<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] Alright, you want difference of squares have some difference of squares
$$(\frac{1+i \sqrt7}{2})^4 + (\frac{1-i \sqrt7}{2})^4=$$
Lets add $0$ cleverly
$$=(\frac{1+i \sqrt7}{2})^4 -2(\frac{1+i \sqrt7}{2})^2(\frac{1-i \sqrt7}{2})^2+ (\frac{1-i \sqrt7}{2})^4 + 2(\frac{1+i \sqrt7}{2})^2(\frac{1-i \sqrt7}{2})^2$$
$$=((\frac{1+i \sqrt7}{2})^2 - (\frac{1-i \sqrt7}{2})^2)^2 + 2(\frac{1+7}{4})^2 $$
The $2(\frac{1+7}{4})^2$ comes from the final term of the second line, it can be viewed as a difference of squares, squared. Next deal with the difference of tsquares in the squared parenthesis
$$ = (\frac{2\cdot2 i \sqrt{7}}{4})^2 + 8 $$
$$ =-7+8=1$$
QED
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