One recent proof of quadratic reciprocity involves computing various rations of the Gauss sum.

In Quadratic reciprocity and the sign of the Gauss sum via the finite Weil representation Gurevich, Hadani and Howe discuss the Gauss sums, which are finite field analogues of the Gamma function $\Gamma(n+1)$:

$$ G_p = \sum_{x \in \mathbb{F}_p} e^{\frac{2\pi i x^2}{p}} = \left\{ \begin{array}{rc} \sqrt{p} & \text{if }p \equiv 1 \mod 4 \\ i\sqrt{p} & \text{if }p \equiv 3 \mod 4\end{array}\right.$$

with the relationship $G_p^2 = (\tfrac{-1}{p})\cdot p$. Using properties of the Weil representation of $SL(2,\mathbb{Z})$, and the fact that the Finite Fourier Transform corresponds to

$$ S = \left( \begin{array}{cr} 0 & -1 \\ 1 & 0 \end{array} \right)$$

they are able to prove quadratic reciprocity and correctly get the sign of the Legendre symbol. In fact, they compute the ratio:

$$ (\tfrac{p}{q}) (\tfrac{q}{p}) = \frac{G_p \cdot G_q}{G_{pq}}$$

The expression on the right looks like a Mobius function for the divisors of $pq$. So I started trying other combinations to see if I got meaningful symbols. Using three variables:

$$ \frac{G_p G_q G_r G_{pqr}}{G_{pq}G_{qr}G_{pr}} \equiv 1$$

Computer experiments showed this ratio always to be $1$. In order to fine more interesting behavior I found:

$$ \frac{G_p G_q G_r }{G_{pqr}} \equiv \pm 1$$

And there is no reason to stop there since Mobius functions can be defined for any lattice using incidence algebra. In particular, the poset of numbers and their divisors.

Is there a name for these generalized symbols, or are these trivial extensions of the original Legendre symbol? Finally, what is the rule determining the sign in the second example?

  • $\begingroup$ Maybe you should mention that generally, the summation for $G_n$ is just over $\mathbb Z/n\mathbb Z$ instead of $\mathbb F_p$ for primes. :) $\endgroup$
    – Wolfgang
    Commented Jan 19, 2015 at 16:31
  • $\begingroup$ Has the sign of $\frac{G_p G_q G_r }{G_{pqr}}$ no correlation with the sign of, say, $i^{pqr}$ or with $\frac{i^{pqr}}{i^{p+q+r}}$? $\endgroup$
    – Wolfgang
    Commented Jan 19, 2015 at 16:41
  • $\begingroup$ @Wolfgang The formula for Gauss sum holds for odd primes. And the factor of 1 or $i$ has to be carefully checked. $G_{pqr} \propto \sqrt{pqr}$ but we dont know which square root of $-1$ to put there. $\endgroup$ Commented Jan 19, 2015 at 16:51
  • $\begingroup$ I know. But I mean you are talking about $G_{pq}$ without defining it. And when I mentioned the sign of $i^{pqr}$, that was just a synonyme for some conjectural link with $p,q,r\mod 4$. (4 cases) $\endgroup$
    – Wolfgang
    Commented Jan 19, 2015 at 16:56
  • $\begingroup$ @Wolfgang it's something like $(-1)^{\frac{p-1}{2}}(-1)^{\frac{q-1}{2}}(-1)^{\frac{r-1}{2}}$ but it could be some other joint parity condition mod 4. For example it must be symmetric upon interchanging $p, q, r$ $\endgroup$ Commented Jan 19, 2015 at 17:27

1 Answer 1


If you look at bit further in the preprint you will see that the authors also state the generalisation \begin{equation*} \frac{G_{n_1} G_{n_2}}{G_{n_1 n_2}} = \left(\frac{n_1}{n_2}\right) \left(\frac{n_2}{n_1}\right) \end{equation*} for any coprime odd numbers $n_1$, $n_2$. From this it easy to see that indeed \begin{equation*} \frac{G_p G_q G_r G_{pqr}}{G_{pq} G_{qr} G_{pr}}=1, \end{equation*} and that \begin{equation*} \frac{G_p G_q G_r}{G_{pqr}}= \begin{cases} +1 & \mbox{if at most one of $p$, $q$ and $r$ is $\equiv 3\pmod{4}$,}\\ -1 & \mbox{otherwise.} \end{cases} \end{equation*} In general the formula also shows that such "balanced" quotients as above always can be converted to a symmetric product of Legendre symbols which in turn only depends on the congruence classes mod $4$ for the primes involved.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.