Part I, Chapter III — The Division of the Circle into Equal Parts¶
Part I, Chapter III (§§ 1–8; Ginn & Co., 1897, pp. 16–23)
§ 1. History of the Problem¶
1. The problem of dividing a given circle into $n$ equal parts has come down from antiquity; for more than two millennia mathematicians knew the possibility of solving it with straightedge and compasses only when $n = 2^h$, $3$, $5$, or the product of any two or three of these numbers (Euclid, Elements, Book IV).
In his Disquisitiones Arithmeticae (1801), Gauss extended this series of numbers at a single stroke by showing that the division into $p$ equal parts is possible for every prime number of the form
and impossible for all other odd prime numbers and their powers.
If in $p = 2^{2^\mu} + 1$ we set $\mu = 0$ and $\mu = 1$, we obtain $p = 3$ and $p = 5$, cases already known to the ancients.
For $\mu = 2$ we obtain $p = 2^{2^2} + 1 = 17$, the celebrated regular heptadecagon completely discussed by Gauss (discovered March 30, 1796).
For $\mu = 3$ we obtain $p = 2^{2^3} + 1 = 257$, likewise a Fermat prime; the regular polygon of $257$ sides is constructible, and Richelot published an extended investigation of it in Crelle’s Journal, Vol. IX (1832, pp. 1–26, 146–161, 209–230, 337–357), entitled De resolutione algebraica aequationis $X^{257} = 1$, sive de divisione circuli per bisectionem anguli septies repetitam in partes $257$ inter se aequales commentatio coronata.
Similarly for $\mu = 4$, since $2^{2^4} + 1 = 65537$ is a prime number. To the regular polygon of $65537$ sides Professor Hermes of Lingen devoted ten years of his life, examining with care all the roots furnished by Gauss’s method; his manuscripts are preserved in the collection of the mathematical seminary at Göttingen (compare a communication of Professor Hermes in No. 3 of the Göttinger Nachrichten for 1894).
On the other hand, $\mu = 5$, $\mu = 6$, and $\mu = 7$ yield no prime numbers (for $\mu = 5$, Euler showed $2^{32} + 1 = 641 \times 6700417$). For $\mu = 8$ no one has yet determined whether $2^{256} + 1$ is prime or composite. The proof that the huge numbers corresponding to $\mu = 5, 6, 7$ are composite has required a great expenditure of labor and ingenuity; it is therefore quite possible that $\mu = 4$ is the last exponent for which a prime solution exists.
§§ 2–4. Reduction to Prime Powers and Gauss’s Prime Numbers¶
2. We may immediately restrict the problem of dividing the circle into $n$ equal parts to the cases where $n$ is a prime number $p$ or a prime power $p^\alpha$. For if $n = \mu\nu$ is composite with $\mu$ and $\nu$ relatively prime ($\gcd(\mu, \nu) = 1$), Euclidean division furnishes integers $a$ and $b$ (one positive, one negative) such that
To divide the circle into $\mu\nu = n$ equal parts, it therefore suffices to superpose $a$ parts of the $\nu$-division with $b$ parts of the $\mu$-division. Thus for $n = 15 = 3 \cdot 5$, we have
which is precisely the arithmetic content of Euclid's construction of the regular pentadecagon (Elements IV.16).
3. As will appear in § 5, division into $p$ equal parts ($p$ prime) is possible only when $p = 2^h + 1$. Let us first prove that a prime number can have the form $p = 2^h + 1$ only when the exponent $h$ is itself a power of $2$, $h = 2^\mu$.
First proof (by Fermat’s Little Theorem): If $p$ is prime and $a$ is an integer not divisible by $p$, then $a^{p-1} \equiv +1 \pmod p$. If $s$ is the least positive exponent satisfying $a^s \equiv +1 \pmod p$, then $s$ is necessarily a divisor of $p - 1$. (In particular, when $s = p - 1$, $a$ is called a primitive root modulo $p$; every prime $p$ possesses primitive roots, a fact we shall use in Chapter IV.) Suppose now that $p = 2^h + 1$ is prime, and let $s$ be the least positive exponent such that
Since $2^h = p - 1 < p < 2^s$, we have $s > h$. Moreover, $p = 2^h + 1$ gives
Dividing $2^s \equiv +1$ by $2^h \equiv -1$ gives $2^{s-h} \equiv -1 \pmod p$, whence $s - h \ge h$, i.e., $s \ge 2h$. Conversely, squaring $2^h \equiv -1 \pmod p$ yields $2^{2h} \equiv +1 \pmod p$, so by minimality of $s$ we have $s \le 2h$, and consequently $s = 2h$. Now $s = 2h$ must divide $p - 1 = 2^h$; therefore $h$ divides $2^{h-1}$ and must itself be a power of $2$, $h = 2^\mu$. Hence primes of the form $2^h + 1$ are necessarily of the form $2^{2^\mu} + 1$.
4. Second proof (by elementary algebraic factorization): Suppose $h$ were divisible by an odd integer $2n + 1 > 1$, so that $h = h'(2n + 1)$. By the identity
evaluated at $x = 2^{h'}$, the number $p = 2^{h'(2n+1)} + 1$ is divisible by the proper factor $2^{h'} + 1 > 1$, and therefore cannot be prime!
§ 5. The Cyclotomic Equation¶
5. We now reach our fundamental proposition: For an odd prime $p$, dividing the circle into $p$ equal parts with straightedge and compasses is impossible unless $p = 2^h + 1 = 2^{2^\mu} + 1$.
Trace in the complex $z$-plane ($z = x + iy$) the circle of radius $1$ centered at the origin. Dividing this circle into $n$ equal parts starting at $z = 1$ is algebraically equivalent to solving $z^n - 1 = 0$. Suppressing the known initial vertex $z = 1$ by dividing out $z - 1$, we obtain the cyclotomic equation:
When $n = p$ is prime, the essential point of the proof is to show that this equation of degree $p - 1$ is irreducible over $\mathbb{Q}$. Once irreducibility is proved, Chapter I, § 15 requires its degree $p - 1$ to be a power of $2$, forcing $p = 2^h + 1 = 2^{2^\mu} + 1$ and explaining why Gauss's primes occupy their unique position.
§ 6. Gauss’s Lemma¶
6. To establish irreducibility, we first prove Gauss’s Lemma: If a monic polynomial with integer coefficients,
can be resolved into two monic factors $f(z)$ and $\varphi(z)$ with rational coefficients,
then all the coefficients $\alpha_i$ and $\beta_j$ must actually be integers.
Proof. Suppose the $\alpha$’s or $\beta$’s were fractional. Reduce the coefficients of $f(z)$ to their least common positive denominator $a_0$, and those of $\varphi(z)$ to their least common positive denominator $b_0$. Multiplying both sides by $a_0 b_0$ yields
where $a_0, a_1, \dots$ are coprime integers ($\gcd(a_0, a_1, \dots) = 1$) and likewise $b_0, b_1, \dots$ are coprime integers. If $a_0 b_0 > 1$, let $q$ be any prime divisor of $a_0 b_0$. Because neither all $a$’s nor all $b$’s are divisible by $q$, let $a_i$ be the first coefficient in $f_1(z)$ not divisible by $q$, and $b_k$ the first coefficient in $\varphi_1(z)$ not divisible by $q$. In the expanded product $f_1(z)\varphi_1(z)$, the coefficient of $z^{m' + m'' - i - k}$ equals
By construction, every term after $a_i b_k$ contains either an $a_{i-r}$ ($r \ge 1$) or a $b_{k-s}$ ($s \ge 1$) divisible by $q$, whereas $a_i b_k$ is not divisible by the prime $q$. Yet on the left-hand side $a_0 b_0 F(z)$, every coefficient is divisible by $a_0 b_0$ and hence by $q$—a contradiction! Therefore $a_0 b_0 = 1$, i.e., $a_0 = b_0 = 1$, and all $\alpha_i, \beta_j$ are integers.
§§ 7–8. Irreducibility of the Cyclotomic Equation for Primes and Prime Squares¶
7. By Gauss’s Lemma, to prove $f(z) = z^{p-1} + z^{p-2} + \dots + z + 1$ irreducible for any prime $p$, we need only show it admits no factorization with integer coefficients. Following Eisenstein's remarkably simple method (Crelle's Journal, Vol. XXXIX, 1850, p. 167), substitute $z = x + 1$:
Every coefficient after the leading term $x^{p-1}$ is divisible by the prime $p$, and the constant term is $p$ itself (not divisible by $p^2$). Any polynomial of this Eisenstein form is irreducible. For if it factored over $\mathbb{Z}$ as
the constant term $a_m b_{m'} = p$ would force one factor—say $a_m = \pm p$—to be divisible by $p$ and the other to be $b_{m'} = \pm 1$. Examining the coefficient of $x$, namely $a_{m-1} b_{m'} + a_m b_{m'-1} \equiv 0 \pmod p$, gives $p \mid a_{m-1}$ since $b_{m'} = \pm 1$; examining $x^2, x^3, \dots$ successively forces $p \mid a_{m-2}, \dots, p \mid a_1$, and finally $p \mid 1$ (the leading coefficient of $x^m$), which is absurd! Thus $\Phi_p(z) = 0$ is irreducible of degree $p - 1$. In particular, for $p = 7$ ($\deg = 6 \neq 2^h$), $p = 11$ ($\deg = 10$), and $p = 13$ ($\deg = 12$), the regular $7$-gon, $11$-gon, and $13$-gon cannot be constructed with straightedge and compasses.
8. Finally, consider a prime power $n = p^\alpha$ ($\alpha \ge 2$) for an odd prime $p > 2$. Since dividing the circle into $p^\alpha$ equal parts automatically divides it into $p^2$ equal parts, it suffices to prove that division into $p^2$ equal parts is impossible whenever $p > 2$. Suppressing from $z^{p^2} - 1 = 0$ all roots belonging to the $p$-division ($z^p - 1 = 0$) leaves the primitive $p^2$-cyclotomic equation:
Substituting $z = x + 1$ across these $p$ summands produces
where $\chi(x)$ is an integer polynomial with constant term $\chi(0) = 1$ (since $(x + 1)^{kp} \equiv x^{kp} + 1 \pmod p$ for every $k$). By § 7, this polynomial is irreducible over $\mathbb{Q}$, and its degree is $p(p - 1)$. Since $p > 2$ is an odd prime, $p(p - 1)$ contains the odd factor $p$ and can never be a power of $2$! Thus even for Gauss’s primes ($p = 3 \implies p^2 = 9$ of degree $6$; $p = 5 \implies p^2 = 25$ of degree $20$), division into $p^2$ or $p^\alpha$ ($\alpha \ge 2$) parts is impossible; only $p = 2$ admits repeated powers $2^\alpha$.