Question

using the fact that : A positive integer n ≥ 3 is constructive if it is possible to construct a regular n-gon by straightedge and compass, it is possible to construct the angle 2π/n. And that if both angles α and β can be constructed by straightedge and compass then so are their sums and differences.The outside angle of a regular n-gon is 2π/n.

1. Suppose that n = p^(α1) ··· p^(αk) where p ,··· , pk are distinct odd primes. Prove that n is constructive if and only if each pi^(αi) is constructive.

2. Prove that 2^α is constructive for any positive integer α ≥ 2

Answer 1

1.

n=p1^(a1)...pk^(ak) where p1,...,pk are distinct odd primes.

Let i be any of 1,2,...,k.

Then, n=(p1^(a1)...pi-1^(a(i-1))pi+1^(a(i+1))...pk^(ak))pi^(ai)=xpi^(ai), where x is a positive integer.

I shall denote $$\pi$$ by (pi).

Then, 2(pi)/n=2(pi)/(xpi^(ai))

=> 2(pi)/(pi^(ai))=2x(pi)/n.

Let us suppose that n is constructive. Then, it is possible to construct the angle 2(pi)/n by straightedge and compass.

Now, it has been said that sums of constructible angles are also constructible, then positive integer multiples of constructible angles are also constructible. Then,

2(pi)/n is constructible

=>  x(2(pi)/n) is constructible

=> 2x(pi)/n is constructible

=>  2(pi)/(pi^(ai)) is constructible

=> pi^(ai) is constructive. This is true for all i=1,2,3,...,k

So, n is constructive => pi^(ai) is constructive -------(A)

Now, the Gauss-Wantzel theorem states that:

"A regular n-gon is constructible with straightedge and compass if and only if n = 2kp1p2...pm where k and m are non-negative integers, and the pi's (when m > 0) are distinct Fermat primes (prime numers of the form $2^{2n}+1$ for non-negative integers n ".

Since n is constructive iff regular n-gon is constructible with straightedge and compass, then n is constructive if and only if n = 2kp1p2...pm where k and m are non-negative integers, and the pi's (when m > 0) are distinct Fermat primes. Then, if each piai is constructive,

each piai is a product of 2kq1q2...qm, where k and m are non-negative integers, and the qi's (when m> 0) are distinct Fermat primes

=> p1^(a1)...pk^(ak) is a product of 2kq1q2...qm, where k and m are non-negative integers, and the qi's (when m> 0) are distinct Fermat primes

=> n is a product of 2kq1q2...qm, where k and m are non-negative integers, and the qi's (when m> 0) are distinct Fermat primes

=> n is constructive.

So, each pi^(ai) is constructive => n is constructive --------(B).

From (A) and (B), our proof is complete.

2.

Now, we know that we can bisect any angle using just a compass. Since 2(pi) is constructible,

=> 2(pi)/2^1 is constructible

=> 2(pi)/2^2 is constructible

=> 2(pi)/2^3 is constructible

.

.

.

=> 2(pi)/2^a is constructible, since we can keep on bisecting an angle further and further. For a formal proof, simply apply induction on 'a'

[2(pi)/2^a is constructible=> 2(pi)/2^(a+1) is constructible].

Then, 2^a is constructive.

So, 2^a is constructive for any positive integer a>=2.