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 F_N = Q[√m₁,√m₂, ...,√m_N]. 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 F_N-1 = Q[√m₁,√m₂,...,√m_N-1]. Let a = a + b√m_N, where a, b, m_N∈F_N-1, satisfies P(a) = 0. By direct substitution, P(a) = A + B√m_N, where A and B both belong to F_N-1.

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

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√m_N) + (a - b√m_N) = -a₂/a₃. This implies u = -a₂/a₃ - 2a∈F_N-1. A contradiction with the minimality of N.

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

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

The substitution v = 2x transforms (2) into

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).

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

• Strange Integers, divisors and primes
  • Divisors and Primes
  • Integral Domains: Fundamental Theorem of Arithmetic
  • Gaussian Integers
  • Remarks and Examples
• Reduction: Constructible Numbers
  • Constructible Numbers: Examples
  • Four Problems of Antiquity
  • A Property of Cubic Equations: Three Problems Of Antiquity
  • Angle Trisection

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