**jce** (06/21/82)

I went to get a pizza the other night, and, since I had six friends over, I asked that he cut it in seven pieces. The pizzaman said he couldn't do it exactly with just straight-edge and compass (it was a greek pizzeria), and I asked him why. He started talking about automorphisms and algebraic fields and other things I don't understand - can anybody explain this in plain english? Is there a general rule?

**djj** (06/21/82)

Regarding the seven slices of pizza -- Could it be related to the fact that the prime factorization of 360 is 2 x 2 x 2 x 5 x 9 and the fact that 7 is prime and not among the factors? Dave Johnson BTL - Piscataway (...!pyuxcc!djj) p.s. I realize that's not a very geometric approach, but it makes sense to me!

**thomas** (06/22/82)

There is a general rule for ruler and compass constructions, which can be stated as follows: When the solution x of the problem is real and can be found by rational operations and (not necessarily real) square roots from the given numbers (line segment lengths) a, b, ..., the number (segment length) x can be constructed using ruler and compass. In the case of dividing a circle into n equal parts, the condition is that the number of divisors of n (phi(n)) is a power of 2. Now, n can be written in the form (2^m)(p1^m1)(p2^m2)...(pl^ml), where the p's are prime. Then phi(n) = (2^(m-1))(pq^(m1-1))...(pl^(ml-1))(p1-1)...(pl-1). Thus, m1 thru ml must all be 1, and (pi-1) must be a power of 2. Now, it turns out that all primes which are one greater than a power of 2 are of the form 2^(2^r)+1 for some integer r. (Note, that not all numbers of this form are prime, however, for example, 2^2^5 is divisible by 641.) So, a circle may be divided into n pieces if n is a product of some power of 2 and primes of this sort (to a most the first power). The first few primes in this sequence are 3, 5, 17, 257, 65537. So, if you and 16 other friends were sharing a pizza, you could divide it into equal pieces using only a straightedge and a compass. The result is due to Gauss, my presentation is a paraphrase of the presentation in Modern Algebra, Vol 1, van der Waerden, pp183-187. =Spencer

**G:shallit** (06/22/82)

Why you would want to explain it in plain English is beyond me. Mathematical notation is so much more precise and clear for this kind of thing. So, after starting thus, here goes an attempt at an explanation. It has absolutely nothing to do with the prime factorization of 360, as was suggested. (How could it? 360 is an artificial number, decided by man, not nature. If we decided on 420 degrees in a circle, would this make it possible to divide a circle into 7 parts? Of course not. But I digress.) Gauss showed that the circle can be divided into n parts if and only if n is 1) a Fermat prime (i. e. of the form 1+2*(2*k) ) or (2) a product of such primes or (3) the product of a power of 2 and a set of Fermat primes. His reasoning was approximately as follows: A quantity can be constructed by straightedge and compass if it can be expressed by a finite number of applications of + - * / sqrt. There are polynomials all of whose roots can be expressed in this way; among such, there is one of minimal degree with a given quantity as a root. The degree of these polynomials is a power of 2. The roots of x^n - 1 = 0 lie on a circle , spaced equally around the circumference. In order to divide the circle we must be able to construct these points. We can assume n is a prime or a power of a prime, since if we can construct the roots of x^p - 1 = 0 and x^q - 1 = 0 we can construct the roots of x^pq - 1 = 0. We can certainly bisect an arc with straightedge and compass. Hence we can divide a circle into 2, 4, 8 , ... parts. If we could divide a circle into k parts (k odd) we could divide it into (2^j)*k parts. Look at x^p - 1 = 0, where p is prime. Divide this by x-1 to get an irreducible (unfactorable) equation of degree p-1. Since it is irreducible, it is of minimal degree and so p-1 must be a power of 2. The only numbers p-1 that are powers of 2 are such that p = 1+2^2^n, a Fermat prime. The only ones known are 3,5, 17, 257, and 65537. Hope this helps a little.

**bj** (06/24/82)

People have done wonderfull jobs showing that you can not divide a pizza into 7 equal wedges, but that does not solve the original problem - can you divide a pizza into seven pieces of equal size and any shape (some people don't like crust anyway)? The answer to that question is yes. Construct a line whose length is sqrt ( 1 / 7r ) where r is the radius of the pizza. Take a circle with this radius from the center of the pizza and give it to somebody. Then divide the ring into six pieces for everyone else.

**jce** (06/24/82)

yale-com!bj's solution to the pizza problem is brilliant, because it wasn't even what I had in mind when I posted the original query. Unfortunately, it was a little off - the correct radius for the inner circle is r / sqrt(7).

**wesw** (06/25/82)

What about a rectangular pizza?

**grunwald** (06/27/82)

#R:floyd:-29500:uiucdcs:10300002:000:427 uiucdcs!grunwald Jun 26 23:18:00 1982 why is it not possible to construct a septagon within the pizza, with the longest radius of the septagon equal to the radius of the pizza? Constructing the septagon involves calculating the length of the outer edge and then setting the compass to that length. Then, start anywhere on the edge of the Zah and go around using the compass to measure off the distance on the perimeter. Is there some non obvious catch to this?