"Let $C_n(x)$ be the polynomial such that the roots of $C_n(x)=0$ are the primitive $n^{th}$ roots of unity. Prove that there are no positive integers $q,r,s$ for which $C_q(x)=C_r(x)C_s(x)$."
My latex skills are non existent so I'll have to explain my proof which needs verifying. Clearly $C_n(x)$ is the product of $(x-\exp(i2\pi(m/n)))$ with $0<m\le n-1$ and fundamentally $m$ and $n$ coprime. So using this expression in the equation we must disprove, note that for $q\ne r,s$ surely no fractions in the exponents in the factors on the LHS could appear on the RHS? (The case $q=r$ would mean $C_s(x)=1$ which is impossible).
Is this a valid proof? Because I then go on to prove that if the pairs of numbers $a,b$ and $c,d$ are coprime with $b\ne d$, then $(a/b)\ne (c/d)$. Many thanks!
If such an equality holds, then each primitive $r$th root of unity $\xi\in\mathbb C$ is also a primitive $q$th root of unity, that is, for such a $\xi$ not only is $r$ the smallest positive integer with $\xi ^r=1$ but also $q$ is the smallest positive integer with $\xi^q=1$. This just means that $r=q$. Likewise $s=q$, so $C_q=C_q^2$, i.e., $C_q=1$ or $C_q=0$, which is absurd.