Showing $\sum_{k}^{}e^{ika\left ( n-m \right )}=N \delta_{nm}$

Question

I would like to show that $\sum_{k}^{}e^{ika\left ( n-m \right )}=N \delta_{nm}$

Despite trying to crack this for hours, I am not anywhere closer to proving this identity.

Note:$ k=\frac{2 \pi \left ( p-\frac{N}{2} \right )}{Na} \forall p \in \left [ 1,N \right ]$

Attempt:

$ika\left ( n-m \right )=\frac{2 i \pi\left ( n-m \right )\left ( p-\frac{N}{2} \right )}{N}$

$\sum_{k}^{}e^{ika\left ( n-m \right )}=\sum_{p}^{}e^{\frac{i 2 \pi \left ( n-m \right )\left ( p-\frac{N}{2} \right )}{N}}$

Note that for $x\in \mathbb{C}$ and $p \in \left [ 1,N \right ]$:

$\sum_{p}^{}x^{p}=\frac{x\left ( x^{N}-1 \right )}{x-1}$

I decided to let $x = e^{\frac{i2 \pi }{N}}$ but am unable to proceed.

Any help to get me further is appreciated.

Answer 1

So you get for $n\ne m$ or better when $N$ does not divide $(n-m)$ $$ \sum_{p=1}^Ne^{2i\pi\cdot(n-m)\cdot \frac pN-i\pi\cdot(n-m)}=(-1)^{n-m}⋅e^{2i\pi\cdot(n-m)\cdot \frac 1N}\frac{e^{2i\pi\cdot(n-m)}-1}{e^{2i\pi\cdot(n-m)\cdot \frac 1N}-1}=0 $$ as $e^{i\pi⋅M}=(-1)^M$, $e^{2i\pi⋅M}=1$ for all integer $M$.

In the case where $(n-m)$ is an integer multiple of $N$, the terms all reduce to $(-1)^{n-m}$, so that the sum is $$...=N⋅(-1)^{n-m}.$$