# Gaussian Periods

As before, let $$\zeta = e^{{2\pi i / 17}}$$ be a primitive 17th root of unity. Define $$x_1, x_2$$ by

$x_1 = \zeta^{3^0} + \zeta^{3^2} +...+ \zeta^{3^{14}}$
$x_2 = \zeta^{3^1} + \zeta^{3^3} +...+ \zeta^{3^{15}}$

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

$(x_1 - x_2)^2 = a_0 + a_1 \zeta^{3^1} + a_2 \zeta^{3^2} + ... + a_{16} \zeta^{3^{16}} .$

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

$(x_2 - x_1)^2 = a_0 + a_1 \zeta^{3^2} + a_2 \zeta^{3^3} + ... + a_{16} \zeta^{3^{17}} .$

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 $$a$$:

$(x_2 - x_1)^2 = a_0 + a(\zeta + ... +\zeta^{16}) = a_0 - a$

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

To compute $$x_1 x_2$$, we count the number of times $$b$$ that $$\zeta^r \zeta^s = 1$$ where $$r$$ is a quadratic residue modulo 17 and $$s$$ is a nonresidue, and then count the number of times $$c$$ that $$\zeta^r \zeta^s = \zeta$$. Since we know the answer is rational, each power of $$\zeta$$ appears exactly $$c$$ times so the answer is $$b - c$$.

Since $$17 = 1 \pmod{4}$$ we know $$-1$$ is a quadratic residue, hence if $$r$$ is a quadratic residue so is $$-r$$. This means $$b = 0$$. Unfortunately I don’t know how to quickly count the number of ways $$r + s = 1$$. From the list of powers of 3, the quadratic residues are

$1, 9, 13, 15, 16, 8, 4, 2 .$

Since $$-r$$ is also a quadratic residue we can count the solutions to $$s - r = 1$$ instead, in other words, if $$r$$ is a quadratic residue how often is $$r + 1$$ a quadratic nonresidue?

We see for $$r = 2, 4, 9, 13$$ that $$r+1$$ is not a quadratic residue, so $$c = 4$$, hence $$x_1 x_2 = -4$$.

## A Loose End

Before we argued that given

$a_0 + a_1 \zeta^{3^1} + ... + a_{16} \zeta^{3^{16}} = a_0 + a_1 \zeta^{3^2} + ... + a_{16} \zeta^{3^{17}}$

we can equate coefficients of the powers of $$\zeta$$. Let us see why. In the above equation, we can move all terms to one side then divide through by $$\zeta$$ to find $$b_i$$ such that

$b_1 + b_2 \zeta + ... + b_{16} \zeta^{15} = 0$

(each $$b_i$$ is the difference between some $$a_j$$ and $$a_k$$). Then consider the polynomial $$g(x) = b_1 + b_2 x + ... + b_{16} x^{15}$$. Now $$\zeta$$ is a root of $$g$$ as well as the polynomial

$f(x) = 1 + x + ... + x^{16} .$

Hence $$\zeta$$ must also be a root of $$d = \gcd(f, g)$$. If $$g \ne 0$$, then the polynomial $$d$$ is a nonzero polynomial dividing $$f$$ with degree at most $$g$$, which is smaller than the degree of $$f$$. This is a contradiction since $$f$$ has no factors with rational coefficients (by Eisenstein).

Thus $$g$$ must be the zero polynomial, and we may equate the coefficients of the powers of $$\zeta$$.

[In abstract algebra, we say all this in one line: $$\mathbb{Q}[\zeta]$$ is a degree 17 extension of $$\mathbb{Q}$$ thus $$g = 0$$.]

Ben Lynn blynn@cs.stanford.edu 💡