lpetrich
Contributor
But there are plenty special cases of higher-degree equations that can be solved by Euclidean techniques.
Some of them occur in constructions of regular n-gons. The size of the angle at the center for each side is (360d)/n, and one has this equation for its cosine:
T(n,x) = 1
For a triangle, one must solve
4x^3 - 3x = 1
This has a trivial solution, x = 1, so we can factor it out, giving
(2x + 1)^2 * (x - 1) = 0
So we find that cos(120d) = -1/2
Implying that cos(60d) = 1/2
A square is easy:
8x^4 - 8x^2 + 1 = 1
One can treat it as a series of two bisections:
2*(2*x^2 - 1)^2 - 1 = 1
One gets x = 1, x = -1 (180d), and two of x = 0 (90d)
A pentagon looks hard, since one has to solve a quintic:
16x^5 - 20x^3 + 5x = 1
But it also reduces to a much easier equation:
(4x^2 + 2x - 1)^2 * (x - 1) = 0
We find from it
cos(72d) = (sqrt(5) - 1)/4
cos(144d) = (-sqrt(5) - 1)/4
Filling in,
cos(36d) = (sqrt(5) + 1)/4
cos(108d) = (-sqrt(5) + 1)/4
Around 1800, Carl Friedrich Gauss, a prolific mathematician, discovered a solution for the 17-gon that involves a sequence of 3 quadratic equations, meaning that it can be constructed with Euclidean techniques. Likewise can find a solution for the 257-gon with 7 quadratic equations, and one for the 65537-gon with 15 quadratic equations.
Likewise, with a Pierpont prime (2^u*3^v+1), one can reduce the problem to solving a sequence of quadratic and cubic equations.
Thus, for a regular heptagon (7-gon), one must solve
(8x^3 + 4x^2 - 4x - 1)^2 * (x - 1) = 0
So one needs to solve a cubic equation.
Some of them occur in constructions of regular n-gons. The size of the angle at the center for each side is (360d)/n, and one has this equation for its cosine:
T(n,x) = 1
For a triangle, one must solve
4x^3 - 3x = 1
This has a trivial solution, x = 1, so we can factor it out, giving
(2x + 1)^2 * (x - 1) = 0
So we find that cos(120d) = -1/2
Implying that cos(60d) = 1/2
A square is easy:
8x^4 - 8x^2 + 1 = 1
One can treat it as a series of two bisections:
2*(2*x^2 - 1)^2 - 1 = 1
One gets x = 1, x = -1 (180d), and two of x = 0 (90d)
A pentagon looks hard, since one has to solve a quintic:
16x^5 - 20x^3 + 5x = 1
But it also reduces to a much easier equation:
(4x^2 + 2x - 1)^2 * (x - 1) = 0
We find from it
cos(72d) = (sqrt(5) - 1)/4
cos(144d) = (-sqrt(5) - 1)/4
Filling in,
cos(36d) = (sqrt(5) + 1)/4
cos(108d) = (-sqrt(5) + 1)/4
Around 1800, Carl Friedrich Gauss, a prolific mathematician, discovered a solution for the 17-gon that involves a sequence of 3 quadratic equations, meaning that it can be constructed with Euclidean techniques. Likewise can find a solution for the 257-gon with 7 quadratic equations, and one for the 65537-gon with 15 quadratic equations.
Likewise, with a Pierpont prime (2^u*3^v+1), one can reduce the problem to solving a sequence of quadratic and cubic equations.
Thus, for a regular heptagon (7-gon), one must solve
(8x^3 + 4x^2 - 4x - 1)^2 * (x - 1) = 0
So one needs to solve a cubic equation.