# 가우스와 정17각형의 작도

(차이) ← 이전 판 | 최신판 (차이) | 다음 판 → (차이)
둘러보기로 가기 검색하러 가기

## 개요

• 가우스는 정17각형이 자와 컴파스로 작도가능함을 증명함
• 대수적으로, $$z^{16}+z^{15}+\cdots+z+1=0$$의 풀이를 반복적인 2차방정식의 풀이로 환원할 수 있는가의 문제
• 16차 방정식을 2차방정식 네번 푸는 문제로 바꾸는 것
• 이 아이디어를 좀더 간단한 예를 통해 이해하기 위해서는 정오각형 항목 중 꼭지점의 평면좌표를 참조

## 증명

• $$\zeta=e^{2\pi i \over 17}$$ 로 두자. 이 값을 대수적으로 구하는 것이 목표.
• $$(3^1, 3^2,3^3, 3^4, 3^5, 3^7, 3^8, 3^9, 3^{10}, 3^{11}, 3^{12}, 3^{13}, 3^{14}, 3^{15}, 3^{16}) \equiv (3, 9, 10, 13, 5, 15, 11, 16, 14, 8, 7, 4,12, 2, 6, 1) \pmod {17}$$
• 이 순서대로 2로 나눈 나머지에 따라서 분류
• $$A_0 = \zeta^{9} + \zeta^{13} + \zeta^{15} + \zeta^{16}+\zeta^{8} + \zeta^{4} + \zeta^{2} +\zeta^{1}$$
• $$A_1 = \zeta^3 + \zeta^{10} + \zeta^{5} + \zeta^{11}+\zeta^{14} + \zeta^{7} + \zeta^{12} +\zeta^{6}$$
• $$A_0+A_1= -1$$, $$A_{0}A_{1} = -4$$, $$A_0>A_1$$
• $$A_0 = \frac{-1 + \sqrt{17}}{2}$$ , $$A_1= \frac{-1 - \sqrt{17}}{2}$$
• 이번에는 4로 나눈 나머지에 따라서 분류
• $$B_0 = \zeta^{13}+ \zeta^{16}+ \zeta^4 + \zeta^1$$
• $$B_1= \zeta^3 + \zeta^5 + \zeta^{14} + \zeta^{12}$$
• $$B_2= \zeta^9 + \zeta^{15} + \zeta^8 +\zeta^2$$
• $$B_3 =\zeta^{10} + \zeta^{11} + \zeta^{7} +\zeta^{6}$$
• $$B_0+B_2=A_0$$, $$B_0B_2= -1$$, $$B_0>0$$
• $$B_0 = \frac{-1 + \sqrt{17} + \sqrt{34 - 2\sqrt{17}}}{4}$$, $$B_2 = \frac{-1 + \sqrt{17} - \sqrt{34 - 2\sqrt{17}}}{4}$$
• $$B_1+B_3=A_1$$, $$B_1B_3= -1$$, $$B_{1}> 0$$
• $$B_1 = \frac{-1 - \sqrt{17} + \sqrt{34 + 2\sqrt{17}}}{4}$$, $$B_3 = \frac{-1 - \sqrt{17} - \sqrt{34 + 2\sqrt{17}}}{4}$$
• 이번에는 8로 나눈 나머지에 따라서 분류
• $$C_0= \zeta^{16}+ \zeta^1$$, $$C_4= \zeta^{13} +\zeta^4$$, $$C_0 > C_1$$
• $$C_0+C_4=B_0$$, $$C_0C_4=B_1$$
• $$C_0= \frac{B_0+\sqrt{B_0^2-4B_1}}{2}= \frac{-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}+ \sqrt{68+12\sqrt{17}-4{\sqrt{170+38\sqrt{17}}}} }{8}$$
• $$C_4= \frac{B_0 - \sqrt{B_0^2-4B_1}}{2}$$
• 이제 마무리
• $$\zeta =\frac{{C_0} + \sqrt{{C_0}^2 - 4}}{2}$$
• $$\cos \frac{2\pi}{17}= \frac{-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}+ \sqrt{68+12\sqrt{17}-4{\sqrt{170+38\sqrt{17}}}} }{16}$$

## 가우스합과의 관계

• 참고로 위에서 $$A_0-A_1$$ 은 가우스합 임을 알 수 있음.
• $$\{3, 10, 5, 11, 14, 7, 12, 6\}$$ 는 $$\pmod {17}$$ 에 대하여 이차비잉여
• $$\{9, 13, 15, 16, 8, 4, 2, 1\}$$는 $$\pmod {17}$$ 에 대하여 이차잉여
• 따라서 $$A_{0}A_{1}$$를 계산하는 대신에 $$A_0-A_1=\sqrt{17}$$ 를 활용할 수도 있음.

## 관련논문

• Garcia, Stephan Ramon, Trevor Hyde, and Bob Lutz. ‘Gauss’ Hidden Menagerie: From Cyclotomy to Supercharacters’. arXiv:1501.07507 [math], 29 January 2015. http://arxiv.org/abs/1501.07507.

## 노트

### 말뭉치

1. Then, on 30 March 1796, the 19 year old Gauss discovered that it was possible to construct the regular heptadecagon (17-gon).[1]
2. One of the nicest actual constructions of the 17-gon is Richmond's (1893), as reproduced in Stewart's "Galois Theory".[1]
3. Gauss was clearly fond of this discovery, and there's a story that he asked to have a heptadecagon carved on his tombstone, like the sphere incribed in a cylinder on Archimedes' tombstone.[1]
4. On the other hand, if proximity to the actual remains is not important, then the heptadecagon on the monument to Gauss in his native town of Brunswick, or even the figure above, may suffice.[1]
5. Gauss's "heptadecagon', a 17-sided polygon that showed the relationship between geometry and algebra.[2]
6. But what’s widely considered his first important discovery is his construction of a 17-sided polygon called a heptadecagon, using only a ruler and a compass.[2]
7. A regular heptadecagon is represented by the Schläfli symbol {17}.[3]
8. Constructing a regular heptadecagon thus involves finding the cosine of 2 π / 17 {\displaystyle 2\pi /17} in terms of square roots, which involves an equation of degree 17—a Fermat prime.[3]
9. The explicit construction of a heptadecagon was given by Herbert William Richmond in 1893.[3]
10. These 4 symmetries can be seen in 4 distinct symmetries on the heptadecagon.[3]
11. Gauss proved in 1796 (when he was 19 years old) that the heptadecagon is constructible with a compass and straightedge.[4]
12. The first explicit construction of a heptadecagon was given by Erchinger in about 1800.[4]
13. You now have points and of a heptadecagon.[4]
15. A regular heptadecagon, or 17 sided polygon, was known to have existed by mathematicians for many years, but creating one proved to be a greater challenge.[5]
16. A regular heptadecagon was first created by 19-year-old Carl Friedrich Gauss in 1796 in a groundbreaking proof.[5]
17. The proof in regards to this 17-gon's construction marked the first major breakthrough in polygon construction in over 2,000 years.[5]
18. Apart from a heptadecagon, there are also heptadecagrams, which are 17 sided star polygons.[5]
19. A regular heptadecagon has internal angles each measuring 158.823529411765 degrees.[6]
20. The regular heptadecagon is a constructible polygon, as was shown by Carl Friedrich Gauss in 1796.[6]
21. Gauß proved in 1796 (when he was 19 years old) that the heptadecagon is Constructible with a Compass and Straightedge.[7]
22. The following elegant construction for the heptadecagon (Yates 1949, Coxeter 1969, Stewart 1977, Wells 1992) was first given by Richmond (1893).[7]
23. The following animation of a heptadecagon editing.[8]
25. He described in his Disquitiones Arithmeticae, a major work on number theory, how to construct a regular 17-gon with Euclidean tools.[10]
26. The eighteen year old Gauss began his scientific diary with his construction of the regular 17-gon.[11]