A Property of Cubic Equations

Wantzel's approach to solving the Angle trisection problem works with other two problems: Doubling the cube and Constructing a regular heptagon. All three, in algebraic terms, reduce to an algebraic equation of degree three. Cubic equations possess a pertinent property which constitutes the contents of a lemma below. After proving the lemma, I shall derive cubic equations for the three problems and show that they satisfy the conditions of the lemma.

Lemma

Consider a cubic polynomial equation with integer coefficients P(x) = a3x3 + a2x2 + a1x + a0 = 0. Then either one of its roots is rational or none of its roots is constructible.

Proof

Assume the equation P(x) = 0 has no rational roots. The lemma then asserts that the equation has no constructible roots. Assume on the contrary that the equation has a constructible root. As we know, every constructible number belongs to an extension field FN = Q[√m1,√m2, ...,√mN]. Among all fields that contain a constructible root of the equation choose the one with the least N. Note that, from the minimality of N, no root of the equation belongs to FN-1 = Q[√m1,√m2,...,√mN-1]. Let a = a + b√mN, where a, b, mN∈FN-1, satisfies P(a) = 0. By direct substitution, P(a) = A + B√mN, where A and B both belong to FN-1.

But P(a + b√mN) = A + B√mN implies P(a - b√mN) = A - B√mN. Therefore, if a + b√mN is a root of P(x) = 0, so is its conjugate a - b√mN.

For a polynomial equation P(x) = a3x3 + a2x2 + a1x + a0 = 0, its three roots add up to -a2/a3, a rational number. Let u denote the third root. Then, u + (a + b√mN) + (a - b√mN) = -a2/a3. This implies u = -a2/a3 - 2a∈FN-1. A contradiction with the minimality of N.

Doubling the cube

The volume of a cube with the unit side is 1. The problem is to construct a cube whose volume is 2. Algebraically, the question is reduced to finding a constructible solution to the equation

(1)

x3 = 2

According to Lemma, to show that the task is impossible, we only have to demonstrate that (1) has no rational solutions. Assume the opposite, and let x = p/q be a rational solution of (1). Then p3 = 2q3. The number of prime factors on the left side of the latter equation is divisible by 3. The number of prime factors on the right side of the equation, when divided by three, leaves a remainder of 1. Therefore, the equality is impossible indeed. A contradiction. The problem of doubling the cube is not solvable with ruler and compass.

(There is an interesting construction that seems to solve the Delian problem. See if you can find a flaw there.)

Trisecting an angle

To reduce the problem of trisecting an angle to an algebraic equation, let's first agree to consider only acute angles. Since we intend to show that the problem is not solvable in general, all we need is a single counterexample. Following Wantzel, I'll show that a 60° angle (which is, of course, acute) is not trisectable.

Another important observation is that the problem of constructing an acute angle is equivalent to that of constructing a right triangle with a given angle. By definition, the sides of a right triangle are expressed in terms of trigonometric functions of its angles. We are especially interested in the cosine. With the help of addition formulas for sine and cosine we get,

cos(3a) = cos(a)cos(2a) - sin(a)sin(2a)
= cos(a)(cos2(a) - sin2(a)) - 2sin2(a)cos(a)
= cos(a)(2cos2(a) - 1) - 2(1 - cos2(a))cos(a)
= 4cos3(a) - 3cos(a)

For a = 20°, cos(3a) = cos(60°) = 1/2 and the equation becomes 8cos3(a) - 6cos(a) - 1 = 0. Replacing cos(a) with x we finally get

(2)

8x3 - 6x - 1 = 0

The substitution v = 2x transforms (2) into

(3)

v3 - 3v - 1 = 0

By Lemma, we only have to show that this equation has no rational roots. Assume on the contrary that v = p/q is a rational solution to (3) in lowest terms. Then, p3 - 3pq2 - q3 = 0. First, rewrite this as p(p2 - 3q2) = q3. It follows that p|q3. If p were not a unity (i.e. ±1) then any prime factor of p would divide q. Since the fraction p/q is irreducible, p ought to be a unity. On the other hand, we may also write q(3pq + q2) = p3, which, by the same argument, shows that q|p, i.e. q = ±1.

It appears that the only rational solution equation (3) might have is either +1 or -1. By direct inspection, neither satisfies (3). By Lemma, (3) has no constructible solutions and neither does (2).

Constructing a regular heptagon

The simplest way to tackle this problem is with the help of complex numbers. Vertices of a regular heptagon divide the circumscribed circle into 7 equal parts corresponding. On the unit circle, the same property is shared by the 7 roots of the cyclotomic equation z7 = 1. The equation is obviously satisfied by z = 1. Factoring out the term (z - 1), we obtain the following equation

(4)

(z7 - 1)/(z - 1) = z6 + z5 + z4 + z3 + z2 + z + 1

Among the roots of this equation is z = cos(360°/7) + i·sin(360°/7) where i is the one of the square roots of -1. We show next that the angle 360°/7 is not constructible.

After the substitution x = (z + 1/z), this equation is reduced to the cubic one (please check)

(5)

x3 + x2 - 2x - 1 = 0

Were (5) to have a rational root x = p/q, we would have, on the one hand, p(p2 + pq - 2q2) = q3; and, on the other, q(q2 - p2 + 2pq) = p3. As before, we would then have p = ±1 and q = ±1 which is impossible as easily established by direct inspection. Therefore, (5) has no constructible roots.

If, nonetheless, (4) had a constructible root z, then x = z + 1/z would also be constructible which, as we just saw, is not possible. Therefore, (4) does not have constructible roots either.

Note: there is an alternative treatment of cubic polynomials that employs the notion of dimension of field expansions.

References

  1. R. Courant and H. Robbins, What is Mathematics?, Oxford University Press, NY, 1996.
  2. M. Lascovitch, Conjecture and Proof, MAA, 2001
  3. I. Stewart, Concepts of Modern Mathematics, Dover, 1995

Constructible Numbers, Geometric Construction, Gauss' and Galois' Theories

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