Sheng Li (gmachine1729) wrote,
Sheng Li
gmachine1729

How to construct with straightedge and compass a regular 17-gon, as did Gauss in 1796 at age 19

Originally published at 狗和留美者不得入内. You can comment here or there.

Straightedge and compass constructions, algebraically speaking, enable one to construct points on the plane whose coordinates are a result of some combination of addition, subtraction, multiplication, division, and square root operator. This I won’t go into detail; it is not difficult to figure out this algebraic correspondence.

Knowing what straightedge and compass construction means algebraically tells us that we can construct a regular 17-gon if and only if the 17th root of unity [公式] which satisfies the equation [公式] can be constructed. Upon constructing that point, obviously we can draw a line between the origin and that point and then perform the same construction again based on that line, eventually covering all the points of the regular 17-gon.

Given that [公式] is a prime, Fermat’s little theorem tells us that [公式] for all integers [公式] . Moreover we note that because [公式] can have at most two square roots and that the multiplicative group [公式] is commutative or Abelian, the group must be cyclic, and in fact it is generated by [公式] . This tells us that we can cycle through all the roots of unity by iteratively taking the third power, starting from [公式] . Let us now define [公式] , noting that [公式] , where [公式] denotes multiplication and [公式] addition, and moreover that [公式] per the aformentioned cyclic multiplicative group.

By the factorization

[公式]

we have that

[公式] which means that if we can express the sum in the form [公式] such that the product [公式] corresponding to the [公式] th order coefficient of a quadratic, we are able to construct [公式] on the complex plane, which in their expression require no more than a square roots adjoined to the rationals. We can then proceed to in a similar manner two-partition [公式] in such a way that the two parts can be expressed by adjoining another square root, repeating the process four times until we obtain a root of unity.

The most symmetric way to two partition would be via

[公式] in which case, defining [公式] ,

[公式]

If across the [公式] elements in the sum wherein [公式] , each of the [公式] values of our multiplicative group is attained an equal number of times, then

[公式] and we are finished. So we proceed to prove that the number of solutions [公式] for

[公式] is the same for all [公式] . To do so, we first fix [公式] and assume some solution [公式] for it. We then notice that the action of the set [公式] , which is the set of roots of unity, on the right hand side gives the set of roots of unity, and moreover, that action on the left hand side gives

[公式] the two exponents of which correspond to a solution of

[公式] (Note how because the two exponents [公式] are shifted by the same value [公式] , the two resulting exponents are such that one is even and the other is odd.) Thus, we have shown that every solution [公式] for some fixed right hand side value corresponds to exactly one solution for every possible right hand side value. This means each right hand side value must admit the same number of solutions. Since every one of the [公式] terms in the product [公式] corresponds to a solution of one of the [公式] possible right hand side cases, the number of solutions for each case must be [公式] , which complete our proof.

The reader can verify himself that similar logic holds for [公式] , and then [公式] and finally [公式] . Each is such that the sum has been precendently verified to be constructible by adjoining a square root and that a split into two results in a product that is also constructible, thereby facilitating via a quadratic equation with constructible coefficients the adjoinment of another square root. It is for this reason along with the existence of a cyclic multiplicative group that constructibility can be generalized to a regular polygon the number of sides of which is a prime number also in the form of [公式] .

I know a math PhD who regards Gauss as a demigod. However, I had asked him if Gauss was so out of the league of all his contemporaries, why didn’t he discover Galois theory? Moreover, why is there no evidence that Gauss ever truly understood Galois theory, which perhaps nobody aside from Galois and Abel did throughout Gauss’s lifetime. Those who have learned Galois theory can easily recognize that in the above case that the Galois group of [公式] is [公式] , which is a cyclic group with three proper subgroups all part of the same composition series, the index of which at every step is [公式] , which by the fundamental theorem of Galois theory must correspond to the adjoinment of a square root for the associated field extension.

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