A little bit different quadratic Gauss sumCould anyone give me a direction on how to demonstrate that∑limitsk=0N−1e−iπN(k+C)2=Ne−iπ4,If N∈N is even, for any C∈Z.Unfortunately, the left side of the above equation cannot be exactly defined as a quadratic Gauss sum since it does not have the term 2 within the argument of e and the summation coefficient is shifted by C.

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Mikayla Swan · Accepted Answer

Let N=2n. Note the following:∑k=04n−1e−i2π4n(k+C)2=∑k=04n−1e−i2π4nk2=(1−i)4n=2e−iπ4NHere we used that e−i2π4n is a primitive 4n-th root of unity and a classical Gauß sum identity (I used Lang&#039;s Algebraic Number Theory as a reference for that part). The former part is given since while k runs through a system of representatives of Z4nZ so does k+C for all C∈Z.Now it only remains to show that the sum on the LHS is twice the sum from k=0 to k=2n−1. For that part notice that((2n+k)+C)2≡(k+C)2mod4nwhich a short computation reveals. Therefore,∑k=04n−1e−i2π4n(k+C)2=∑k=02n−1e−i2π4n(k+C)2+∑k=2n4n−1e−i2π4n(k+C)2=2∑k=02n−1e−i2π4n(k+C)2which concludes the proof (well, up to the knowledge of the &quot;classical Gauß sum identity&quot;).

A little bit different quadratic Gauss sum Could anyone give me a direction on how to demonstrat

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