I am interested in the sums
$$g(a,k)=\sum\_{n=0}^{p-1}e^{2\pi i a n^k/p}$$
where $p\equiv1\mod k$ is a prime and $a$ is coprime with $p$.


When $k=2$, it is a [classical fact](http://en.wikipedia.org/wiki/Quadratic_Gauss_sum) that $g(a,2)=\chi(a)g(1,2)$ where $\chi$ is the Legendre symbol.
My question is: Is there an analogue of this formula for higher values of $k$?
More concretely, is there a Dirichlet character $\chi\_k$ such that
$$g(a,k)=\chi\_k(a)g(1,k)$$
for all $a$?



Extension of a formula for the quadratic Gauss sums
No, the relation is not as simple in this case. For example, if $k = 3$ and $p = 7$, the three different cubic Gauss sums are roots of $y^{3} - 21y - 7$, and the three roots of this polynomial do not have equal absolute values.


In general, the $k$ different Gauss sums $g(a,k)$ (where $a$ runs over representatives of the cosets of the $k$th powers in $\mathbb{F}\_{p}^{\times}$) will be roots of a polynomial of degree $k$, and the coefficient of $x^{k-1}$ in that polynomial will be zero, because
$$
\sum\_{a \in \mathbb{F}\_{p}^{\times}/\left(\mathbb{F}\_{p}^{\times}\right)^{k}} g(a,k)
= k \sum\_{n=0}^{p-1} e^{2 \pi i n / p} = 0.
$$
When $k = 2$, this forces both quadratic Gauss sums to be roots of a polynomial of the form $y^{2} - a$ and this makes the quadratic Gauss sums be negatives of each other, but this is no longer the case when $k > 2$.


