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# Constructible polygons

(Redirected from Constructible odd-sided polygons)

## 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  αi , 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 .