^4+b^4=(-3a+2)+(-3b+2)=-3(a+b)+4=-3\cdot 1+4=1\,$.
