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 $Q$?
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 $f(0)=\# \{ n \in F_q^r, Q(n)=0\}=q^{r-1}$.