<s>[INST] 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?
 [/INST] Let $x=(1+i\sqrt 7\;)/(2\sqrt 2\;).$  Since $x\bar x=x^2\bar x^2=1$ and $x+\bar x=1/\sqrt 2\;$ we have $$((1+i\sqrt 7\;)/2)^4+(1-i\sqrt 7\;)/2)^4=4(x^4+ \bar x^4)=$$     $$=4((x^2+\bar x^2)^2-2x^2\bar x^2)=4((x^2+\bar x^2)^2-2)=$$    $$=4( ((x+\bar x)^2-2x\bar x)^2-2)=$$ $$=4((1/\sqrt 2\;)^2-2)^2-2)=4((1/2-2)^2-2)=4(9/4-2)=1.$$
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