20211018, 08:01  #12 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2×7×157 Posts 
Similarly for 3*17=51 gon
1/33/17= 8/51 of a circle. Then you can bisect 3 times to get 1/51 of a circle. Last fiddled with by a1call on 20211018 at 08:02 
20211018, 08:40  #13 
Jun 2003
2×5×17×31 Posts 

20211018, 08:46  #14 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2×7×157 Posts 

20211018, 08:53  #15  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
6,329 Posts 
Quote:
Since there are 5 known Fermat primes then you can take all the 5bit binary numbers from 00000 to 11111 (32 in total) and replace each 0 with nothing, and each 1 with a Fermat prime. And excluding the all zeros, then that leaves 31 possible combinations. 

20211018, 09:55  #16 
Dec 2012
The Netherlands
3×587 Posts 
The mathematical background is that you can construct the regular polygon with n sides using ruler and compass
if and only if ϕ(n) is a power of 2 (where ϕ is Euler's function). And that is true if and only if the prime factorization of n consists of 0 or more 2's and 0 or more distinct Fermat primes. 
20211018, 12:10  #17 
Feb 2017
Nowhere
1010011101010_{2} Posts 
Let f(x) be a monic irreducible polynomial in Z[x]. Let G be the Galois group of f(x). The order of G is always divisible by the degree d of f(x). The roots of f(x) = 0 are constructible with compass and straightedge if, and only if, the Galois group G of f(x) is a 2group; that is, if the order of G is a power of 2.
Let n be a positive integer. The minimum polynomial for the primitive n^{th} roots of unity (cyclotomic polynomial) has degree , the number of positive integers not exceeding n which are relatively prime to n. In PariGP this is eulerphi(n). If p is a prime factor of n, then p1 divides eulerphi(n). If p^2 divides n, then p divides eulerphi(n). Thus, the degree eulerphi(n) is a power of 2 if, and only if, for each odd prime factor p of n, p  1 is a power of 2, and p^2 does not divide n. Luckily, the Galois group of the cyclotomic polynomial has order equal to the degree; in fact so if every odd prime factor p of n is a Fermat prime, and p^2 does not divide n, the Galois group is in fact a 2group if the degree is a power of 2, and the n^{th} roots of unity are constructible with compass and straightedge. 
20211018, 13:32  #18  
"Rashid Naimi"
Oct 2015
Remote to Here/There
100010010110_{2} Posts 
And to think that Gauss had to reach the ripe old age of 19 before he could even prove that 17gon was constructible.
Quote:


20211018, 13:45  #19 
Aug 2002
2^{3}×1,051 Posts 

20211018, 14:32  #20 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2·7·229 Posts 
You can try to construct F33 (=2^(2^33)+1)sided polygon, in fact, if you calculated cos(2*pi/F33) and....
* If this number only contain square roots and no roots beyond square roots, then F33 is prime. * If this number contain roots other than square roots, then F33 is composite. 
20211018, 16:43  #21  
Feb 2017
Nowhere
2×2,677 Posts 
Quote:
Proposing to attempt the construction without knowing whether F_{33} is prime, is epically inane. The cosine can be calculated fairly precisely, however, using cos(x) = 1  x^{2}/2 approximately, if x is small and in radians. If I did the arithmetic right, cos(2*pi/F_{33}) = 1  2.128362442349688837*10^{5171655945}, approximately. 

20211018, 17:14  #22  
"Matthew Anderson"
Dec 2010
Oregon, USA
2^{4}×3^{2}×7 Posts 
Quote:
You are correct Retina. The Binary counting from 00000 to 11111 and subtract the 0 case is an easy way to see that 31 is the right answer. 1+2+4+8+16 = 31. Thanks for that. Regards, Matt 
