Show that the Kloosterman Sum $S(a,b,p^k)$ vanishes under the given condition. Let $a,b,k\in \mathbb{N}$, $k>1$. $p $ be an odd prime number and $p$ does not divide $a,b$. The Kloosterman sum is given by
$$S(a,b,p^k)=\sum_{x\in (\mathbb{Z}/p^k \mathbb{Z})^\times}e\Big(\frac{a x+b \overline{x}}{p^k}\Big)$$ where $e(x)=e^{2\pi i x}$. The given condition is when $(\frac{a}{p})\neq (\frac{b}{p})$.
Where $x\overline{x}=1 \in (\mathbb{Z}/p^k \mathbb{Z})^\times$ and $(\frac{m}{n})$ denotes the Legendre symbol.
So far I have been able to show that
$S(a,b,p^k)=S(1,ab,p^k)$. I am not able to proceed afterwards.
 A: To prove this we are going to consider two cases, one when $k$ is even, and one when $k$ is odd.
For the first case let $k = 2n$. Now we consider $u(1+vp^n)$, where $u$ ranges through the units modulo $p^k$ and $v$ ranges freely modulo $p^n$. It is not very hard to see that every primitive class modulo $p^k$ will be covered $p^n$ times. Moreover, note that the multiplicative inverse of $u(1+vp^n)$ modulo $p^k$ is $\overline{u}(1-vp^n)$ Therefore:
$$p^n \cdot S(a,b,p^k) = \sum_{u \pmod{p^k}^*} \quad\sum_{v \pmod{p^n}} e\left(\frac{au(1+vp^n) + b\overline{u}(1-vp^n)}{p^k}\right)$$
$$=\sum_{u \pmod{p^k}^*}  e\left(\frac{au+b\overline{u}}{p^k}\right) \cdot \sum_{v \pmod{p^n}} e\left(v \cdot \frac{au-b\overline{u}}{p^n}\right) $$
The second sum is a sum of $p^n$-th roots of unity, so as $n \ge 1$ it will be non-zero if and only if $au \equiv b\overline{u} \pmod{p^n}$. However, if the last congruence relation is satisfied we have that $au \equiv b\overline{u} \pmod p$ and in particular $au^2 \equiv b \pmod p$, which contradicts the fact that $\left(\frac ap\right) \neq \left(\frac bp\right)$. Hence the second sum is always $0$, and so $S(a,b,p^k) = 0$
The second case when $k=2n+1$ is solved analogously. We consider $u(1+vp^{n+1})$ as $u$ runs through the units modulo $p^k$, while $v$ runs freely modulo $p^{n}$. Again each primitive class modulo $p^k$ will be covered exactly $p^n$ times. Also the multiplicative inverse of $u(1+vp^{n+1})$ modulo $p^k$ is $\overline{u}(1-vp^{n+1})$. So:
$$p^{n} \cdot S(a,b,p^k) = \sum_{u \pmod{p^k}^*} \quad\sum_{v \pmod{p^n}} e\left(\frac{au(1+vp^{n+1}) + b\overline{u}(1-vp^{n+1})}{p^k}\right)$$
$$=\sum_{u \pmod{p^k}^*}  e\left(\frac{au+b\overline{u}}{p^k}\right) \cdot \sum_{b \pmod{p^n}} e\left(v \cdot \frac{au-b\overline{u}}{p^n}\right) = 0$$
where we use the same argument as above. Also note that we used $n \ge 1$, which is why you want $k \ge 2$. Indeed, the claim is not true for $k=1$.
