History of Mathematics

Transcription

1 History of Mathematics Paul Yiu Department of Mathematics Florida Atlantic University Spring B: Gauss Construction of the regular 17-gon

2 Gauss first diary entry (1796) The principles upon which the division of the circle depend, and geometrical divisibility of the same into seventeen parts, etc. March 30 Brunswick ζ 6 ζ 5 ζ 4 ζ 3 ζ 2 ζ 7 ζ ζ 8 ζ ζ 16 ζ 10 ζ 11 ζ 12 ζ 13 ζ 14 ζ 15 The details appeared in Section VII of his Disquisitiones Arithmeticae (1801). 1

3 Construction of the roots of a 16 degree equation ζ 6 ζ 5 ζ 4 ζ 3 ζ 2 ζ 7 ζ ζ 8 ζ ζ 16 ζ 10 ζ 11 ζ 12 ζ 13 ζ 14 ζ,ζ 2,...,ζ 15,ζ 16 are the roots of the equation ζ 15 x 17 1 x 1 = x16 + x x +1=0. Gauss construction depends on two key ideas. 2

4 (1) Solution of quadratic equations by ruler and compass B A b M b a a B K b (o) x 2 = ax + b 2 : x 1 = KA, x 2 = KB A (p) x 2 + b 2 = ax : x 1 = AM, x 2 = BM 3

5 (2) A primitive root for the prime 17 Gauss observed that, modulo 17, all the residues are powers of 3 (which is a primitive root for the prime 17): k k k k

6 Solution of x 16 + x x +1=0 The 16 roots being ζ k, k =1, 2,..., 16, If we take then y 1 + y 2 = 1. y 1 = ζ + ζ 9 + ζ 13 + ζ 15 + ζ 16 + ζ 8 + ζ 4 + ζ 2, y 2 = ζ 3 + ζ 10 + ζ 5 + ζ 11 + ζ 14 + ζ 7 + ζ 12 + ζ 6, More importantly, the product y 1 y 2 does not depend on the choice of ζ. 5

7 The product y 1 y 2 y 1 = ζ + ζ 9 + ζ 13 + ζ 15 + ζ 16 + ζ 8 + ζ 4 + ζ 2, y 2 = ζ 3 + ζ 10 + ζ 5 + ζ 11 + ζ 14 + ζ 7 + ζ 12 + ζ 6. This table gives ζ h ζ k = ζ m, m h + k (mod 17) Each of 1, 2,...,16 appears exactly 4 times. y 1 + y 2 = ζ + ζ ζ 16 = 1, y 1 y 2 = 4(ζ + ζ ζ 16 )= 4. 6

8 A quadratic equation with roots y 1, y 2 y 1 + y 2 = ζ + ζ ζ 16 = 1, y 1 y 2 = 4(ζ + ζ ζ 16 )= 4. It follows that y 1 and y 2 are the roots of and are constructible. We may take y 1 = y 2 + y 4=0,, y 2 =

9 The second stage Recall y 1 = ζ + ζ 9 + ζ 13 + ζ 15 + ζ 16 + ζ 8 + ζ 4 + ζ 2, y 2 = ζ 3 + ζ 10 + ζ 5 + ζ 11 + ζ 14 + ζ 7 + ζ 12 + ζ 6. Now separate each of y 1, y 2 into two groups of four: Then, z 1 = ζ + ζ 13 + ζ 16 + ζ 4, z 2 = ζ 9 + ζ 15 + ζ 8 + ζ 2 ; z 3 = ζ 3 + ζ 5 + ζ 14 + ζ 12, z 4 = ζ 10 + ζ 11 + ζ 7 + ζ 6. z 1 z 2 = (ζ + ζ 13 + ζ 16 + ζ 4 )(ζ 9 + ζ 15 + ζ 8 + ζ 2 ) = ζ + ζ ζ 16 = 1; z 3 z 4 = 1. Numbers quadratic equation z 1,z 2 z 2 y 1 z 1=0 z 3,z 4 z 2 y 2 z 1=0 8

10 The third stage Further separating z 1 = ζ + ζ 13 + ζ 16 + ζ 4 into we obtain t 1 = ζ + ζ 16, t 2 = ζ 13 + ζ 4, t 1 + t 2 = z 1, t 1 t 2 = (ζ + ζ 16 )(ζ 13 + ζ 4 ) = ζ 14 + ζ 5 + ζ 12 + ζ 3 = z 3. Therefore, t 1 = ζ + ζ 16 and t 2 = ζ 13 + ζ 4 are the roots of and are constructible. t 2 z 1 t + z 3 =0, 9

11 The fourth stage Finally, ζ and ζ 16 are the roots of z 2 t 1 z +1=0, and ζ = t 1 + t , ζ 16 = t 1 t

12 An explicit construction of a regular 17-gon cos 2π 17 = 1 ( ζ + ζ 16 ) = ( ) Circle (O) with perpendicular diameters PQand RS. (1) A on OR with OA = 1 4 OP. (2) Construct the bisectors OB and OC of angle OAP. (3) D = OP C(A) and E = OP B(A) R A Q C O D B E P 11

13 An explicit construction (continued) (4) M = midpoint M of QD. (5) F = OS M(Q). (6) G on semicircle on OE, with OG = OF. (7) H = OP E(G). (8) P 1 H OP. A R P 1 Q M C O D B E H P F G S 12

14 Which regular n-gons are constructible? Gauss: A regular polygon of n sides can be constructed with ruler and compass only if and only if n is the product of distinct Fermat primes and a power of 2. A Fermat prime is a prime of the form F k := 2 2k +1. Examples: F 0 = =3, F 1 =2 2 +1=5, F 2 =2 4 +1=17, F 3 = = 257, F 4 = = are all prime numbers. Fermat had wrongly conjectured that F k is always prime. This was disproved by Euler: F 5 = = is divisible by

15 Construction of regular polygons According to H. S. M. Coxeter, Introduction to Geometry, p.27: Richelot and Schwendenheim constructed the regular 257-gon about O. Hermes spent ten years on the regular gon and deposited the manuscript in a large box in the University of Göttingen, where it may still be found. 14