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a1call 2021-10-18 08:01

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.

axn 2021-10-18 08:40

[QUOTE=a1call;590936]The 2 statements from Wikipedia are contradictory IMHO. The 2nd statement does not seem to be correct[/QUOTE]

I assume from the subsequent posts that you no longer consider the wikipedia statements as contradictory?

a1call 2021-10-18 08:46

[QUOTE=axn;590945]I assume from the subsequent posts that you no longer consider the wikipedia statements as contradictory?[/QUOTE]

Nah, I misunderstood the 1st statement. I interpreted distinct as distinct in general and not per polygon. :smile:

retina 2021-10-18 08:53

[QUOTE=MattcAnderson;590934]So add 5+10+6+5+1 = 26 errrrr I seem to have missed five somewhere. It should be 31.[/QUOTE]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.

Nick 2021-10-18 09:55

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.

Dr Sardonicus 2021-10-18 12:10

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 n[sup]th[/sup] roots of unity (cyclotomic polynomial) has degree [tex]\varphi(n)[/tex], 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 [i]not[/i] divide n.

Luckily, the Galois group of the cyclotomic polynomial has order equal to the degree; in fact [tex]G\;=\;\(\mathbb{Z}/n\mathbb{Z}\)^{\times}[/tex] 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 n[sup]th[/sup] roots of unity are constructible with compass and straightedge.

a1call 2021-10-18 13:32

And to think that Gauss had to reach the ripe old age of 19 before he could even prove that 17-gon was constructible. :no:

[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]
[/QUOTE]

[url]https://en.wikipedia.org/wiki/Carl_Friedrich_Gauss[/url]

:no:

Xyzzy 2021-10-18 13:45

[url]https://upload.wikimedia.org/wikipedia/commons/d/d1/Regular_Heptadecagon_Inscribed_in_a_Circle.gif[/url]

:mike:

sweety439 2021-10-18 14:32

[QUOTE=xilman;590789]A 257-gon can also be so constructed. Never seen anyone do so.[/QUOTE]

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.

Dr Sardonicus 2021-10-18 16:43

[QUOTE=sweety439;591001]You can try to construct F33 (=2^(2^33)+1)-sided polygon, in fact, if you calculated cos(2*pi/F33) and....
<snip>[/QUOTE]:missingteeth: :missingteeth: :missingteeth: :missingteeth:

Proposing to attempt the construction without knowing whether F[sub]33[/sub] is prime, is epically inane.

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

cos(2*pi/F[sub]33[/sub]) = 1 - 2.128362442349688837*10[sup]-5171655945[/sup], approximately.

MattcAnderson 2021-10-18 17:14

[QUOTE=retina;590949]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.[/QUOTE]

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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