TITLE: 24 different complex numbers
QUESTION [11 upvotes]: There are $24$ different complex numbers $z$ such that $z^{24}=1$. For how many of these is $z^6$ a real number?

This is one of the AMC problems from this year. I've been trying to solve it, but I couldn't and a peek at the answers (not recommended, I know) talked about Euler's theorem etc., which I haven't learnt yet. Is there an 'easier' way to solve this problem?

REPLY [4 votes]: If you know something a bit more general than state in the opening sentence, namely that for any nonzero $a\in\Bbb C$ and any integer $n>0$ the equation $z^n=a$ has exactly $n$ solutions, then this is easy.
The condition is really not about $z$, but about $z^6$, so it is convenient to view the $24$ solutions to $z^{24}$ as obtained as follows: find two solutions to $x^2=1$, then for each find two solutions to $y^2=x$, then for each find $6$ solutions to $z^6=y$. Simple substitution says that $z^{24}=1$ for each such $z$, and we must have obtained all solutions this way. Now clearly your solutions for $x$ were $x=1$ and $x=-1$; taking $x=1$ you got the same solutions for $y$, which are real, but for $x=-1$ there are no real solutions for $y$.
In conclusion, half of the $24$ triples $(x,y,z)$ found will have $y$ real.