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

Fearne Castro

Fearne Castro

Answered question

2022-03-03

A little bit different quadratic Gauss sum
Could anyone give me a direction on how to demonstrate that
limitsk=0N1eiπN(k+C)2=Neiπ4,
If NN is even, for any CZ.
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.

Answer & Explanation

Mikayla Swan

Mikayla Swan

Beginner2022-03-04Added 9 answers

Let N=2n. Note the following:
k=04n1ei2π4n(k+C)2=k=04n1ei2π4nk2=(1i)4n=2eiπ4N
Here we used that ei2π4n is a primitive 4n-th root of unity and a classical Gauß sum identity (I used Lang'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 CZ.
Now it only remains to show that the sum on the LHS is twice the sum from k=0 to k=2n1. For that part notice that
((2n+k)+C)2(k+C)2mod4n
which a short computation reveals. Therefore,
k=04n1ei2π4n(k+C)2=k=02n1ei2π4n(k+C)2+k=2n4n1ei2π4n(k+C)2=2k=02n1ei2π4n(k+C)2
which concludes the proof (well, up to the knowledge of the "classical Gauß sum identity").

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