Part I, Chapter IV — The Construction of the Regular Polygon of 17 Sides¶
Part I, Chapter IV (§§ 1–9; Ginn & Co., 1897, pp. 24–41)
§§ 1–3. Primitive Roots Modulo 17 and Gauss’s Periods¶
1. Having seen in Chapter III that division of the circle into an odd prime number of equal parts by straightedge and compasses is possible only for Gauss's Fermat primes, we shall now show in an elementary way how to actually inscribe the regular polygon of $17$ sides in the unit circle.
Since we possess as yet no construction of the regular $17$-gon derived from purely geometric considerations, we follow the algebraic path indicated by our general theory: we investigate the $16$ roots of the cyclotomic equation
and translate the resulting nested square-root expression into a geometric construction. In transcendental form the roots are
with $\varepsilon_k = \varepsilon_1^k$, represented geometrically in the complex plane by the $16$ vertices different from $1$ of the regular $17$-gon inscribed in the unit circle centered at the origin. Once $\varepsilon_1$ is chosen as initial vertex, the angle belonging to $\varepsilon_k$ is $k$ times that of $\varepsilon_1$.
2. The fundamental idea of Gauss's solution (Disquisitiones Arithmeticae, 1801; see also Bachmann, Die Lehre von der Kreisteilung, Leipzig, 1872) is to arrange the $16$ roots into a single cycle by means of a primitive root modulo $17$—an integer $a$ for which the least positive exponent satisfying $a^s \equiv +1 \pmod{17}$ is $s = 17 - 1 = 16$. The number $a = 3$ possesses this property:
Ordering the roots $\varepsilon_k$ by these remainders gives the cycle
in which each root is the cube of its predecessor ($\varepsilon_{r'} = \varepsilon_r^3$). Gauss decomposes this $16$-cycle into sums containing $8$, $4$, $2$, and $1$ roots—corresponding to the divisors of $16$—called Gaussian periods, which can be calculated successively as roots of quadratic equations.
3. Replacing each root $\varepsilon_k$ symbolically by its exponent $k \pmod{17}$ and alternating by steps of $2$, we form:
Two $8$-periods ($x_1$ from even positions in the cycle, $x_2$ from odd positions):
\[\begin{split}\begin{aligned} x_1 &= 9 + 13 + 15 + 16 + 8 + 4 + 2 + 1, \\ x_2 &= 3 + 10 + 5 + 11 + 14 + 7 + 12 + 6. \end{aligned}\end{split}\]Four $4$-periods (alternating inside $x_1$ and $x_2$):
\[\begin{split}\begin{alignedat}{2} y_1 &= 13 + 16 + 4 + 1, &\qquad y_2 &= 9 + 15 + 8 + 2, \\ y_3 &= 10 + 11 + 7 + 6, &\qquad y_4 &= 3 + 5 + 14 + 12. \end{alignedat}\end{split}\]Eight $2$-periods (alternating inside $y_1, y_2, y_3, y_4$):
\[\begin{split}\begin{alignedat}{4} z_1 &= 16 + 1, &\quad z_2 &= 13 + 4, &\quad z_3 &= 15 + 2, &\quad z_4 &= 9 + 8, \\ z_5 &= 11 + 6, &\quad z_6 &= 10 + 7, &\quad z_7 &= 5 + 12, &\quad z_8 &= 3 + 14. \end{alignedat}\end{split}\]
§§ 4–5. Reality and Relative Magnitude of the Periods¶
4. In every $2$-period $z_m$, the two exponents sum to $17$, pairing $\varepsilon_r$ with its complex conjugate $\varepsilon_{17-r} = \overline{\varepsilon_r}$. Hence all the periods $z_m$, $y_j$, $x_i$ are real numbers:
with $y_1 = z_1 + z_2$, $y_2 = z_3 + z_4$, $y_3 = z_5 + z_6$, $y_4 = z_7 + z_8$, $x_1 = y_1 + y_2$, and $x_2 = y_3 + y_4$.
5. To determine unambiguously which sign belongs to each square root in our quadratic equations, we establish the relative magnitudes of the periods by an elementary geometric artifice. Divide the upper semicircle of unit radius into $17$ equal arcs at $A_1, A_2, \dots, A_{17}$ starting from $O_{\text{arc}} = A_0$, and denote the chord lengths $A_0 A_k$ by $S_1, S_2, \dots, S_{17}$ (with diameter $S_{17} = 2$):
Matching $2\cos\frac{2h\pi}{17} = 2\cos\frac{4h\pi}{34}$ against $S_k$ for $h = 1, 2, \dots, 8$ gives $k = 17 - 4h$, i.e.:
Since $0 < S_1 < S_2 < \dots < S_{17} = 2$, we read off immediately
Furthermore, in $\triangle A_0 A_k A_{k+p}$ the triangle inequality gives $S_{k+p} < S_k + S_p$. Forming pairwise differences of $y_1, y_2, y_3, y_4$ yields:
Hence the four $4$-periods obey the strict inequality chain
§ 6. The Tower of Four Quadratic Equations and Normal Radical Form¶
6. Using the symbolic product rule $\varepsilon_k \cdot \varepsilon_p = \varepsilon_{k+p \pmod{17}}$ and $\sum_{k=1}^{16} \varepsilon_k = -1$, we compute the elementary symmetric functions at each level:
First quadratic $(\xi)$ for $x_1, x_2$: $x_1 + x_2 = -1$, and in the $64$-term expansion of $x_1 x_2$ every primitive root $\varepsilon_1, \dots, \varepsilon_{16}$ occurs exactly $4$ times, so $x_1 x_2 = -4$. Thus $x_1 > x_2$ solve
\[(\xi) \qquad x^2 + x - 4 = 0 \implies x_1 = \frac{-1 + \sqrt{17}}{2}, \quad x_2 = \frac{-1 - \sqrt{17}}{2}.\]Second quadratic $(\eta)$ for $y_1, y_2$: $y_1 + y_2 = x_1$, and expanding $y_1 y_2 = (13 + 16 + 4 + 1)(9 + 15 + 8 + 2)$ yields each of the $16$ roots once, so $y_1 y_2 = -1$. Since $y_1 > y_2$:
\[(\eta) \qquad y^2 - x_1 y - 1 = 0 \implies y_1 = \frac{x_1 + \sqrt{x_1^2 + 4}}{2}, \quad y_2 = \frac{x_1 - \sqrt{x_1^2 + 4}}{2}.\]Third quadratic $(\eta')$ for $y_4, y_3$: Similarly $y_3 + y_4 = x_2$ and $y_3 y_4 = -1$. Since $y_4 > y_3$:
\[(\eta') \qquad y^2 - x_2 y - 1 = 0 \implies y_4 = \frac{x_2 + \sqrt{x_2^2 + 4}}{2}, \quad y_3 = \frac{x_2 - \sqrt{x_2^2 + 4}}{2}.\]Fourth quadratic $(\zeta)$ for $z_1, z_2$: $z_1 + z_2 = y_1$, and $z_1 z_2 = (16 + 1)(13 + 4) = 12 + 3 + 14 + 5 = y_4$. Since $z_1 > z_2$:
\[(\zeta) \qquad z^2 - y_1 z + y_4 = 0 \implies z_1 = \frac{y_1 + \sqrt{y_1^2 - 4y_4}}{2}, \quad z_2 = \frac{y_1 - \sqrt{y_1^2 - 4y_4}}{2}.\]
Thus $z_1 = 2\cos\frac{2\pi}{17}$ appears to involve four square roots: $\sqrt{17}$, $\sqrt{x_1^2 + 4}$, $\sqrt{x_2^2 + 4}$, and $\sqrt{y_1^2 - 4y_4}$. Are these four radicals algebraically independent? Expanding symbolically $(y_1 - y_2)(y_4 - y_3)$ across all $64$ signed terms gives $2(x_1 - x_2)$, i.e.:
Hence $\sqrt{x_2^2 + 4}$ is rationally expressible in terms of $\sqrt{17}$ and $\sqrt{x_1^2 + 4}$ (confirming that $z_1$ satisfies an irreducible equation of degree $8 = 2^3$, not $16$). Substituting the numerical values yields the explicit nested radical chain:
§ 7. Historical Excursion on Restricted Instruments (Mascheroni, Brianchon, Poncelet, Steiner)¶
7. Before translating these four quadratic equations into geometry, we insert a brief historical excursion on constructions with straightedge and compasses:
In ancient geometry, straightedge and compasses were always employed side by side without distinguishing whether individual steps used one instrument or the other; the geometer’s art lay simply in arranging the parts of a figure so as to avoid superfluous lines.
In 1797, the Italian geometer Lorenzo Mascheroni succeeded in executing all Euclidean constructions with the compasses alone in his Geometria del compasso, arguing that for instrument-makers and mechanics divider-arcs are practically more accurate than ruled lines. (Because Mascheroni’s original treatise is prolix, Hutt provided a concise German summary, Die Mascheroni’schen Constructionen, Halle, 1880.)
Conversely, in France the disciples of Carnot (Géométrie de position) strove to accomplish as much as possible with the straightedge alone (compare also Lambert, Freie Perspective, 1774, and Brianchon, Les applications de la théorie des transversales, 1818, motivated by field surveying). Algebra answers immediately what pure straightedge lines can achieve: with the straightedge alone we can construct precisely those algebraic expressions whose form is rational.
Finally, Poncelet first conceived in his Traité des propriétés projectives (1822, Vol. I, Nos. 351–357) the fundamental theorem—developed systematically by Steiner in his celebrated monograph Die geometrischen Constructionen, ausgeführt mittels der geraden Linie und eines festen Kreises (1833)—that a single fixed circle with given center, together with the straightedge alone, suffices to construct every algebraic expression built from rational operations and square roots (the Poncelet–Steiner theorem).
§§ 8–9. The von Staudt–Schröter Single-Circle Construction of the 17-Gon¶
8. To construct the regular $17$-gon with a single fixed circle and straightedge alone, we follow the elegant projective method of Karl Georg Christian von Staudt (Crelle's Journal, Vol. XXIV, 1842), as modified by Heinrich Schröter (Crelle's Journal, Vol. LXXV, 1872). (A Mascheroni compass-only construction of the regular $17$-gon was subsequently published by L. Gérard in Mathematische Annalen, Vol. XLVIII, 1896, pp. 390–392.)
Let the fixed unit circle have diameter $OA$ along the $Y$-axis ($O = (0, 0)$, center $C_0 = (0, 1)$, upper pole $A = (0, 2)$), with lower horizontal tangent $y = 0$ (the $X$-axis) and upper horizontal tangent $y = 2$:
To solve any quadratic equation $x^2 - px + q = 0$ having real roots $x_1, x_2$:
Lay off on the upper tangent $y = 2$ (from $A$) the abscissa $\frac{4}{p}$, and on the lower tangent $y = 0$ (from $O$) the abscissa $\frac{q}{p}$.
Connect $\left(\frac{4}{p}, 2\right)$ and $\left(\frac{q}{p}, 0\right)$ by the transversal line 3, meeting the fixed circle in two points.
Project those two circle intersections from the upper pole $A(0, 2)$ by lines 1 and 2 onto the $X$-axis $y = 0$: the intercepts cut off on the $X$-axis are precisely $x_1$ and $x_2$!
Proof. The projecting rays $1$ and $2$ from $A(0, 2)$ to $(x_1, 0)$ and $(x_2, 0)$ have equations $2x + x_1(y - 2) = 0$ and $2x + x_2(y - 2) = 0$. Multiplying them gives the degenerate conic
Subtracting the circle $x^2 + y(y - 2) = 0$ produces another conic through the four common points of the line-pair and the circle; factoring out the double-point tangent $y - 2 = 0$ at $A$ leaves the secant line 3:
Setting $y = 2$ gives $x = \frac{4}{p}$, and setting $y = 0$ gives $x = \frac{q}{p}$. $\blacksquare$
9. We now apply this single-circle solver successively to $(\xi)$, $(\eta)$, $(\eta')$, and $(\zeta)$, needing only $x_1, x_2, y_1, y_4$, and $z_1$. On the upper tangent $y = 2$ and lower tangent $y = 0$ we require the successive abscissas:
Step 1 — Solving $(\xi): x^2 + x - 4 = 0$ ($p = -1, q = -4$): Connect $(+4, 0)$ below with $(-4, 2)$ above (which passes directly through the circle center $(0, 1)$). Projecting its two intersections with the fixed circle from $A(0, 2)$ cuts off $x_1 > 0$ and $x_2 < 0$ on the $X$-axis.
Projective Inversion $x' = \frac{4}{x}$ (Fig. 5): For any abscissa $x$ on $y = 0$, the ray from $A(0, 2)$ to $(x, 0)$ meets the circle again at $\left(\frac{4x}{x^2 + 4}, \frac{2x^2}{x^2 + 4}\right)$, and projecting that circle point from $O(0, 0)$ onto the upper tangent $y = 2$ cuts off $x' = \frac{4}{x}$! Projectively, $x \mapsto x'$ is a perspective one-to-one correspondence sending $x = \infty \leftrightarrow x' = 0$ and fixing $x = x' = 2$, whence $x' = \frac{4}{x}$. Thus projecting the circle points already belonging to $x_1$ and $x_2$ from $O(0, 0)$ onto $y = 2$ immediately yields $\frac{4}{x_1}$ and $\frac{4}{x_2}$!
Step 2 — Solving $(\eta): y^2 - x_1 y - 1 = 0$ (Fig. 6): Join $(-4, 2)$ above to $(+1, 0)$ below ($5y + 2x = 2$), cutting the vertical diameter $OA$ at the fixed division point $H = \left(0, \frac{2}{5}\right)$. Joining $\left(\frac{4}{x_1}, 2\right)$ through $\left(0, \frac{2}{5}\right)$ ($5y - 2x_1 x = 2$) meets $y = 0$ at $-\frac{1}{x_1}$! Now connect $\left(-\frac{1}{x_1}, 0\right)$ to $\left(\frac{4}{x_1}, 2\right)$ and project the upper circle intersection from $A(0, 2)$ onto $y = 0$ to obtain $y_1$.
Step 3 — Solving $(\eta'): y^2 - x_2 y - 1 = 0$ (Fig. 7): Likewise join $\left(\frac{4}{x_2}, 2\right)$ through the same division point $\left(0, \frac{2}{5}\right)$ to cut off $-\frac{1}{x_2}$ on $y = 0$. Connect $\left(-\frac{1}{x_2}, 0\right)$ to $\left(\frac{4}{x_2}, 2\right)$ and project the positive-quadrant circle intersection from $A(0, 2)$ onto $y = 0$ to obtain $y_4$.
Step 4 — Solving $(\zeta): z^2 - y_1 z + y_4 = 0$ and Inscribing the $17$-Gon (Figs. 8–9): Project the circle intersection on ray $A y_1$ from $O(0, 0)$ onto $y = 2$ to get $\frac{4}{y_1}$. Join $(+4, 2)$ above to $(y_4, 0)$ below, meeting the vertical axis $x = 0$ at $\left(0, \frac{2y_4}{y_4 - 4}\right)$; then join that axial point to $\left(\frac{4}{y_1}, 2\right)$ ($2y_1 x + (y_4 - 4)y = 2y_4$) to cut off $\frac{y_4}{y_1}$ on $y = 0$! Connecting $\left(\frac{y_4}{y_1}, 0\right)$ to $\left(\frac{4}{y_1}, 2\right)$ and projecting the upper circle intersection from $A(0, 2)$ cuts off $z_1 = 2\cos\frac{2\pi}{17}$ on $y = 0$, and simultaneously cuts off $\cos\frac{2\pi}{17}$ directly on the horizontal diameter $y = 1$! Erecting the perpendicular chord at $x = \cos\frac{2\pi}{17}$ immediately marks the first and sixteenth vertices $V_1, V_{16}$ of the regular $17$-gon on the fixed circle!