Let $Q(n_1,n_2,\ldots,n_r)$ be a quadratic form of several variables with integral coefficients. Let $q$ be a prime and let $0<a<q$. We are interested in the associated Gauss sum $$S(a,q)=\sum_{n_1,n_2,\ldots,n_r=0,\ldots,q-1}e\left(\frac{aQ(n_1,n_2,\ldots,n_r)}{q}\right).$$

Now suppose $a$ is a quadratic residue mod $q$. Writing $a\equiv b^2$ for some $b$ coprime to $q$, we see that $$S(a,q)=\sum e\left(\frac{Q(bn_1,bn_2,\ldots,bn_r)}{q}\right)=S(1,q).$$

What happens when $a$ is a quadratic nonresidue? When the quadratic form $Q$ is a sum of squares, we do know that $$S(a,q)=\left(\frac{a}{q}\right)^rS(1,q)$$ where $\left(\frac{a}{q}\right)$ stands for the Legendre symbol. My question is, is it true for a general $Q$?


1 Answer 1


Yes https://en.wikipedia.org/wiki/Quadratic_form#Equivalence_of_forms

So $$S(a,q)=\left(\frac{a}{q}\right)^R S(1,q)$$ with $R$ the number of non-zero terms in the sum of squares, it is $r$ when the symmetric matrix of the quadratic form has full rank.

When $R$ is odd and $a$ is not a square, together with $S(0,q)+ \frac{q-1}{2} S(1,q)+\frac{q-1}{2}S(a,q)= \sum_{k=0}^{q-1} S(k,q)=q f(0)$ it gives $f(0)=\# \{ n \in F_q^r, Q(n)=0\}=q^{r-1}$.

  • $\begingroup$ But does it matter here we are talking about integral coefficients rather than rational coefficients? The summing over the integers 0,1,...,q-1 would be changed into over some rationals, then I am not sure if the same relations for S(a,q) holds... $\endgroup$ Nov 24, 2020 at 7:10
  • 1
    $\begingroup$ We are summing over $n$ in the finite field with $q$ elements (with $q>2$ since otherwise it is trivial) and $Q$ is a quadratic form over $F_q$, no rationals here. The theorem is saying that for some $P\in GL_r(F_q), Q'(n) = Q(Pn)$ is a sum of squares quadratic form. $\endgroup$
    – reuns
    Nov 24, 2020 at 14:01
  • $\begingroup$ What happens when $q$ is not necessarily prime, say a prime power? Can we still somehow make the diagonalization work? Since now we do not have a field anymore. (The general case of $q$ can be reduced to the prime power case since S is multiplicative.) $\endgroup$ Nov 24, 2020 at 16:05
  • $\begingroup$ For $p$ odd and $k\ge 2$ we have $\sum_{n=0}^{p^k-1} e^{2i\pi n^2/p^k}=\frac1p \sum_{m=0}^{p-1}\sum_{n=0}^{p^k-1} e^{2i\pi n^2(1+mp^{k-1})/p^k} =1$ because each $1+mp$ is a square, the same holds for $S(a,p^k)=\frac1p \sum_{m=0}^{p-1} S(a(1+mp^{k-1}),p^k)=p^{r(k-1)} f(0)$. $\endgroup$
    – reuns
    Nov 24, 2020 at 16:10
  • $\begingroup$ Your calculation seems to contradict en.wikipedia.org/wiki/Quadratic_Gauss_sum where it says that $S(a,p^k)=p S(a,p^{k-2})$ for the classical Gauss sum? $\endgroup$ Nov 24, 2020 at 16:34

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