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! Thanks, Sione 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 Check out our web site! http://mathforum.org/dr.math/ |
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