Gaussian Periods

Let \(\zeta = e^{{2\pi i / 17}}\) be a primitive 17th root of unity. Define \(x_1, x_2\) by

The exponents of the terms of \(x_1\) are the quadratic residues and those of \(x_2\) are the nonresidues.

These are examples of Gaussian periods. We stated before that \(x_1 x_2\) is a rational that is easy to compute. Why?

Write \((x_1 - x_2)^2\) as

for integers \(a_i\). Replacing \(\zeta\) with \(\zeta^3\) merely swaps \(x_1\) and \(x_2\) due to their construction, whence

Thus looking at the coefficients of the powers of \(\zeta\) we have \(a_{16} = a_1, a_1 = a_2, ..., a_{15} = a_{16}\), that is, they are all equal to some integer \(a\):

Thus \(x_1 x_2 = ((x_1+x_2)^2 - (x_1-x_2)^2)/4\) is some rational number. (Recall \(x_1 + x_2 = -1\).)

Each of \(x_1\) and \(x_2\) consists of 8 terms so their product \(x_1 x_2\) has 64 terms of the form \(\zeta^k\) for some \(k\). Since \(17 = 1 \pmod{4}\) we know \(-1\) is a quadratic residue, hence if \(r\) is a quadratic residue so is \(-r\). As each \(k\) can be viewed as the sum of a residue and a nonresidue modulo 17, this means \(k\) is never zero.

Hence if we write:

we must have \(b_0 = 0\), and the 64 terms must be evenly distributed among \(\zeta ,..., \zeta^{16}\), that is, \(b = 4\). Therefore \(x_1 x_2 = -4\). (Thanks to Dennis Westra for bringing this argument to my attention.)

Exercise: what happens when \(-1\) is a quadratic nonresidue?

We can generalize these statements.