The proofs of quadratic reciprocity based on quadratic Gaussian sums involve sums of the form

$$\sum_{k=1}^{p-1}\left(\frac k p \right)\zeta^{ak}$$

Where $\left(\frac k p \right)$ is a Legendre symbol and $\zeta$ is a $p$-th root of unity. Where were these sums considered (if in the DA, which section) ? Can someone summarize what motivated these seemingly arbitrary expressions?

  • $\begingroup$ Wikipedia references 1811 and 1818 papers rather than Disquitiones Arithmetica en.wikipedia.org/wiki/… $\endgroup$
    – Conifold
    Commented Apr 6, 2015 at 23:37
  • 2
    $\begingroup$ Gauss sums are introduced in Disq. Arith., §356. $\endgroup$ Commented Apr 7, 2015 at 15:29
  • $\begingroup$ Related Math.SE question $\endgroup$
    – Danu
    Commented May 10, 2015 at 7:49

2 Answers 2


One answer is that Gauss sums are special cases of Lagrange resolvents; these in turn are expressions that Lagrange used for analysing the solution of equations by radicals.

Another source are Euler's and Lagrange's factorizations of the cyclotomic equations $4(x^p-1)/(x-1)$ in the form $Y^2 - p^*Z^2$, where $p^* = (-1)^{\frac{p-1}2}p$, for $p = 3, 5, 7$ in connection with problems on special cases of the quadratic reciprocity law. Both of them were unable to find the factorization for $p = 11$, but Gauss realized that the roots of the linear factors of the quadratic polynomials are roots of unity $\zeta^r$ and $\zeta^s$, respectively, where $r$ and $n$ denote the quadratic residues and nonresidues modulo $p$. Once you start writing down the polynomials $(X - \zeta)(X-\zeta^4)(X-\zeta^9)\cdots$ and look at the coefficient of the second largest power of $X$, you will discover quadratic Gauss sums.

  • $\begingroup$ Yes, Lagrange resolvents! Exactly. If we were wanting to express roots of unity in radicals, we would look at these. :) $\endgroup$ Commented Jun 16, 2022 at 23:31

Can someone summarize what motivated these seemingly arbitrary expressions?

They are far from arbitrary.

Because quadratic reciprocity is about whether a prime $p$ is a square modulo another prie $q$, it is natural to look for an actual square-root when it is (an idea that was already familiar to Euler). The use of quadratic Gauss sums is to construct the desired square-root (but possibly in a finite extension). With that explicit goal in mind, it is in fact not so hard to see patterns emerging when doing computations, especially if you have been thinking long and hard about Legendre's symbols and even if you don't know the answer beforehand. I have sometimes asked students to guess them for instance and they do succeed (occasionally).

But though the students do not know the result, they have a tremendous advantage: they are conducting the guessing in a course on finite extension of finite fields or algebraic number theory. That idea, i.e the idea of looking for the square-root in extensions (of the integers or of finite fields) and of studying the behavior of primes in such extensions, was absolutely groundbreaking.

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    $\begingroup$ So is the sum in the OP a representation of a square root? $\endgroup$
    – Conifold
    Commented Apr 8, 2015 at 0:04
  • $\begingroup$ @Conifold A square root of $\pm p$ depending on whether $p$ is congruent to 1 or 3 modulo 4, yes. Gauss then worked hard to determine which square root. $\endgroup$
    – Olivier
    Commented May 6 at 7:08

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