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Old 2021-10-18, 08:01   #12
a1call
 
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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
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Old 2021-10-18, 08:40   #13
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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?
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Old 2021-10-18, 08:46   #14
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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.
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Old 2021-10-18, 08:53   #15
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Quote:
Originally Posted by MattcAnderson View Post
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.
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Old 2021-10-18, 09:55   #16
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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.
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Old 2021-10-18, 12:10   #17
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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.
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Old 2021-10-18, 13:32   #18
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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

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Old 2021-10-18, 13:45   #19
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https://upload.wikimedia.org/wikiped...n_a_Circle.gif

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Old 2021-10-18, 14:32   #20
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Quote:
Originally Posted by xilman View Post
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.
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Old 2021-10-18, 16:43   #21
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Quote:
Originally Posted by sweety439 View Post
You can try to construct F33 (=2^(2^33)+1)-sided polygon, in fact, if you calculated cos(2*pi/F33) and....
<snip>


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.
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Old 2021-10-18, 17:14   #22
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Quote:
Originally Posted by retina View Post
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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