20211018, 20:17  #23  
"Robert Gerbicz"
Oct 2005
Hungary
5FB_{16} Posts 
Quote:
Quote:
So we can make a 2*Pi/n and 2*Pi/m angle. We assumed that gcd(m,n)=1 so with extended Euclidean algorithm there exists x and y integers: n*x+m*y=1 divide this equation by m*n x/m+y/n=1/(m*n) multiplie by 2*Pi x*(2*Pi/m)+y*(2*Pi/n)=2*Pi/(m*n), what we needed. 

20211020, 05:11  #24 
"Matthew Anderson"
Dec 2010
Oregon, USA
2^{4}×3^{2}×7 Posts 
Hi all,
From before, According to Wikipedia (constructible polygon article), there are infinitely many constructible polygons, but only 31 with an odd number of sides are known. 5 Fermat primes are known. I worked out why 31 different regular polygons with an odd number of sides are constructible. We have nCk, read n choose k, defined as nCk = n!/(k!*(nk)!) So we want combinations of 5 things taken 1,2,3,4, and 5 at a time without repetition. Hence 5C1 = 5 5C2 = 10 5C3 = 10 5C4 = 5 and 5C5 = 1 So 5+10+10+5+1 = 31. So we see that there are 31 ways of, among 5 things, taking 1,2,3,4 or all 5 of them without repetition. And all is right with the world. Regards, Matt 
20211020, 17:54  #25 
Aug 2002
Buenos Aires, Argentina
2·3^{2}·79 Posts 
If you open my Polynomial factorization and roots calculator, enter x^2551 and press Factor, you will see after a few seconds the 255 roots of that polynomial, and as explained before, only square roots are needed, because 255 = 3 * 5 * 17, which is the product of three different Fermat primes.
