TITLE: Doubt :$\prod_{b=1}^n(1+\zeta^{ab})=2^k$, where $\zeta = e^{2\pi i/n}$ , $n$ is an odd integer, gcd$(a,n)=k$.
QUESTION [1 upvotes]: $\prod_{b=1}^n(1+\zeta^{ab})=2^k$, where $\zeta = e^{2\pi i/n}$ , $n$ is an odd integer, gcd$(a,n)=k$.

Can someone explain me why this is true, I got that $\prod_{b=1}^n(1+\zeta^{b})=2$ where $\zeta = e^{2\pi i/n}$
It's from https://artofproblemsolving.com/community/c7h1171025p5624333 post 3.
Thanks in advance!

REPLY [0 votes]: Okay so we will use two facts:

FunFact 1: $\zeta^{cb}=\zeta^{ab}$ if $c\equiv a\pmod{n}$ where $\zeta$ is the $n^{th}$ root of unity.


FunFact 2: $\prod_{b=1}^n (1+\zeta^b)=2$ where $\zeta$ is the $n^{th}$ root of unity.

As, $\gcd(a,n)=k\implies a=kA, n=kN$ for some co-prime natural numbers $A$ and $N$. Thus,
$$\zeta^{ab}=e^{\frac{2\pi i\cdot kAb}{kN}}=e^{\frac{2\pi iAb}{N}}=\xi^{Ab}$$where $\xi$ is the $N^{th}$ root of unity and $\gcd(A,N)=1$. Thus, from FunFact 1. we get
$$\zeta^{ab}=\xi^{cb}$$where $A\equiv c\pmod{N}$ such that $0<c<N$. Also, $\gcd(A,N)=1\implies \gcd(c,N)=1$ and thus, $$\mathcal S:=\{bc\pmod{N}: b\in\mathbb N, 1\leq b\leq N\}$$forms a complete residue class modulo $N$. Thus,
$$\prod_{b=1}^N(1+\zeta^{ab})=\prod_{b=1}^N(1+\xi^{cb})=\prod_{i=1}^N(1+\xi^{i})=2$$where the last equality follows from FunFact 2. Now, note that, from FunFact 1., we also know that
$$\prod_{b=1}^{N}(1+\zeta^{ab})=\prod_{b=N+1}^{2N}(1+\zeta^{ab})=\cdots = \prod_{b=N(k-1)+1}^{kN}(1+\zeta^{ab})$$and thus, as $kN=n$, we have,
$$\prod_{b=1}^{n}(1+\zeta^{ab})=\left(\prod_{b=1}^{N}(1+\zeta^{ab})\right)^k=2^k$$completing the proof.
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