Estimates for certain double-Kloosterman sums Sorry to disturb. I encounter a double-Kloosterman sum, which needs some help from the experts here.
For any $q\in \mathbb{N}^+$, how can we estimate the type of sum
$$ \sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{x,y \bmod q} \left( \frac{x+\overline{x}+y+\overline{y}+\overline{x+y}}{q}\right)?$$
It can be certain that this double sum is bounded by $q^{1+\varepsilon}$. I searched many papers, but haven't found a relevant description involving proving this.
So, if some expert has seen this type of sum in the question or leans something how to show the bound above, please give some comments or guide a reference.
Great thanks in advance!
 A: I can do it for $q$ prime, and thus $q$ squarefree, by a long but elementary manipulation followed by the Weil bound. Until the end, the elementary manipulations will work for $q$ arbitrary.
Taking the simplest interpretation of your formula in the case $x+y$ not coprime to $q$ (that the term should be $0$ for such $x,y$),
$$ \sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{x,y \bmod q} \left( \frac{x+\overline{x}+y+\overline{y}+\overline{x+y}}{q}\right) = \sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{\substack{ x,y,z \bmod q\\ x+y =z }} \left( \frac{x+\overline{x}+y+\overline{y}+\overline{z}}{q}\right) $$
$$= \frac{1}{ G }  \sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{\substack{ x,y,z ,w \bmod q\\ x+y =z }}  \left( \frac{w}{q} \right)  e\left( \frac{w( x+\overline{x}+y+\overline{y}+\overline{z})}{q}\right)$$
(where $G$ is a Gauss sum and we now multiply the Legendre symbol by an additive character)
$$= \frac{1}{ qG }  \sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{\substack{ x,y,z ,w \bmod q}} \sum_{u \bmod q}  \left( \frac{w}{q} \right)  e\left( \frac{w( x+\overline{x}+y+\overline{y}+\overline{z}) + xu + yu -zu }{q}\right)$$
(detecting $x+y=z$ by additive characters)
Now I introduce the change of variables $u =u' \overline{w}, x= x'\overline{w}, y= y'\overline{w}, z=z'\overline{w} $ and then the new variable $a = w^2+ u'$ seemingly at random.
$$= \frac{1}{ qG }  \sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{\substack{ x',y',z' ,w \bmod q}} \sum_{u' \bmod q}  \left( \frac{w}{q} \right)  e\left( \frac{(\overline{x}'+\overline{y}'+\overline{z}') + w^2(x'+y') +(x'+y'-z') u^\prime }{q}\right)$$
$$ = \frac{1}{ qG }  \sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{\substack{ x',y',z' ,w \bmod q}} \sum_{ \substack{ a, u' \bmod q\\ a =w^2+ u' } }  \left( \frac{w}{q} \right)  e\left( \frac{(\overline{x}'+\overline{y}'+\overline{z}') + a(x'+y') - z'u'   }{q}\right)$$
$$ = \frac{1}{ q^2G }  \sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{\substack{ x',y',z' ,w \bmod q}} \sum_{ \substack{ a, u',c \bmod q } }  \left( \frac{w}{q} \right)  e\left( \frac{(\overline{x}'+\overline{y}'+\overline{z}') + a(x'+y') - z'u' + c a - cw^2 - cu'  }{q}\right)$$
Now we can finally start eliminating variables. The sum over $u'$ of the terms involving $u'$ gives $q$ if $z'=-c$ and $0$ otherwise, so we get
$$ = \frac{1}{ qG }  \sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{\substack{ x',y',c ,w \bmod q}} \sum_{ \substack{ a \bmod q } }  \left( \frac{w}{q} \right)  e\left( \frac{(\overline{x}'+\overline{y}'-\overline{c}) + a(x'+y')  + c a - cw^2   }{q}\right)$$
We have $\sum_{ \substack{ w \bmod q\\ w^2=t }} \left(\frac{w}{q}\right) = \sum_{\substack{ \chi: (\mathbb Z/q)^\times \to \mathbb C^\times \\ \chi^2 = \left(\frac{\cdot}{q} \right) }} \chi(t) $ and since $\sideset{_{}^{}}{^{\ast}_{}}\sum_t \chi(t) e \left( \frac{-ct}{q} \right) = \chi(\overline{c})$ times a Gauss sum $G_\chi$, we have
$$ =\frac{1}{ qG } \sum_{\substack{ \chi: (\mathbb Z/q)^\times \to \mathbb C^\times \\ \chi^2 = \left(\frac{\cdot}{q} \right) }}  G_\chi \sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{\substack{ x',y',c\bmod q}} \sum_{ \substack{ a \bmod q } }    e\left( \frac{(\overline{x}'+\overline{y}'-\overline{c}) + a(x'+y')  + c a   }{q}\right) \chi( \overline{c}). $$
Now the sum over $a$ vanishes unless $y = -x' -c$ and equals $q$ in that case, so we get
$$\frac{1}{ G } \sum_{\substack{ \chi: (\mathbb Z/q)^\times \to \mathbb C^\times \\ \chi^2 = \left(\frac{\cdot}{q} \right) }}  G_\chi \sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{\substack{ x',c\bmod q \\ \gcd( x'+c, q)=1}}   e\left( \frac{\overline{x}'+\overline{-c-x'}-\overline{c}  }{q}\right) \chi( \overline{c}). $$
The change of variables $x' = c v$ gives
$$\frac{1}{ G } \sum_{\substack{ \chi: (\mathbb Z/q)^\times \to \mathbb C^\times \\ \chi^2 = \left(\frac{\cdot}{q} \right) }}  G_\chi \sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{\substack{ v,c\bmod q \\ \gcd( v+1, q)=1}}   e\left( \frac{\overline{v c}-\overline{(v+1) c}-\overline{c}  }{q}\right) \chi( \overline{c}). $$
The sum in $c$ is now another Gauss sum, whose value is $G_\chi \overline{\chi} ( 1 + \overline{(v+1)} - \overline{v} )$ except in the case that $( 1 + \overline{(v+1)} - \overline{v} )$ is not coprime to $q$, in which case it is zero unless some $\chi$ is imprimitive. Since this never occurs in the $q$ squarefree case I will ignore it.
$$\frac{1}{ G } \sum_{\substack{ \chi: (\mathbb Z/q)^\times \to \mathbb C^\times \\ \chi^2 = \left(\frac{\cdot}{q} \right) }}  G_\chi  \sum_{ \substack{ v \bmod q \\ \gcd(v,q) =\gcd(v+1,q)=1}}  \overline{\chi} ( 1 + \overline{(v+1)} - \overline{v})$$
For $q$ prime, the Weil bound for the inner sum is $2 \sqrt{q} + 1$. The Gauss sums all have size $q$, and there are at most two characters $\chi$, so your sum is at most $2\sqrt{q} ( 2 \sqrt{q} +1)$.
Multiplying, this gives $O( q^{ 1+ \epsilon})$ for $q$ squarefree.
