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# Constructible 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 $\scriptstyle n \,$-gon can be constructed with straightedge and compass if $\scriptstyle n \,$ 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, {$\scriptstyle {\rm F_0} \,$, $\scriptstyle {\rm F_1} \,$, $\scriptstyle {\rm F_2} \,$, $\scriptstyle {\rm F_3} \,$, $\scriptstyle {\rm F_4} \,$} = {3, 5, 17, 257, 65537}, (Cf. A019434) then there are

$\binom{5}{0} + \binom{5}{1} + \binom{5}{2} + \binom{5}{3} + \binom{5}{4} + \binom{5}{5} = 1 + 5 + 10 + 10 + 5 + 1 = 32 = 2^5 \,$

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

Divisors of $\scriptstyle 2^{32} - 1 \,$ (polygons with an odd number of sides constructible with ruler and compass). (Cf. A004729)

{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}

Odd values of $\scriptstyle n \,$ for which a regular $\scriptstyle n \,$-gon can be constructed by compass and straightedge. (Cf. A045544)

{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 Sierpinski'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.

Sierpinski's triangle (Pascal's triangle mod 2) rows converted to decimal. (Cf. A001317$\scriptstyle (n),\, n \,\ge\, 0 \,$)

{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, ...}
Sierpinski's triangle rows converted to base 10
$n \,$ $[{\rm Row}(n)]_{10} \,$ Factorization into Fermat numbers
0 1
1 3 $\scriptstyle 3 \,=\, {\rm F}_0 \,$
2 5 $\scriptstyle 5 \,=\, {\rm F}_1 \,$
3 15 $\scriptstyle 3 \,\cdot\, 5 \,=\, {\rm F}_0 {\rm F}_1 \,$
4 17 $\scriptstyle 17 \,=\, {\rm F}_2 \,$
5 51 $\scriptstyle 3 \,\cdot\, 17 \,=\, {\rm F}_0 {\rm F}_2 \,$
6 85 $\scriptstyle 5 \,\cdot\, 17 \,=\, {\rm F}_1 {\rm F}_2 \,$
7 255 $\scriptstyle 3 \,\cdot\, 5 \,\cdot\, 17 \,=\, {\rm F}_0 {\rm F}_1 {\rm F}_2 \,$
8 257 $\scriptstyle 257 \,=\, {\rm F}_3 \,$
9 771 $\scriptstyle 3 \,\cdot\, 257 \,=\, {\rm F}_0 {\rm F}_3 \,$
10 1285 $\scriptstyle 5 \,\cdot\, 257 \,=\, {\rm F}_1 {\rm F}_3 \,$
11 3855 $\scriptstyle 3 \,\cdot\, 5 \,\cdot\, 257 \,=\, {\rm F}_0 {\rm F}_1 {\rm F}_3 \,$
12 4369 $\scriptstyle 17 \,\cdot\, 257 \,=\, {\rm F}_2 {\rm F}_3 \,$
13 13107 $\scriptstyle 3 \,\cdot\, 17 \,\cdot\, 257 \,=\, {\rm F}_0 {\rm F}_2 {\rm F}_3 \,$
14 21845 $\scriptstyle 5 \,\cdot\, 17 \,\cdot\, 257 \,=\, {\rm F}_1 {\rm F}_2 {\rm F}_3 \,$
15 65535 $\scriptstyle 3 \,\cdot\, 5 \,\cdot\, 17 \,\cdot\, 257 \,=\, {\rm F}_0 {\rm F}_1 {\rm F}_2 {\rm F}_3 \,$
16 65537 $\scriptstyle 65537 \,=\, {\rm F}_4 \,$
17 196611 $\scriptstyle 3 \,\cdot\, 65537 \,=\, {\rm F}_0 {\rm F}_4 \,$
18 327685 $\scriptstyle 5 \,\cdot\, 65537 \,=\, {\rm F}_1 {\rm F}_4 \,$
19 983055 $\scriptstyle 3 \,\cdot\, 5 \,\cdot\, 65537 \,=\, {\rm F}_0 {\rm F}_1 {\rm F}_4 \,$
20 1114129 $\scriptstyle 17 \,\cdot\, 65537 \,=\, {\rm F}_2 {\rm F}_4 \,$
21 3342387 $\scriptstyle 3 \,\cdot\, 17 \,\cdot\, 65537 \,=\, {\rm F}_0 {\rm F}_2 {\rm F}_4 \,$
22 5570645 $\scriptstyle 5 \,\cdot\, 17 \,\cdot\, 65537 \,=\, {\rm F}_1 {\rm F}_2 {\rm F}_4 \,$
23 16711935 $\scriptstyle 3 \,\cdot\, 5 \,\cdot\, 17 \,\cdot\, 65537 \,=\, {\rm F}_0 {\rm F}_1 {\rm F}_2 {\rm F}_4 \,$
24 16843009 $\scriptstyle 257 \,\cdot\, 65537 \,=\, {\rm F}_3 {\rm F}_4 \,$
25 50529027 $\scriptstyle 3 \,\cdot\, 257 \,\cdot\, 65537 \,=\, {\rm F}_0 {\rm F}_3 {\rm F}_4 \,$
26 84215045 $\scriptstyle 5 \,\cdot\, 257 \,\cdot\, 65537 \,=\, {\rm F}_1 {\rm F}_3 {\rm F}_4 \,$
27 252645135 $\scriptstyle 3 \,\cdot\, 5 \,\cdot\, 257 \,\cdot\, 65537 \,=\, {\rm F}_0 {\rm F}_1 {\rm F}_3 {\rm F}_4 \,$
28 286331153 $\scriptstyle 17 \,\cdot\, 257 \,\cdot\, 65537 \,=\, {\rm F}_2 {\rm F}_3 {\rm F}_4 \,$
29 858993459 $\scriptstyle 3 \,\cdot\, 17 \,\cdot\, 257 \,\cdot\, 65537 \,=\, {\rm F}_0 {\rm F}_2 {\rm F}_3 {\rm F}_4 \,$
30 1431655765 $\scriptstyle 5 \,\cdot\, 17 \,\cdot\, 257 \,\cdot\, 65537 \,=\, {\rm F}_1 {\rm F}_2 {\rm F}_3 {\rm F}_4 \,$
31 4294967295 $\scriptstyle 3 \,\cdot\, 5 \,\cdot\, 17 \,\cdot\, 257 \,\cdot\, 65537 \,=\, {\rm F}_0 {\rm F}_1 {\rm F}_2 {\rm F}_3 {\rm F}_4 \,$
32 4294967297* 4294967297* $\scriptstyle \,=\, {\rm F}_5 \,$
* $\scriptstyle {\rm F}_5 \,$ = 4294967297 = 641 * 6700417 is the first composite Fermat number.

Rows $\scriptstyle 2^n,\, n \,\ge\, 0, \,$ give $\scriptstyle F_{n} \,$, the $\scriptstyle n \,$th Fermat number.

For ($\scriptstyle n \,$ = 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 $\scriptstyle n \,$ = 0 and 1, we get 1 (empty product) and 3 respectively, the products of distinct Fermat numbers in $\scriptstyle \{ F_0 \} \,=\, \{ 3 \}; \,$
• Inductive hypothesis: Assume all rows from $\scriptstyle n \,=\, 0 \,$ to $\scriptstyle 2^n - 1 \,$ are products of distinct Fermat numbers in $\scriptstyle \{ {\rm F}_0, \ldots, {\rm F}_{n-1} \}; \,$.

If row $\scriptstyle k,\, 0 \,\le\, k \,\le\, 2^n - 1, \,$ is

${\rm row}(k) = \prod_{i=0}^{n-1} {{\rm F}_i}^{\alpha_i},\, {\alpha_i} \in \{ 0,\, 1 \}, \,$

then by the fractal property of Sierpinski's triangle, row $\scriptstyle 2^n + k,\, 0 \,\le\, k \,\le\, 2^n - 1, \,$ is

${\rm row}(2^n + k) = {\rm row}(k) ~ {\rm F}_n. \,$