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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 17gon 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 gon can be constructed with straightedge and compass if 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 oddsided 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
() + () + () + () + () + () = 1 + 5 + 10 + 10 + 5 + 1 = 32 = 2 5 
known constructible oddsided 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
for which a regular
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 nonempty 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



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
give
, the
^{th} Fermat number.
For
1 to
31, we get all the nonempty products of the
5 known
Fermat primes, giving the number of sides of
constructible oddsided polygons. If there are no other Fermat primes, there are then no more constructible (with straightedge and compass) oddsided 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 0 and 1, we get 1 (empty product) and 3 respectively, the products of distinct Fermat numbers in ;
 Inductive hypothesis: Assume all rows from to are products of distinct Fermat numbers in .
If row
is

row (k ) = 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 . 