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Let $r$ be a primitive root of unity mod $p$. $x \in \mathbb{F}_p$ with $1 + x \not\equiv 0$ (mod $p$). I recently came across the sum:

$$ \sum_{y = 0}^{p-1} r^{y(1 + x)} = p $$

How do I derive this identity? The geometric sum would equal:

$$ \frac{r^{p(1+x)} - 1}{r^{1+x} - 1} $$

I'm stuck how this equal $p$. Any hints that help me here?

user7802048
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    I'm fairly sure that in the sum you are supposed to have $y$ range from $1$ to $p-1$ or from $0$ to $p-2$. Otherwise one term is repeated. And $r^{p-1}\equiv1$, so with that fix in place your numerator will be $\equiv0\pmod p$. Also you want $x$ not to be a multiple of $p-1$. After all the exponents should be integers, not elements of $\Bbb{F}_p$. – Jyrki Lahtonen Dec 05 '17 at 13:45
  • Similar to https://math.stackexchange.com/questions/1051380/how-to-prove-sum-of-powers-property-of-roots-of-unity – lhf Dec 05 '17 at 13:52

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