Worked solution: The Gauss–Wantzel theorem: constructible regular polygons via cyclotomic fields

Frequently asked questions

Why split the roots using powers of

3

specifically?

Because $3$ happens to be a primitive root modulo $17$ : its powers cycle through every nonzero residue exactly once, giving a single natural cyclic ordering of the $16$ roots. Splitting a cyclic list into alternating halves, quarters, and so on is what makes each stage's sum and product come out as clean rational numbers — any other ordering would not have this property.

Source

Carl Friedrich Gauss (sufficiency, 1796/1801); Pierre Laurent Wantzel (necessity, 1837), 1837

Disquisitiones Arithmeticae

Step 2 of 8: Gauss's trick: grouping

16

roots into two periods

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In plain words

Gauss's trick is to not solve for each of the $16$ nontrivial roots individually, but first lump them into two big groups of $8$ , called periods, by picking every other term from a cleverly ordered list. Remarkably, the sum and the product of these two groups both turn out to be ordinary whole numbers, so the two periods are simply the two roots of an everyday quadratic equation with integer coefficients — solvable with a single square root, exactly the kind of step a compass can perform.

This is the heart of the whole construction: turn one hard degree- $16$ problem into a short chain of easy degree- $2$ problems.

\eta_{1,0}+\eta_{1,1}=-1,\quad \eta_{1,0}\eta_{1,1}=-4 \implies \eta_{1,0},\eta_{1,1} \text{ are roots of } t^2+t-4=0

Detailed analysis

Since $3$ is a primitive root modulo $17$ (its powers $3^0,3^1,\ldots,3^{15}$ run through every nonzero residue mod $17$ exactly once), the $16$ nontrivial roots of unity can be listed in the cyclic order $\zeta^{3^0},\zeta^{3^1},\ldots,\zeta^{3^{15}}$ , where $\zeta=\zeta_{17}$ . Gauss defined the two-term-index periods $\eta_{1,0}=\sum_{j\text{ even}}\zeta^{3^j}$ and $\eta_{1,1}=\sum_{j\text{ odd}}\zeta^{3^j}$ , each a sum of $8$ roots of unity.

Because the cyclotomic equation gives $1+\zeta+\zeta^2+\cdots+\zeta^{16}=0$ , all $16$ nontrivial roots sum to $-1$ , so $\eta_{1,0}+\eta_{1,1}=-1$ . A direct (if tedious) computation using the multiplicative structure of the exponents shows $\eta_{1,0}\eta_{1,1}=-4$ . Knowing both the sum and the product of two numbers pins them down as the two roots of $t^2-(\eta_{1,0}+\eta_{1,1})t+\eta_{1,0}\eta_{1,1}=0$ , i.e. $t^2+t-4=0$ — an ordinary quadratic with rational (in fact integer) coefficients, solvable by the quadratic formula, which needs only $\sqrt{17}$ .

So the field generated by $\eta_{1,0}$ has degree $2$ over $\mathbb{Q}$ : the first rung of the ladder Gauss needs to climb from $\mathbb{Q}$ up to $\mathbb{Q}(\zeta_{17})$ using only square roots.

Terms in this step

Primitive root modulo $p$: An integer $g$ whose powers $g^0, g^1, \ldots, g^{p-2}$ run through every nonzero residue modulo a prime $p$ exactly once; for $p=17$ , the number $3$ works.
Gaussian period: A sum of a subset of roots of unity, chosen using the cyclic structure given by a primitive root, so that periods of a given size satisfy a polynomial equation of predictable low degree over the previous stage.

Knowledge used in this step

Prime numbers