A Kloosterman sum is defined as

$$K(a,b;m)=\sum_{\substack{ 0\leq x \leq m-1\\\gcd(x,m)=1}} e^{2\pi \mathcal{i} (ax+bx^*)/m}$$

where $a,b,m \in \mathbb{N}$ and $x^*$ is the inverse of $x$ modulo $m$.

How can I show that

$$K(1,1;16n)\neq 0$$

for odd $n$?

  • $\begingroup$ I don't know if it can be useful, but you can try to use the trasformation (if $\left(a,m\right)=1$) $$K\left(a,a;m\right)=\underset{r=0}{\overset{m-1}{\sum}}\left(\frac{r^{2}-4a}{m}\right)e^{2\pi ir/m}$$where $\left(\frac{a}{b}\right)$ is the Jacobi symbol. $\endgroup$ Commented Dec 15, 2014 at 12:44
  • $\begingroup$ I believe this only works for $m$ prime? $\endgroup$
    – ruadath
    Commented Dec 15, 2014 at 20:10
  • $\begingroup$ I've found on wikipedia that works if $(a,m)=1$ and this is your situation. $\endgroup$ Commented Dec 15, 2014 at 20:13


You must log in to answer this question.

Browse other questions tagged .