Download Constructible Polygons Now that we have zeroed in on the

Constructible Polygons
Now that we have zeroed in on the constructible numbers, let’s apply the knowledge to
determine which regular polygons can be constructed.
Let’s determine which regular polygons are constructible.
First, we note that the problem is really just finding an angle of 360◦ /n, because if
we mark this angle and all its multiples around a circle, we’ll have constructed a regular
n-gon. So, for instance, when n = 6 we get a 60◦ angle, which we know we can construct
(equilateral triangle!), so we can construct a regular hexagon. We also know how to construct
the equilateral triangle, square, and regular pentagon.
Next, we note that if we can construct an angle of α, we can bisect this angle to get an
angle of α/2. So if we can construct an n-gon, we can construct a 2n-gon. So this allows us
to construct the regular octagon, decagon, dodecagon, 16-gon, 20-gon, 24-gon, and so forth.
It should also be clear that we can make the regulat 15-gon. Construct a triangle and
pentagon so that they share a vertex. Since the difference 2/5 − 1/3 = 1/15 there is a vertex
of the triangle and one of the pentagon that are 1/15 of the circle apart from each other. In
fact, if we can construct an a-gon and a b-gon, and a and b have no common factors, we can
use this trick to make an ab-gon.
So the first few numbers we don’t have yet are 7, 9, 11, 13, 14, 17, 18, 19, 21, . . . . Of
course, if we can get the heptagon (7-sided) we can also get the 14, 21, 28, 35, 42. But
not necessarily the 49—we can’t have both a and b equalling 7, because a and b can’t have
common factors! Similarly, if we can get the 9, we could get the 18, though not necessarily
the 27. And so forth.
Since the problem seems to come down to factors, let’s factor the number of sides we
want: n = 2a1 3a2 5a3 · · · . We know we can get all the factors of 2 we want by bisecting angles.
Which of the odd primes (and their powers) can we get?
One thing we (might) remember form MI-4 is that the roots of unity in the complex
plane form a regular polygon around the unity circle. In other words, the n complex number
roots of the equation z n = 1 are the vertices of a regular n-gon. So now the question
basically becomes: when can we solve this equation using nothing but addition, subtraction,
multiplication, division, and square roots?
Let’s try. If we take z 3 − 1 = 0 we can factor it as (z − 1)(z 2 + z + 1) = 0. Clearly z = 1
is a root, and the other two roots come from a quadratic—which can be solved by taking
square roots. So the 3-gon (triangle) can be constructed.
Try n = 4. Well, z 4 − 1 = (z 2 − 1)(z 2 + 1) = (z − 1)(z + 1)(z 2 + 1). Two real roots, 1
and −1, and two complex roots i and −i, giving us a square.
Try n = 5. This time, z 5 − 1 = (z − 1)(z 4 + z 3 + z 2 + z + 1). Now it may not look
like it, but the second term actually factors as (z 2 + φz + 1)(z 2 + (1 − φ)z + 1) where
φ = (1 + Sqrt[5])/2—which is a number that can be found from rationals by taking square
roots. Since both of these are quadratics, we can solve them with square roots. Note that the
solutions have square roots of things with square roots—they are in a degree four extension
of the rationals.
For n = 6 we get z 6 − 1 = (z 3 − 1)(z 3 + 1) = (z − 1)(z 2 + z + 1)(z + 1)(z 2 − z + 1). We’ve
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actually seen all but the very last term before, and the last term is a quadratic, so can be
solved. So as we already knew, we can construct a regular hexagon.
It looks like factoring these polynomials z n −1 is really imporant. Do they all break down
into quadratics like these have? Which ones factor into pieces we’ve seen before, and what
are the pieces that we haven’t seen before? Let’s make a definition. The nth cyclotomic
polynomial, denoted Φn (z) is the piece we haven’t seen before. We’ll start with n = 1 and
the polynomial z − 1, and n = 2 with z 2 − 1 = (z − 1)(z + 1). So we have:
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2
3
4
5
6
n
Φn (z) z − 1 z + 1 z 2 + z + 1 z 2 + 1 z 4 + z 3 + z 2 + z + 1 z 2 − z + 1
So far, we’ve managed to factor all of these into quadratic polynomials, perhaps (as in
the case of n = 5) with quadratic numbers as coefficients.
Now let’s try n = 7. Clearly z 7 − 1 = (z − 1)(z 6 + z 5 + z 4 + z 3 + z 2 + z + 1). The sixth
degree polynomial is new, so that must be Φ7 (z). And it is not factorable, even if we use
square roots, or square roots of square roots, or whatever! It is sixth degree, which means its
roots are in an extension of degree six—which is not a power of two! So the regular heptagon
is not constructible with compass and straightedge!
So now the question shifts to: which numbers—in particular which primes and powers of
primes—have cyclotomic polynomials that have degrees that are a power of two? It turns
out the answer to this is amazingly easy!
The beginning of the answer is: the degree of the cyclotomic polynomial is the number of
numbers between 1 and n that have no common factor with n. Why is that? Well, let a divide
evenly into n, say n = ab. Then we can factor z n −1 = (z a −1)(z ( n−a)+z ( n−2a)+· · ·+z a +1).
Since a < n, we have seen z a − 1 and its factors before. So we can eliminate from z n − 1 all
number which are divisor of n (and their multiples—all numbers that have some common
factor with n) and the only things left are the powers which have no common multiples with
n. How many of these are there? It’s the number of numbers between 1 and n which have
no common factor with n. This number is called φ(n) and is studied a lot in number theory.
Another way to think about all this is that we actually know all the roots of z n − 1 = 0.
They are 1(= cis(0)) and all the numbers cis(2π/n), cis(2 · 2π/n), cis(3 · 2π/n), . . ., upto
cis((n − 1) · 2π/n). But if the numerator shares a common factor with n, we can remove that
common factor from numerator and denominator, and we get something that came from an
earlier polygon. We are only left with the numerators that have no common factor with n,
and these leftovers must be the roots of the cyclotomic polynomial.
OK, so next question. For a prime or prime power, what is φ(pk )? The answer is that
since p is a prime, none of the numbers less than pk share a factor with it unless they are
multiple of p. There are p( k − 1) of those, so we learn that φ(pk ) = pk−1 (p − 1).
Now which of these are powers of two? Well, if k > 1 then there is a factor of p
involved, so it can’t be a power of two. So we need to find primes, p, where p − 1 is a
power of two. In other words, p = 2a + 1. But not every power of two works! For instance,
25 + 1 = 33 is not prime. In fact, you might know the factoring trick: if a is odd, then
xa + y a = (x + y)(xa−1 − xa−2 y + xa−3 y 2 − · · · + x2 y a−3 − xy a−2 + y a−1 ). We need a to be odd
so that the pluses and minuses come out right in the second factor. So since 2a + 1 = 2a + 1a
can be factored if a is odd—or itself has an odd factor—the number 2a + 1 can’t be a prime
unless a itself is a power of two!
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n
So let’s look at numbers Fn = 22 + 1, called the Fermat numbers. When n = 0 we
get 21 + 1 = 3, and we already know we can construct the 3-gon. With n = 1 we end up
with 22 + 1 = 5, and the pentagon is constructible. Next is 24 + 1 = 17. Amazingly, the
regular 17-gon is constructible! Next, for n = 3 we get 28 + 1 = 257 is prime, and (gulp!)
with enough patience you can construct—with compass and straightedge—a regular 257-gon.
And, yes, F4 = 65537 is prime, and the polygon can be constructed (there is a manuscript
that explains how to do so, which took the person who wrote it about ten years to complete).
At least in theory it can be constructed. If you actually tried, and you made the sides, say,
1 centimeter long, you’d have a figure with a perimeter of 65537. Since it pretty much a
slightly bumpy cirlce at this point, this circle would have a diameter of 65537/π, or over two
thousand centimeters—over twenty meters. That’s a big piece of paper!
Can we go on, in theory at least? Fermat thought so. After getting here, he looked
at F5 = 4294967297. This ten-digit number sure looks prime. In fact, Fermat conjectured
that all the Fermat numbers were prime. But it turns out he was wrong. In fact, F5 isn’t.
Euler first found a factor, 641, quite a few years later. These numbers get harder and harder
to factor, as each time n goes up by one the number of digits roughly doubles. But using
more modern techniques (take number theory!) you can show that many of these number
have factors, even if you can’t find them. In fact, modern mathematicians conjecture that
Fermat was as wrong as possible—that after F4 there are no more primes among the Fermat
numbers. But this hasn’t been proven!
So far, we’ve been able to prove that if an n-gon is constructible, then n must have the
form 2a f1 f2 . . . fk where the f ’s are all distinct Fermat primes. That is, if a number doesn’t
have this exact form, it cannot be constructed.
To complete the story, which already involves Euclid, Archimedes, Fermat, and Euler,
we’ll bring in the greatest mathematician of all time, Gauss. He finished it off by proving
that all the numbers of that form do have constructible polygons. Note that we didn’t show
that if you have a Fermat prime, whose cyclotomic polynomial has degree a power of two,
necessarily has a constructible polygon. For instance, there are fourth degree polynomials
whose roots can’t be found by just taking square roots—you might need to jump directly to
fourth roots or something—which would mess up constructibility. Gauss proved that doesn’t
happen, that if you have a cyclotomic polynomial that is a power of two it can always be
solved by square roots, and square roots of square roots, and square roots of those, etc. So
it does stay constructible. This final piece of the proof is a little too complicated for us.
So we’ve gone about as far as we can go with compass and straightedge. Care to break
the rules a little, and add an extra drawing tool to see how far it can get us?
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