A routine Fourier-analytic calculation Erdos discrepancy problem

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I am reading the awesome paper Erdos discrepancy problem by Terence Tao and my Fourier-analytic calculations are a bit rusty. On page 11 the author claims that the following (left hand side of (2.3)):

$$\frac{1}{M^r} \sum_{x \in (\mathbb{Z}/M\mathbb{Z})^r} \left\| \sum_{j=1}^n F(x + \pi(j)) \right\|_H^2$$

can be rewritten as

$$\sum_{\xi \in (\mathbb{Z}/M\mathbb{Z})^r} \|\hat F(\xi)\|_H^2 \left|\sum_{j=1}^n e\left(\frac{\pi(j)\cdot \xi}{M}\right)\right|^2$$

by a routine calculation (using the Plancherel identity).

I have tried to do these calculations myself, but I got stuck after expressing $F(x+\pi(j))$ as the Fourier expansion and rewriting the norm as a scalar product. I can see that if the scalar products of different $x, y \in (\mathbb{Z}/M\mathbb{Z})^r$ sum to zero then the rewriting can be done. I do not see that the scalar products sum to zero though.

How can one show that the products will sum to zero? Is there another easier way to do these computations?

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The Fourier transform of $x\mapsto F(x+\pi(j))$ is $\xi\mapsto \hat F(\xi) e(\frac{\pi(j)\cdot\xi}M).$

So the Fourier fransform of $x\mapsto \sum_{j=1}^nF(x+\pi(j))$ is $\xi\mapsto \sum_{j=1}^n\hat F(\xi) e(\frac{\pi(j)\cdot\xi}M).$

By Plancherel, the sum over $x$ of $\|\sum_{j=1}^nF(x+\pi(j))\|_H^2,$ multiplied by a normalization constant $1/M^r,$ is the sum over $\xi$ of $\|\sum_{j=1}^n\hat F(\xi) e(\frac{\pi(j)\cdot(\xi)}M)\|_H^2.$