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Can we prove the following statement about Kloosterman sums? Recall that a Kloosterman sum is given by:

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

Where $x^*$ is the modular inverse of $x \mod m$.

Let $b=nk$. Then I want to say that if $4|n$, then:

$$K\left(\frac{b}{4k},\frac{b}{4k},2b\right)=0$$

if and only if $k\neq 2\mod 4$ and $8\nmid k$.

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  • $\begingroup$ Is this true? Meaning, did you check it numerically for small values? $\endgroup$ – Alex Degtyarev Nov 30 '14 at 12:37
  • $\begingroup$ Yes, I did #fill $\endgroup$ – ruadath Nov 30 '14 at 12:42
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That should be doable by combining

(1) the Chinese Remainder Theorem to reduce to $m$ being a prime power (namely the Kloosterman sum factors as a product of Kloosterman sums module the primes powers dividing $m$; the other arguments do change, so this is called "twisted multiplicativity" in the literature).

(2) For $m=p^k$ a prime power, there are two cases:

(2.1) it is known (looking at congruences modulo the ideal $(1-e(1/p))$ of the $p$-th cyclotomic field) that the Kloosterman sum is non-zero if $k=1$;

(2.2) if $k\geq 2$, there is an exact formula in terms of the roots of the equation $x^2=ab$ modulo $p$, at least if $p$ is odd (I don't know what happens for $p=2$, but in principle it should be something similar). For instance, see Exercise 1 in Chapter 12 of Iwaniec-Kowalski. This formula shows exactly which Kloosterman sums vanish in that case.

It seems that for the specific question, the only tricky part is to handle what happens for the prime $2$.

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