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

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.

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 $17=1(mod4)$ 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

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 $x_1{x}_2=-4$.

We can generalize these statements.

Before we argued that given

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

(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

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 $g=0$.]