Can you construct a regular pentadecagon (15-sided) polygon with compass and straightedge? What’s the restriction on *n*≥3 such that a regular *n*-gon is constructible?

Constructing an equilateral triangle, *n*=3, is very simple, as are constructing a square, *n*=4, and a hexagon, *n*=6. The method of constructing a regular pentagon, *n*=5, has been known since the days of Euclid.

In addition, if one can construct a regular *n*-gon, then by bisecting its sides one can construct a 2*n*-gon; by repeating, the polygons are thus constructible for 4*n*, 8*n*, etc.

Further, suppose that we can construct a regular *n*-gon and *d*≥3 is a divisor of *n*. Then by connecting every vertices of the *n*-gon, one produces a *d*-gon. So, suppose that the regular *k*-gon is **not** constructible. Then we can conclude that a polygon with a number of sides that is any multiple of *k* is also not constructible.

Now, suppose we have that a regular *a*-gon is constructible, and the regular *b*-gon is also constructible, with *a* and *b* coprime (gcd(*a*,*b*)=1). Suppose we construct these two polygons inscribed in the same circle with one (and only one, since *a* and *b* are coprime) vertex in common. Then imagine the regular polygon with *ab* sides. One can form a regular *a*-gon by connecting every *b* vertices, and a regular *b*-gon by connecting every *a* vertices. Now, Bézout’s identity tells us that for any two non-zero integers *a* and *b*, there exist integers *u* and *v* such that , which means that for our coprime *a* and *b*, there exist integers *u* and *v* for which , and thus there exist integer multiples of *a* and *b* which differ by exactly one. This means that in our construction of the *a*-gon and *b*-gon in the same circle with common vertex, there exist a vertex of the *a*-gon and a vertex of the *b*-gon which form one side of the *ab*-gon, and from that side and the circle in which the polygons are inscribed, one can construct the entire *ab*-gon. Thus, if the regular *a*-gon and regular *b*-gon are both constructible with *a* and *b* coprime, then the regular *ab*-gon is constructible.

So since 15=3*5, and the equilateral triangle and pentagon are both constructible, the regular pentadecagon is thus constructible. To use this as an example of the argument given above, we have that 3*2+5*(-1)=1. Thus, traveling around from the common vertex, if we take two sides of the pentagon and one side of the triangle, the resulting vertices are the sixth and fifth around of the pentadecagon, respectively, and so they form one side of the pentadecagon.

The question of what values of *n* allow the regular *n*-gon to be constructible was studied by many mathematicians; Gauss developed a sufficient condition in 1796 which he conjectured was also necessary; Pierre Wantzel later proved that it was so. The proof arises from analytic geometry, and is best expressed in terms of field theory and, in particular, quadratic extensions. The result, however, is that the regular *n*-gon is constructible if and only if *φ*(*n*) is a power of 2, where *φ*(*n*) is Euler’s totient function. Our result about *ab* with coprime *a* and *b* thus follows from *φ*(*n*) being a multiplicative function.

Since for prime *p*, we see that , again confirming that if *n* gives us a constructible polygon, *n* times any power of two also gives us a constructible polygon. As for prime *p*>2, we see that is divisible by *p* for *k*>1, and is thus not a power of 2 in that case; so, no odd prime may appear as a factor with multiplicity greater than 1; thus the enneagon, *n*=9=3^{2}, is not constructible. Further, a prime factor *p* will be able to give a power of 2 only if is a power of 2, which means that *p* is of the form . I showed a proof last week that the only primes of this form are the Fermat primes, primes of the form . Thus, the above is equivalent to Gauss’ phrasing that a regular *n*-gon is constructible with compass and straightedge if (and only if) *n* is the product of a power of 2 (including 2^{0}=1) and any number (including zero) of distinct Fermat primes.

So we see that the first few constructible regular polygons are

*n*=3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, …

while the first few non-constructible regular polygons are

*n*=7, 9, 11, 13, 14, 18, 19, …

Since the only known Fermat primes are *F*_{0}=3, *F*_{1}=5, *F*_{2}=17, *F*_{3}=257, and *F*_{4}=65537, their product, 4,294,967,295, is thus the largest odd number of sides for which the regular polygon is known to be constructible.

Tags: Bézout's Identity, Compass and Straightedge, Construction, Fermat Prime, Gauss, Math, Monday Math, Regular Polygon, Totient

## Leave a Reply