## 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 $g = 0$.]