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Drawing Regular N-gons (Compass and Straightedge)

Date: 11/17/97 at 04:16:45
From: Sione Smythe
Subject: Regular n-gons and compass & st. edge only

Dear Sir,

I can't remember where I read it, but I'm certain there was a 
statement made that the only regular n-gons that can be drawn using 
ONLY a straight edge and compass are those with the number of sides 
equal to a Fermat Prime or a product of Fermat Primes. Is this right, 
and why is this?  

Also, why is it true that a prime squared and then 1 subtracted is 
divisable by 24?  (Prime to the fourth minus 1 is divisable by 120?)

Any help will really be appreciated!


Date: 11/17/97 at 15:28:31
From: Doctor Rob
Subject: Re: Regular n-gons and compass & st. edge only

1) You are almost right. The only regular n-gons that can be drawn
using only a straightedge and compass are those with the number of
sides equal to a power of 2 times a product of distinct Fermat primes.
This is because to solve the construction problem, you have to be able
to construct a line segment whose length is the cosine of 2*Pi/n.
With straightedge and compass you can construct line segments whose
lengths are gotten by rational operations (adding, subtracting,
multiplying, or dividing by nonzero rational numbers) and taking
square roots. Any other operations are not possible. These numbers 
are solutions to algebraic equations whose degrees are powers of 2.
Cos(2*Pi/n) is the solution to an algebraic equation whose degree is
Phi(n), where Phi(n) is the Euler phi-function or totient. It is a
power of 2 if and only if n has the form given above.

Proving all of the above is difficult, and usually is reserved for a
graduate-level abstract algebra course, where one can talk about field
extensions and their degrees.

2) p^2-1 = (p+1)*(p-1). Since among any three consecutive numbers,
one must be a multiple of 3, and for p > 3, p is not a multiple of 3,
it must be that either p+1 or p-1 is. Similarly, among any two
consecutive even numbers, one is a multiple of 4. Both p-1 and p+1
are even for p > 2, so one is divisible by 2 and the other by 4, so
their product is divisible by 8. A number divisible by 3 and 8 must
be divisible by 24, since 3 and 8 are relatively prime. Your
statement is true for primes bigger than 3, but false for p = 2 or
p = 3.

3) p^4-1 = (p-1)*(p+1)*(p^2+1). Use a similar analysis to show that
one of these numbers must be divisible by 5 for p > 5. Since 5 and 24
are relatively prime, 5*24 = 120 must divide p^4-1 for p > 5. It is
false for primes p < 7.

-Doctor Rob,  The Math Forum
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Associated Topics:
College Constructions
College Modern Algebra
College Triangles and Other Polygons

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