24 different complex numbers 
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?
 A: Since $|z|=1$ for all $z$ such that $z^{24}=1,$ the only two real numbers that $z^{6}$ can possibly be are the ones with magnitude $1,$ specifically $1$ and $-1.$ Since there are $6$ solutions to $z^{6}=1$ and $6$ more to $z^{6}=-1,$ there must be $\boxed{12}$ total $z$ satisfying the conditions.

Alternatively, consider that $(z^6)^4=z^{24}=1.$ Now, if we set $z^6=w,$ then for every $w$ there are $6$ solutions in $z.$ Since the solutions for $w$ are $\pm 1, \pm i,$ half of the $24$ solutions for $z$ satisfy $w$ being a real number, so the answer is again $\boxed{12}.$
A: 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.
A: Hint:
$$z^{24}=1=e^{2m\pi i}$$ where $m$ is any integer
$$z=e^{2m\pi i/24}$$
$$z^6=e^{m\pi/2}=\cos\frac{m\pi}2+i\sin\frac{m\pi}2$$ where $0\le m\le3$
So, we need $$\dfrac{m\pi}2\equiv0\pmod\pi\iff2|m$$ 
A: Let's say $w=z^6$. We know that $w^4=1$, so $w=\pm 1,\pm i$. Each of these four numbers has $6$ distinct sixth roots.
Does that help?
A: The $n$th complex roots of $1$ are the points on the complex unit circle with angles of $\frac{2\pi k}n, k\in \Bbb N$. This looks something like this (for $n=7$):

Now imagine this picture with $n=24$, and $n=6$. Where do they overlap?
A: The solutions to $z^{24}=1$ are $z=\sqrt[24]{|1|}e^{(\arg(1)+2k\pi)i/24}$ where $k\in\{0,1,2,\ldots,23\}$.
So we are concerned with seeing when $z^6=e^{(2k\pi)i/4}=e^{k\pi i/2}$ is in $\Bbb R$ for $k\in\{0,1,2,\ldots,23\}$. This occurs when $\sin\left(\frac{k\pi }{2}\right)=0$.
A: No, there isn't a smart/easy way to solve this problem in my opinion. Either you have seen a lecture on roots of unity, and then you know how to solve it, or you haven't, and then it's extremely difficult to come up with the right bits of theory during a competition.
For this reason we prefer to avoid this kind of problems in many low-level mathematical competitions. (Source: I write problems for the ItalianMO.)
