On the upper-bound for a type of quintuple Kloosterman sums Sorry to disturb, dear experts here. I have a question involving the quintuple Kloosterman sum, and expect some hints to show the upper-bound.
My question is, for any $x,y,z,w,\delta \in \mathbb{Z}$ and $q\in \mathbb{N}^+$, how can we estimate the sum
$$ \sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{x,y,z,w,\delta \bmod q} \left( \frac{x+\overline{x}\cdot w\delta+ y+\overline{y}\cdot \overline{w(\delta +1)}+z+\overline{z}\cdot \overline{\delta}
}{q}\right)?$$
Certainly the upper-bound should be $q^{5/2+\varepsilon}$. And it can be sure some stuff involving $l$-adic cohomology can be put into use; see for example Adolphson-Sperber's paper (https://www.researchgate.net/profile/Steven-Sperber-2/publication/38390736_Exponential_sums_and_Newton_polyhedra/links/558c275708ae591c19d9efe8/Exponential-sums-and-Newton-polyhedra.pdf) or Denef-Loeser's paper (https://webusers.imj-prg.fr/~francois.loeser/inv91.pdf).
Unfortunately, I am not really familiar with the $l$-adic cohomology, and I really can not figure out the Newton polyhedron $\Delta_\infty(f)$ of the polynomial $$f(x_1,x_2,x_3,x_4,x_5)=x_1+\overline{x_1}x_4x_5+x_2+\overline{x_2x_4(x_5+1)}+x_3+\overline{x_3x_5}$$ to verify whether or not $f$ is non-degenerate with respect to $\Delta_\infty(f)$.
So, if some expert has seen this type of sum in the question or leans something how to show the Newton polyhedron nondegeneracy, please give some comments or guide a reference.
Great great thanks in advance!
Your help is highly appreciated!!
 A: The Newton polyhedron method for your sum is as follows:
We first replace the quadratic character with an additive character and a simpler quadratic character using a Gauss sum, then introduce a variable equal to $\delta+1$, and finally detect $\alpha = \delta+1$ by additive characters.
$$ \sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{x,y,z,w,\delta \bmod q} \left( \frac{x+\overline{x}\cdot w\delta+ y+\overline{y}\cdot \overline{w(\delta +1)}+z+\overline{z}\cdot \overline{\delta}
}{q}\right)$$
$$=  \frac{1}{G} \sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{x,y,z,w,\delta, \lambda \bmod q} \left( \frac{\lambda}{q} \right) e\left( \frac{\lambda (x+\overline{x}\cdot w\delta+ y+\overline{y}\cdot \overline{w(\delta +1)}+z+\overline{z}\cdot \overline{\delta}
) }{q}\right)$$
$$=  \frac{1}{G} \sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{\substack{x,y,z,w,\delta, \lambda, \alpha \bmod q\\ \alpha=\delta+1}} \left( \frac{\lambda}{q} \right) e\left( \frac{\lambda (x+\overline{x}\cdot w\delta+ y+\overline{y}\cdot \overline{w} \cdot  \overline{\alpha}+z+\overline{z}\cdot \overline{\delta}
) }{q}\right)$$
$$=  \frac{1}{qG} \sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{\substack{x,y,z,w,\delta, \lambda, \alpha \bmod q}} \sum_{ \gamma \bmod q} \left( \frac{\lambda}{q} \right) e\left( \frac{\lambda (x+\overline{x}\cdot w\delta+ y+\overline{y}\cdot \overline{w} \cdot  \overline{\alpha}+z+\overline{z}\cdot \overline{\delta}
) + \gamma ( \alpha- \delta-1) }{q}\right)$$
Now for $q$ prime, this is almost a sum we can apply the Newton polyhedron methods to - we just need to break up into the $\gamma = 0$ and $\gamma \neq 0$ cases, getting
$$ \frac{1}{qG} \sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{\substack{x,y,z,w,\delta, \lambda, \alpha , \gamma \bmod q}}\left( \frac{\lambda}{q} \right) e\left( \frac{\lambda (x+\overline{x}\cdot w\delta+ y+\overline{y}\cdot \overline{w} \cdot  \overline{\alpha}+z+\overline{z}\cdot \overline{\delta}
) + \gamma ( \alpha- \delta-1) }{q}\right)$$
$$+  \frac{1}{qG} \sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{\substack{x,y,z,w,\delta, \lambda, \alpha \bmod q}}\left( \frac{\lambda}{q} \right) e\left( \frac{\lambda (x+\overline{x}\cdot w\delta+ y+\overline{y}\cdot \overline{w} \cdot  \overline{\alpha}+z+\overline{z}\cdot \overline{\delta}
) }{q}\right)$$
Now for each of these terms we can apply Proposition 0.1 of Weights of Twisted Exponential Sums by Lei Fu. I'll explain how to do the first term as the second is simpler. We have the Laurent polynomial
$$ \lambda (x+\overline{x}\cdot w\delta+ y+\overline{y}\cdot \overline{w} \cdot  \overline{\alpha}+z+\overline{z}\cdot \overline{\delta}
) + \gamma ( \alpha- \delta-1) = \lambda( x+ x^{-1} w \delta + y + y^{-1} w^{-1} \alpha^{-1} + z + z^{-1} \delta^{-1} ) + \gamma (\alpha-\delta-1) $$
which setting $X_1=x, X_2 =y, X_3 = z, X_4=w, X_5=\delta, X_6= \lambda, X_7 = \alpha, X_8 = \gamma$ is
$$ X_6( X_1+ X_1^{-1} X_4X_5 + X_2 + X_2^{-1} X_4^{-1} X_7^{-1} + X_3 + X_3^{-1} X_5^{-1} ) + X_8 (X_7-X_5-1)$$
$$=  X_6 X_1+ X_1^{-1} X_4X_5X_6 + X_2X_6 + X_2^{-1} X_4^{-1}X_6 X_7^{-1} + X_3X_6 + X_3^{-1} X_5^{-1} X_6 + X_7 X_8 -X_5X_8-X_8 $$
Now for each monomial we look at the exponents of $X_1,\dots, X_8$ and make them the entries of a vector, getting the following vectors
$$ \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{pmatrix},\begin{pmatrix} -1 \\ 0 \\ 0 \\ 1 \\ 1 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ -1 \\ 0 \\ -1 \\ 0 \\ 1 \\ -1 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ -1 \\ 0 \\ -1 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 1 \end{pmatrix},\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ 1 \end{pmatrix},\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}$$
Then $\Delta_\infty$ is defined to be the convex hull of those eight vectors together with the zero vector. Now you "just" need to calculate every face of that polytope, except the ones containing zero, find the associated polynomials, and check that their derivatives don't all vanish, probably using a computer algebra system.
Actually the second term is not so hard as you can eliminate the variables in the order $\alpha, y, w, x, \delta, z, \lambda$ and don't need to apply étale cohomology.
