Constructible polygons

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Constructible polygons (with straightedge and compass)

A constructible polygon is a regular polygon that can be constructed with straightedge and compass.

Carl Friedrich Gauss proved the constructibility of the regular 17-gon in 1796. Five years later, he developed the theory of Gaussian periods in his Disquisitiones Arithmeticae. This theory allowed him to formulate a sufficient condition for the constructibility of regular polygons

A regular

n

-gon can be constructed with straightedge and compass if

n

is the product of a power of 2 and any number of distinct Fermat primes.

Gauss stated without proof that this condition was also necessary, but never published his proof. A full proof of necessity was given by Pierre Wantzel in 1837.

Constructible odd-sided polygons (with straightedge and compass)

Since there are 5 known Fermat primes,

{ F0 , F1 , F2 , F3 , F4 }

= {3, 5, 17, 257, 65537} (see A019434), then there are

     

( 5 0 ) + ( 5 1 ) + ( 5 2 ) + ( 5 3 ) + ( 5 4 ) + ( 5 5 )  =  1 + 5 + 10 + 10 + 5 + 1  =  32  =  2 5

known constructible odd-sided polygons (actually 31 since a polygon has at least 3 sides).

A004729 Divisors of 2 32  −  1 (polygons with an odd number of sides constructible with ruler and compass).

{1, 3, 5, 15, 17, 51, 85, 255, 257, 771, 1285, 3855, 4369, 13107, 21845, 65535, 65537, 196611, 327685, 983055, 1114129, 3342387, 5570645, 16711935, 16843009, 50529027, 84215045, 252645135, 286331153, 858993459, 1431655765, 4294967295}

A045544 Odd values of

n

for which a regular

n

-gon can be constructed by compass and straightedge.

{3, 5, 15, 17, 51, 85, 255, 257, 771, 1285, 3855, 4369, 13107, 21845, 65535, 65537, 196611, 327685, 983055, 1114129, 3342387, 5570645, 16711935, 16843009, 50529027, 84215045, 252645135, 286331153, 858993459, 1431655765, 4294967295, ?}

If one interprets the rows of Sierpiński’s triangle as a binary number, the first 32 rows give the 32 products of distinct Fermat primes, 1 (empty product) for row 0 and the 31 non-empty products of distinct Fermat primes.

A001317 Sierpiński’s triangle (Pascal’s triangle mod 2) rows converted to decimal.

{1, 3, 5, 15, 17, 51, 85, 255, 257, 771, 1285, 3855, 4369, 13107, 21845, 65535, 65537, 196611, 327685, 983055, 1114129, 3342387, 5570645, 16711935, 16843009, 50529027, 84215045, 252645135, 286331153, 858993459, 1431655765, 4294967295, 4294967297, ...}

Sierpiński’s triangle rows

n	[row(n)]2	[row(n)]10	Factorization into Fermat numbers
0	1	1
1	11	3	(11)2 = 3 = F0
2	101	5	(101)2 = 5 = F1
3	1111	15	(11)2 ⋅ (101)2 = 3 ⋅ 5 = F0 F1
4	10001	17	(10001)2 = 17 = F2
5	110011	51	(11)2 ⋅ (10001)2 = 3 ⋅ 17 = F0 F2
6	1010101	85	(101)2 ⋅ (10001)2 = 5 ⋅ 17 = F1 F2
7	11111111	255	(11)2 ⋅ (101)2 ⋅ (10001)2 = 3 ⋅ 5 ⋅ 17 = F0 F1 F2
8	100000001	257	(100000001)2 = 257 = F3
9	1100000011	771	(11)2 ⋅ (100000001)2 = 3 ⋅ 257 = F0 F3
10	10100000101	1285	(101)2 ⋅ (100000001)2 = 5 ⋅ 257 = F1 F3
11	111100001111	3855	(11)2 ⋅ (101)2 ⋅ (100000001)2 = 3 ⋅ 5 ⋅ 257 = F0 F1 F3
12	1000100010001	4369	(10001)2 ⋅ (100000001)2 = 17 ⋅ 257 = F2 F3
13	11001100110011	13107	(11)2 ⋅ (10001)2 ⋅ (100000001)2 = 3 ⋅ 17 ⋅ 257 = F0 F2 F3
14	101010101010101	21845	(101)2 ⋅ (10001)2 ⋅ (100000001)2 = 5 ⋅ 17 ⋅ 257 = F1 F2 F3
15	1111111111111111	65535	(11)2 ⋅ (101)2 ⋅ (10001)2 ⋅ (100000001)2 = 3 ⋅ 5 ⋅ 17 ⋅ 257 = F0 F1 F2 F3
16	10000000000000001	65537	(10000000000000001)2 = 65537 = F4
17	110000000000000011	196611	(11)2 ⋅ (10000000000000001)2 = 3 ⋅ 65537 = F0 F4
18	1010000000000000101	327685	(101)2 ⋅ (10000000000000001)2 = 5 ⋅ 65537 = F1 F4
19	11110000000000001111	983055	(11)2 ⋅ (101)2 ⋅ (10000000000000001)2 = 3 ⋅ 5 ⋅ 65537 = F0 F1 F4
20	100010000000000010001	1114129	(10001)2 ⋅ (10000000000000001)2 = 17 ⋅ 65537 = F2 F4
21	1100110000000000110011	3342387	(11)2 ⋅ (10001)2 ⋅ (10000000000000001)2 = 3 ⋅ 17 ⋅ 65537 = F0 F2 F4
22	10101010000000001010101	5570645	(101)2 ⋅ (10001)2 ⋅ (10000000000000001)2 = 5 ⋅ 17 ⋅ 65537 = F1 F2 F4
23	111111110000000011111111	16711935	(11)2 ⋅ (101)2 ⋅ (10001)2 ⋅ (10000000000000001)2 = 3 ⋅ 5 ⋅ 17 ⋅ 65537 = F0 F1 F2 F4
24	1000000010000000100000001	16843009	(100000001)2 = (10000000000000001)2 = 257 ⋅ 65537 = F3 F4
25	11000000110000001100000011	50529027	(11)2 ⋅ (100000001)2 = (10000000000000001)2 = 3 ⋅ 257 ⋅ 65537 = F0 F3 F4
26	101000001010000010100000101	84215045	(101)2 ⋅ (100000001)2 = (10000000000000001)2 = 5 ⋅ 257 ⋅ 65537 = F1 F3 F4
27	1111000011110000111100001111	252645135	(11)2 ⋅ (101)2 ⋅ (100000001)2 = (10000000000000001)2 = 3 ⋅ 5 ⋅ 257 ⋅ 65537 = F0 F1 F3 F4
28	10001000100010001000100010001	286331153	(10001)2 ⋅ (100000001)2 = (10000000000000001)2 = 17 ⋅ 257 ⋅ 65537 = F2 F3 F4
29	110011001100110011001100110011	858993459	(11)2 ⋅ (10001)2 ⋅ (100000001)2 = (10000000000000001)2 = 3 ⋅ 17 ⋅ 257 ⋅ 65537 = F0 F2 F3 F4
30	1010101010101010101010101010101	1431655765	(101)2 ⋅ (10001)2 ⋅ (100000001)2 = (10000000000000001)2 = 5 ⋅ 17 ⋅ 257 ⋅ 65537 = F1 F2 F3 F4
31	11111111111111111111111111111111	4294967295	(11)2 ⋅ (101)2 ⋅ (10001)2 ⋅ (100000001)2 = (10000000000000001)2 = 3 ⋅ 5 ⋅ 17 ⋅ 257 ⋅ 65537 = F0 F1 F2 F3 F4
32	100000000000000000000000000000001	4294967297*	(100000000000000000000000000000001)2 = 4294967297* = F5

* F5 = 4294967297 = 641 ⋅ 6700417 is the first composite Fermat number. From here on no other Fermat number is known to be prime.
Rows

2 n, n   ≥   0,

give

Fn

, the

n

^th Fermat number. For

n =

1 to 31, we get all the non-empty products of the 5 known Fermat primes, giving the number of sides of constructible odd-sided polygons. If there are no other Fermat primes, there are then no more constructible (with straightedge and compass) odd-sided polygons.

It can easily be proved by induction that the rows of Sierpinski’s triangle interpreted as a binary number are products of distinct Fermat numbers.

Inductive proof:

• Base case: For rows

  n =

  0 and 1, we get 1 (empty product) and 3 respectively, the products of distinct Fermat numbers in

  {F0 } = {3}

  ;

• Inductive hypothesis: Assume all rows from

  n = 0

  to

  2 n  −  1

  are products of distinct Fermat numbers in

  {F0 , ..., Fn  − 1}

  .

If row

k, 0   ≤   k   ≤   2 n  −  1,

is

row (k )  =    n  − 1 ∏   i  = 0    Fi   α i , αi ∈ {0, 1},

then by the fractal property of Sierpinski’s triangle, row

2 n + k, 0   ≤   k   ≤   2 n  −  1,

is

row (2 n + k )  =  row (k ) Fn .