Marcos
Posted on Sunday, 17 August, 2003 - 02:50 pm:
For which n Î N is is true that cos(2p/n) is expressible using only +, -, /, × and
  
Ö
 

(only square roots)?

Is there anything we can say about such an n? I've tried several approaches, none of which seem to lead to even an elimination of cases...

Thanks,
Marcos

Sarah Sarah
Posted on Monday, 18 August, 2003 - 12:42 am:
Hello, I've had a rather vague think about this and haven't really got anywhere. In case you haven't tried these, though, I'll mention them:

You can use the observation that if z=cost+isint then z+z-1=2cost and zn+z-n=2cosnt to express cos2p = 1 as a polynomial in cos(2p/n).

I don't see immediately how you would go on to find values of n for which the polynomial has roots of the required form.

One other thing that occurred to me was that there might be some sort of nifty algebraic manipulation you could do, multiplying things of the form (a+Öb) by (a-Öb) (or adding) so that you're left finding under what circumstances a polynomial has rational roots, which could be easier... again, a vague thought, but magic man Marcos might find a way.

Love Sarah

Demetres Christofides
Posted on Monday, 18 August, 2003 - 08:34 am:
This is quite difficult to show. It uses Galois theory (3rd year course in Cambridge). The answer is any number of the form
2n p1 ...pr where the pi are distinct Fermat primes. Note that this solves the problem of the constructibility of a reqular n-gon.

I may try to give some hints later. (It is easier to show that no other number will do than to show that the above numbers have the property)

Demetres
Marcos
Posted on Friday, 22 August, 2003 - 09:31 am:
Thanks Sarah and Demetres for your replies (Sorry I didn't get back to you sooner, I was out of town).

Sarah, it's nice to know we both used the same approach, I had basically attacked the problem as you suggested but I hadn't really got far...

Demetres, in fact the reason I thought of this problem was to find which regular polygons are constructible.
I hope there's a way that doesn't require Galois theory (although I'm gonna go read Dan Goodman's article again)...
I'll go have a shot at it. Right now, I can only show that if p has the property then so does 2n p.

Marcos
Demetres Christofides
Posted on Friday, 22 August, 2003 - 10:40 am:
I will prove that if cos(2p/p) is constructible and p is prime, then it must be a Fermat prime. (By constructible, I mean expressible using a finite number of algebraic operations, and extracting square roots. The proper definition is somewhat different but is in some sense equivalent.)

The only proof I know of the converse, that if p is a Fermat prime then cos(2p/p) is, requires a lot of Galois Theory so I'll omit it.

By solving a quadratic equation, show that cos(2p/p) is constructible if and only if x=exp(2pi/p) is.

Note that xp=1 and since x ¹ 1, then xp-1+xp-2+¼+1=0.

A basic step is to show that the polynomial f(X)=Xp-1+¼+X+1 is irreducible, i.e. it cannot be written as a product of two polynomials with integer coefficients and degree less than p.

Before proving this, I will explain where we are heading for.

Suppose that a complex number z is the zero of a polynomial with rational coefficients. Let f be such a polynomial which is monic (i.e. leading coefficient 1) and of minimum degree. Show that f is unique with these properties.

f is called the minimum/minimal polynomial of z and the degree of f is called the degree of z.

Examples

  1. The minimum polynomial of 1/2 is f(X)=X-1/2
  2. The min. poly. of i is f(X)=X2+1
  3. We'll show that the min. poly. of exp(2pi/p) is f(X)=Xp-1+¼+X+1.
The main theorem is the following: A number is constructible only if its minimum polynomial has degree a power of 2.

This, together with example 3 show that p-1 is a power of 2 and elementary number theory shows that p is a Fermat prime.

More examples of main theorem

  1. The min. poly of
    3 æ
    Ö
    2
     

    is f(X)=X3-2 which has degree 3. From the main theorem we deduce that the 'duplication of the cube' is impossible.
  2. Use a formula for cos3x where x=p/9 to find the min. poly of cos(p/9). Use this to show that the angle of 60 degrees cannot be trisected.
  3. p is not the zero of any equation with rational coefficients (REALLY HARD theorem). Thus 'squaring the circle' is impossible.

We want to show that f(X)=Xp-1+¼+X+1 is irreducible. It is enough to show that
g(X)=f(X+1)= (X+1)p-1
(X+1)-1
 =Xp-1+pXp-2+¼+ æ
ç
è
p
n
 ö
÷
ø 		Xp-n-1+¼+p
is irreducible.

Suppose it is reducible, say g(X)=a(X)b(X) where a(X)=a0+a1 X + ¼ + an Xn and b(X)=b0+b1 X + ¼+ bm Xm.

We now argue inductively:

a0 b0=p so without loss of generality, a0 is a multiple of p while b0 is not.

Then a0 b1+a1b0 is a multiple of p, hence so is a1.

Continue inductively to show that an is a multiple of p, a contradiction.

It remains to prove the main theorem, of course!

Demetres

Demetres Christofides
Posted on Friday, 22 August, 2003 - 10:45 am:
By the way, I've seen somewhere that

LaTeX Image

I never checked if it's correct.
There is a rumour of somebody trying to find an analogous expression for LaTeX Image.

Demetres