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As part of a combinatorics question I was able to simplify it into evaluating $(1+x)(1+x^2)\ldots (1+x^{p-1})$ at pth roots of unity for odd primes p.

I have guessed and checked with a computer that this product is always $1$ except for when we use the root of unity $1$, in which case it's $2^{p-1}$ (the latter case is trivial)

I'm not sure how to prove that the product is $1$ for non $1$ pth roots of unity.

What I've tried:

Let $z$ be a pth root of unity that isn't equal to $1$.

Hence the product equals $$ \prod_{k=1}^{\frac{p-1}2}(1+z^k)(1+z^{-k}) \quad\quad\text{(pairing opposite factors)}$$ $$ = \prod_{k=1}^{\frac{p-1}2}\left(2\cdot\cos\left(\frac{k\pi}{p}\right)\right)^2$$

However I see no way to go forward from here and prove the product equals $1$.

ViktorStein
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Karan Elangovan
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1 Answers1

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First note that since $p$ is prime for any p-th root of unity $x$ we know that $-x$ is a p-th root of $-1$. Furthermore we know $X^p-1=\prod_{i=0}^{p-1}(X-x^i)$, thus we have $X^p+1=\prod_{i=0}^{p-1}(X+x^i)$ as by the above mentioned we know all roots of $-1$ hence we know how the polynomial factors(alternatively multiply the equation with $-1$ to obtain $-X^p+1=-\prod_{i=0}^{p-1}(X-x^p)=-(-1)^p\prod_{i=0}^{p-1}(-X+x^i)=\prod_{i=0}^{p-1}(-X+x^i)$ as p is odd, and presume as below with the only difference evaluating at $-1$ and adjusting the sign of $X$). Now we observe that $x^0=1$ hence we have $X^p+1=(X+1)\prod_{i=1}^{p-1}(X+x^i)$. Hence evaluating at $X=1$ yields $2=1^p+1=2\prod_{i=1}^{p-1}(1+x^i)$ i.e. $1=\prod_{i=1}^{p-1}(1+x^i)$.

Mo145
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