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 2021-10-18, 08:01 #12 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 2×7×157 Posts Similarly for 3*17=51 gon 1/3-3/17= 8/51 of a circle. Then you can bisect 3 times to get 1/51 of a circle. Last fiddled with by a1call on 2021-10-18 at 08:02
2021-10-18, 08:40   #13
axn

Jun 2003

2×5×17×31 Posts

Quote:
 Originally Posted by a1call The 2 statements from Wikipedia are contradictory IMHO. The 2nd statement does not seem to be correct
I assume from the subsequent posts that you no longer consider the wikipedia statements as contradictory?

2021-10-18, 08:46   #14
a1call

"Rashid Naimi"
Oct 2015
Remote to Here/There

2×7×157 Posts

Quote:
 Originally Posted by axn I assume from the subsequent posts that you no longer consider the wikipedia statements as contradictory?
Nah, I misunderstood the 1st statement. I interpreted distinct as distinct in general and not per polygon.

2021-10-18, 08:53   #15
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

6,329 Posts

Quote:
 Originally Posted by MattcAnderson So add 5+10+6+5+1 = 26 errrrr I seem to have missed five somewhere. It should be 31.
To make the set enumerating easier for you next time:

Since there are 5 known Fermat primes then you can take all the 5-bit 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.

 2021-10-18, 09:55 #16 Nick     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.
 2021-10-18, 12:10 #17 Dr Sardonicus     Feb 2017 Nowhere 10100111010102 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 2-group; that is, if the order of G is a power of 2. Let n be a positive integer. The minimum polynomial for the primitive nth roots of unity (cyclotomic polynomial) has degree $\varphi(n)$, the number of positive integers not exceeding n which are relatively prime to n. In Pari-GP this is eulerphi(n). If p is a prime factor of n, then p-1 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 $G\;=\;$$\mathbb{Z}/n\mathbb{Z}$$^{\times}$ 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 2-group if the degree is a power of 2, and the nth roots of unity are constructible with compass and straightedge.
2021-10-18, 13:32   #18
a1call

"Rashid Naimi"
Oct 2015
Remote to Here/There

1000100101102 Posts

And to think that Gauss had to reach the ripe old age of 19 before he could even prove that 17-gon was constructible.

Quote:
 His breakthrough occurred in 1796 when he showed that a regular polygon can be constructed by compass and straightedge if the number of its sides is the product of distinct Fermat primes and a power of 2.[a]
https://en.wikipedia.org/wiki/Carl_Friedrich_Gauss

 2021-10-18, 13:45 #19 Xyzzy     Aug 2002 23×1,051 Posts
2021-10-18, 14:32   #20
sweety439

"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

2·7·229 Posts

Quote:
 Originally Posted by xilman A 257-gon can also be so constructed. Never seen anyone do so.
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.

2021-10-18, 16:43   #21
Dr Sardonicus

Feb 2017
Nowhere

2×2,677 Posts

Quote:
 Originally Posted by sweety439 You can try to construct F33 (=2^(2^33)+1)-sided polygon, in fact, if you calculated cos(2*pi/F33) and....

Proposing to attempt the construction without knowing whether F33 is prime, is epically inane.

The cosine can be calculated fairly precisely, however, using cos(x) = 1 - x2/2 approximately, if x is small and in radians. If I did the arithmetic right,

cos(2*pi/F33) = 1 - 2.128362442349688837*10-5171655945, approximately.

2021-10-18, 17:14   #22
MattcAnderson

"Matthew Anderson"
Dec 2010
Oregon, USA

24×32×7 Posts

Quote:
 Originally Posted by retina To make the set enumerating easier for you next time: Since there are 5 known Fermat primes then you can take all the 5-bit 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.
Hi all,

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

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