Given that $n$ is a positive integer and $\omega=e^{2\pi i/n}$, how do you prove that $a_0+a_1\omega+a_2\omega^2+...+a_{n-1}\omega^{n-1}=0$ is only true when all the $a$'s are equal?
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2I must be missing something. What if $n = 4$ and $a_0 = a_2 = 1$ and $a_1 = a_3 = 2$? – Brian Tung Jun 03 '22 at 03:49
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Amended result. If $n$ is prime and $\omega=e^{2\pi i/n}$ and $a_0,\ldots,a_{n-1}$ are integers and $$a_0+a_1\omega+a_2\omega^2+\cdots+a_{n-1}\omega^{n-1}=0\ ,$$ then $a_0=a_1=a_2=\cdots=a_{n-1}$.
Proof. Under the stated conditions, $\omega$ is a root of the integer polynomial $$f(z)=a_0+a_1z+a_2z^2+\cdots+a_{n-1}z^{n-1}\ .$$ Therefore $f(z)$ is a (polynomial) multiple of the minimal polynomal of $\omega$, which is $$m(z)=1+z+z^2+\cdots+z^{n-1}\ .$$ Since $f(z)$ has the same degree as $m(z)$, it must in fact be a constant multiple of $m(z)$, and the result follows.
See also this book, problem 3.26 (shameless free plug).
David
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+1. I'm chagrined not to have realized that's what was probably meant. – Brian Tung Jun 03 '22 at 04:53