This site is supported by donations to The OEIS Foundation.

# 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 $n\,$ -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, {${\rm {F_{0}}}\,$ , ${\rm {F_{1}}}\,$ , ${\rm {F_{2}}}\,$ , ${\rm {F_{3}}}\,$ , ${\rm {F_{4}}}\,$ } = {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 $2^{32}-1\,$ (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 $n\,$ for which a regular $n\,$ -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$(n),\,n\,\geq \,0\,$ )

{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 $3\,=\,{\rm {F}}_{0}\,$ 2 5 $5\,=\,{\rm {F}}_{1}\,$ 3 15 $3\,\cdot \,5\,=\,{\rm {F}}_{0}{\rm {F}}_{1}\,$ 4 17 $17\,=\,{\rm {F}}_{2}\,$ 5 51 $3\,\cdot \,17\,=\,{\rm {F}}_{0}{\rm {F}}_{2}\,$ 6 85 $5\,\cdot \,17\,=\,{\rm {F}}_{1}{\rm {F}}_{2}\,$ 7 255 $3\,\cdot \,5\,\cdot \,17\,=\,{\rm {F}}_{0}{\rm {F}}_{1}{\rm {F}}_{2}\,$ 8 257 $257\,=\,{\rm {F}}_{3}\,$ 9 771 $3\,\cdot \,257\,=\,{\rm {F}}_{0}{\rm {F}}_{3}\,$ 10 1285 $5\,\cdot \,257\,=\,{\rm {F}}_{1}{\rm {F}}_{3}\,$ 11 3855 $3\,\cdot \,5\,\cdot \,257\,=\,{\rm {F}}_{0}{\rm {F}}_{1}{\rm {F}}_{3}\,$ 12 4369 $17\,\cdot \,257\,=\,{\rm {F}}_{2}{\rm {F}}_{3}\,$ 13 13107 $3\,\cdot \,17\,\cdot \,257\,=\,{\rm {F}}_{0}{\rm {F}}_{2}{\rm {F}}_{3}\,$ 14 21845 $5\,\cdot \,17\,\cdot \,257\,=\,{\rm {F}}_{1}{\rm {F}}_{2}{\rm {F}}_{3}\,$ 15 65535 $3\,\cdot \,5\,\cdot \,17\,\cdot \,257\,=\,{\rm {F}}_{0}{\rm {F}}_{1}{\rm {F}}_{2}{\rm {F}}_{3}\,$ 16 65537 $65537\,=\,{\rm {F}}_{4}\,$ 17 196611 $3\,\cdot \,65537\,=\,{\rm {F}}_{0}{\rm {F}}_{4}\,$ 18 327685 $5\,\cdot \,65537\,=\,{\rm {F}}_{1}{\rm {F}}_{4}\,$ 19 983055 $3\,\cdot \,5\,\cdot \,65537\,=\,{\rm {F}}_{0}{\rm {F}}_{1}{\rm {F}}_{4}\,$ 20 1114129 $17\,\cdot \,65537\,=\,{\rm {F}}_{2}{\rm {F}}_{4}\,$ 21 3342387 $3\,\cdot \,17\,\cdot \,65537\,=\,{\rm {F}}_{0}{\rm {F}}_{2}{\rm {F}}_{4}\,$ 22 5570645 $5\,\cdot \,17\,\cdot \,65537\,=\,{\rm {F}}_{1}{\rm {F}}_{2}{\rm {F}}_{4}\,$ 23 16711935 $3\,\cdot \,5\,\cdot \,17\,\cdot \,65537\,=\,{\rm {F}}_{0}{\rm {F}}_{1}{\rm {F}}_{2}{\rm {F}}_{4}\,$ 24 16843009 $257\,\cdot \,65537\,=\,{\rm {F}}_{3}{\rm {F}}_{4}\,$ 25 50529027 $3\,\cdot \,257\,\cdot \,65537\,=\,{\rm {F}}_{0}{\rm {F}}_{3}{\rm {F}}_{4}\,$ 26 84215045 $5\,\cdot \,257\,\cdot \,65537\,=\,{\rm {F}}_{1}{\rm {F}}_{3}{\rm {F}}_{4}\,$ 27 252645135 $3\,\cdot \,5\,\cdot \,257\,\cdot \,65537\,=\,{\rm {F}}_{0}{\rm {F}}_{1}{\rm {F}}_{3}{\rm {F}}_{4}\,$ 28 286331153 $17\,\cdot \,257\,\cdot \,65537\,=\,{\rm {F}}_{2}{\rm {F}}_{3}{\rm {F}}_{4}\,$ 29 858993459 $3\,\cdot \,17\,\cdot \,257\,\cdot \,65537\,=\,{\rm {F}}_{0}{\rm {F}}_{2}{\rm {F}}_{3}{\rm {F}}_{4}\,$ 30 1431655765 $5\,\cdot \,17\,\cdot \,257\,\cdot \,65537\,=\,{\rm {F}}_{1}{\rm {F}}_{2}{\rm {F}}_{3}{\rm {F}}_{4}\,$ 31 4294967295 $3\,\cdot \,5\,\cdot \,17\,\cdot \,257\,\cdot \,65537\,=\,{\rm {F}}_{0}{\rm {F}}_{1}{\rm {F}}_{2}{\rm {F}}_{3}{\rm {F}}_{4}\,$ 32 4294967297* 4294967297* $\,=\,{\rm {F}}_{5}\,$ * ${\rm {F}}_{5}\,$ = 4294967297 = 641 * 6700417 is the first composite Fermat number.

Rows $2^{n},\,n\,\geq \,0,\,$ give $F_{n}\,$ , 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 $\{F_{0}\}\,=\,\{3\};\,$ • Inductive hypothesis: Assume all rows from $n\,=\,0\,$ to $2^{n}-1\,$ are products of distinct Fermat numbers in $\{{\rm {F}}_{0},\ldots ,{\rm {F}}_{n-1}\};\,$ .

If row $k,\,0\,\leq \,k\,\leq \,2^{n}-1,\,$ 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 $2^{n}+k,\,0\,\leq \,k\,\leq \,2^{n}-1,\,$ is

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