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This is really interesting.. does anyone who knows more about Gauss's proof know why you can construct a 5 sided polygon with ruler and compass, but not a 7 or

by zengid 2y ago

This is really interesting.. does anyone who knows more about Gauss's proof know why you can construct a 5 sided polygon with ruler and compass, but not a 7 or 11 sided polygon? Why do some primes work and others not?

lanternfish 2y ago
It has to do with Fermat primes.
- gus_massa 2y ago
  This was downvoted. I guess someone though it was a joke. But it's part of the correct answer. More details in https://en.wikipedia.org/wiki/Fermat_number https://en.wikipedia.org/wiki/Fermat_number
plg94 2y ago
For 17, Gauss noticed that cos(360°/17) can be written only with elementary operations, see https://www.heise.de/imgs/18/2/1/2/3/3/6/4/siebzehneck-b95b58b916934efb.png https://www.heise.de/imgs/18/2/1/2/3/3/6/4/siebzehneck-b95b5... Later he proved that all n-gons with $n=2^k*p_1…*p_r$ where the p_i are Fermat-primes (2^(2^m)+1 prime, today we only know of 3, 5, 17, 257, 65537) are constructible. The opposite direction, i.e. all other n are not constructible, was only a few years later proved. Look up "Theorem of Gauss-Wantzel". I only skimmed the proof, but it seems to generalize the concept of constructing the cos of the angle with "Galois-Theory". (edit: or see https://en.wikipedia.org/wiki/Constructible_polygon https://en.wikipedia.org/wiki/Constructible_polygon)
- zengid 2y ago
  Very cool, thanks!
gus_massa 2y ago
I can give a very fast and incomplete explanation, but you must trust me. In the complex numbers, the vertices of a pentagon are z^5-1=0. You can factorize it as (z^4+z^3+z^2+z+1)*(z-1)=0. The hard part is solving z^4+z^3+z^2+z+1=0. Now that equation can't be factorized, and has degree 4. It's important that the solutions have a property that is related to the degree of the equation so they have a property that is 4. With a compass and a straightedge you can solve only equations of degree 2, that is like taking a square root. If you repeat the process you can solve (some) equations of degree 4. So after a few tricks, you can solve the equation and draw the pentagon. For 17, the equation is z^16+z^15+...+z+1=0. So the property is 16 and you must use the square root a few times. Each time the solutions double their property, so you get 1 -> 2 -> 4 -> 8 -> 16. Near the bottom of the article is the formula, and it's possible to see a lot of nested and repeated square roots. For 7, the equation is z^6+z^5+...+z+1=0. So the property of the solutions is 6. With the square root you can only double the property, so you get 1 -> 2 -> 4 -> 8 -> 16 -> 32 ... but you can never get a solution which has a property equal to 6. (There are more technicals details. You can solve some equations of degree 16, for example to draw the 17agon, but you can't solve every one of them.)
- glial 2y ago
  Interesting, so in general you could construct polygons with 2^n+1 sides?
  - clausecker 2y ago
    Yes, but only if that number is a prime.
    - gus_massa 2y ago
      I agree. More details: For example with 9, you can factorize z^9-1=0 as (z^6+z^3+1)*(z^2+z+1)*(z-1)=0, and now the property to calculate is 6*2 instead of 8, so it's not a power of 2 and the polygon is impossible to construct.
paulpauper 2y ago
cyclotomic polynomials and Galois theory a 17-gon reduces to a 4th-degree polynomial, and a 2nd degree one, which can be solved in radicals, by studying the permutations and multiplicities of the roots of this polynomial, in which the solutions are multiples of n*pi/17.
dadoum 2y ago
If you are interested and have the time you can watch the 2 videos from the YouTube channel Another Roof[1] on that subject. He spends some time on easy stuff too to allow the general public to relatively understand the bases, so don't be surprised if the videos are quite long. [1]: https://youtube.com/@anotherroof https://youtube.com/@anotherroof
seanhunter 2y ago
This is described in quite an approachable way in the video I posted in another comment. https://www.youtube.com/watch?v=EX7U0DGBmbM https://www.youtube.com/watch?v=EX7U0DGBmbM

by zengid 2y ago

lanternfish 2y ago

It has to do with Fermat primes.

gus_massa 2y ago
This was downvoted. I guess someone though it was a joke. But it's part of the correct answer. More details in https://en.wikipedia.org/wiki/Fermat_number https://en.wikipedia.org/wiki/Fermat_number

plg94 2y ago

For 17, Gauss noticed that cos(360°/17) can be written only with elementary operations, see https://www.heise.de/imgs/18/2/1/2/3/3/6/4/siebzehneck-b95b58b916934efb.png https://www.heise.de/imgs/18/2/1/2/3/3/6/4/siebzehneck-b95b5... Later he proved that all n-gons with $n=2^k*p_1…*p_r$ where the p_i are Fermat-primes (2^(2^m)+1 prime, today we only know of 3, 5, 17, 257, 65537) are constructible. The opposite direction, i.e. all other n are not constructible, was only a few years later proved. Look up "Theorem of Gauss-Wantzel". I only skimmed the proof, but it seems to generalize the concept of constructing the cos of the angle with "Galois-Theory". (edit: or see https://en.wikipedia.org/wiki/Constructible_polygon https://en.wikipedia.org/wiki/Constructible_polygon)

zengid 2y ago
Very cool, thanks!

gus_massa 2y ago

I can give a very fast and incomplete explanation, but you must trust me. In the complex numbers, the vertices of a pentagon are z^5-1=0. You can factorize it as (z^4+z^3+z^2+z+1)*(z-1)=0. The hard part is solving z^4+z^3+z^2+z+1=0. Now that equation can't be factorized, and has degree 4. It's important that the solutions have a property that is related to the degree of the equation so they have a property that is 4. With a compass and a straightedge you can solve only equations of degree 2, that is like taking a square root. If you repeat the process you can solve (some) equations of degree 4. So after a few tricks, you can solve the equation and draw the pentagon. For 17, the equation is z^16+z^15+...+z+1=0. So the property is 16 and you must use the square root a few times. Each time the solutions double their property, so you get 1 -> 2 -> 4 -> 8 -> 16. Near the bottom of the article is the formula, and it's possible to see a lot of nested and repeated square roots. For 7, the equation is z^6+z^5+...+z+1=0. So the property of the solutions is 6. With the square root you can only double the property, so you get 1 -> 2 -> 4 -> 8 -> 16 -> 32 ... but you can never get a solution which has a property equal to 6. (There are more technicals details. You can solve some equations of degree 16, for example to draw the 17agon, but you can't solve every one of them.)

glial 2y ago
Interesting, so in general you could construct polygons with 2^n+1 sides?
- clausecker 2y ago
  Yes, but only if that number is a prime.
  - gus_massa 2y ago
    I agree. More details: For example with 9, you can factorize z^9-1=0 as (z^6+z^3+1)*(z^2+z+1)*(z-1)=0, and now the property to calculate is 6*2 instead of 8, so it's not a power of 2 and the polygon is impossible to construct.

paulpauper 2y ago

cyclotomic polynomials and Galois theory a 17-gon reduces to a 4th-degree polynomial, and a 2nd degree one, which can be solved in radicals, by studying the permutations and multiplicities of the roots of this polynomial, in which the solutions are multiples of n*pi/17.

dadoum 2y ago

If you are interested and have the time you can watch the 2 videos from the YouTube channel Another Roof[1] on that subject. He spends some time on easy stuff too to allow the general public to relatively understand the bases, so don't be surprised if the videos are quite long. [1]: https://youtube.com/@anotherroof https://youtube.com/@anotherroof

seanhunter 2y ago

This is described in quite an approachable way in the video I posted in another comment. https://www.youtube.com/watch?v=EX7U0DGBmbM https://www.youtube.com/watch?v=EX7U0DGBmbM