@mpresic3: in case of possible interest, here is my presentation on the topic of impossible constructions, to my class of bright 8,9,10 year olds.
Remarks on impossible constructions:
I want to say something about why the only regular p-gons that can be constructed with
prime p, are for the Fermat primes p = 2^(2^n) + 1, or equivalently p – 1 = 2^(2^n).
First of all, note that a number of form 2^(nm)+1 where m is odd is never prime, because it
equals (2^n)^m + 1, and any number of form x^m+1 where m is odd can be factored.
E.g., you probably know the basic example of x^3+1 = (x+1)(x^2+x+1). Then x^5 +1 =
(x+1)(x^4+x^3+x^2+x+1), and so on…. Wait a minute, why am I going to so much
trouble? Probably you know the factor theorem from algebra implies that a polynomial
f(x) is divisible by x+1 if and only if f(-1) = 0. Since plugging x = -1 into x^m + 1 does
give zero when m is odd, x+1 always divides x^m+1 for m odd. So 2^n +1 always
divides (2^n)^m + 1 when m is odd, by taking x = 2^n.
Thus if a prime has form 2^n + 1, then it has form 2^(2^n) + 1. Hence it suffices to show
that every constructible prime must have form 2^n + 1.
Theorem: If p is a constructible prime, then it has form p = 2^n+1.
Description of proof: Notice that every construction involves finding points by
intersecting lines and circles. This is the key to understanding what constructions are
possible.
Assume we are in an Archimedean geometry, i.e. Euclid’s geometry plus the extra
axiom of Archimedes, that says any segment can be laid off repeatedly until it reaches
any other point. Then we can introduce real numbers as coordinates, by introducing a
pair of perpendicular axes in the plane. Then every point of the plane can be described
by a pair of real coordinates (x,y). Conversely we may assume that every pair of real
coordinates (x,y) represents a point of the plane, but they need not all be constructible.
We want to examine which ones among the many points of the plane are constructible
by ruler and compass.
We start from only two points (0,0) and (1,0), and ask what points can be constructed
from these. E.g. we can lay off as many copies of these as we wish along the x axis, so
we get all points of form (n,m) where n,m are integers. Then we can also subdivide the
interval between (0,0) and (1,0) into n equal parts for every natural number n, and then
lay off copies of those, so we also get all rational points on the x-axis of form (n/m, 0)
where n,m, are integers, and m ≠ 0.
Since we can construct the perpendicular to the x-axis we also get the y axis, and then
we can lay off rational points on the y axis. Now we can construct perpendiculars to
both x and y axes and intersect them, so we also get all “rational points” of the plane,
i.e. all points of form (n/m, a/b), with a,b,n,m, integers and mb ≠ 0. What else can we
get?
Well we also get points that arise from intersecting lines determined by two rational
points, with other such lines, or with circles with rational centers and a rational point on
the circumference, or two such circles.
Now intersecting sets defined by equations means solving the equations
simultaneously, so we want to know what kind of numbers occur as simultaneous
solutions of equations of lines and circles. In fact all we need to know is the degree of
the equations. Solving two linear equations with rational coefficients just gives rational
solutions, so intersecting such rational lines does not give any new points. Moreover
intersecting two rational circles gives two points which are also obtained by intersecting
one circle with a rational line (subtract the equation for the two circles to get the
equation of the line). So we look at points obtained by intersecting rational lines and
rational circles.
An equation for a line through two rational points looks like ax+by = c, where a,b,c, are
rational numbers. An equation for a circle determined by a rational center and rational
point on its circumference looks like (x-u)^2 + (y-v)^2 = w^2, where u,v,w, are rational
numbers. To solve ax+by = c, and (x-u)^2 + (y-v)^2 = w^2, simultaneously, we solve
the linear equation for y, getting: by = c – ax, then y = c/b – ax/b, and substitute this for y
in the quadratic equation. The result is some complicated quadratic equation in x,
Maybe (x-u)^2 + ([c/b – ax/b] – v)^2 = w^2. I don’t care what it is exactly, as I am only
interested in its degree, or the fact that it is quadratic, i.e. of degree two. We call
solutions of quadratic equations with rational coefficients “quadratic” numbers.
Conversely we know how to use Pythagoras to solve quadratic equations whose
coefficients are segments, or numbers, that we have already constructed. So we can
construct points in the plane whose coefficients are solutions of quadratic equations with
rational coefficients, and any algebraic combination of those numbers. E.g. we can
construct the point (sqrt(2)-sqrt(5), 1+sqrt(7/3)). Similarly we can construct all quadratic
points, i.e. points whose coefficients are quadratic numbers. These quadratic points are
the points that only require one “quadratic step” to construct, i.e. one use of the
compass.
What next? Well once we have those quadratic guys we can intersect more lines and
circles. So now we are solving quadratic equations whose coefficients are quadratic
numbers. We call these biquadratic numbers. Thus we can construct all biquadratic
points I claim the solution of such an equation is also the solution of an equation of
degree 4, but with rational coefficients. I.e. we claim all biquadratic numbers are also
quartic numbers, or degree 4 numbers.
E.g. if we have an equation like X^2 – sqrt(3).X + 1 = 0, with quadratic coefficients, we
can rewrite it as X^2 + 1 = sqrt(3)X, and square both sides, to get (X^2+1)^2 = 3X^2, or
X^4 + 2XC^2 + 1 = 3X^2, which becomes X^4 –X^2 + 1 = 0. Thus our number X
becomes a solution of a 4th degree equation with rational coefficients. In general every
biquadratic number is a quartic number. The idea is that biquadratic numbers are those
that only require two quadratic steps to construct, or two uses of the compass. As in
this example, they are all quartic numbers, i.e. they satisfy degree 4 equations with
rational coefficients. Now what about points that require three quadratic steps to
construct?
If we have X^2 – 2^(1/4)X -3 = 0, where the coefficient 2^(1/4) is biquadratic hence
quartic, we get X^2 -3 = 2^(1/4)X, and raising both sides to the 4th power gives (X^2-
3)^4 = 2X^4, which is an equation of degree 8 with rational coefficients. So the
triquadratic number X, which requires three uses of the compass, is of degree 8, or an
“octic” number.
I don’t know how to make this entirely clear, but as we go on what happens is that the
points we get are solutions of equations of degree 2,4,8,16,32,….., i.e. degree 2^n, with
rational coefficients. And that is all we can get. More precisely, a point that can be
constructed in n steps satisfies an equation with rational coefficients, and has degree
dividing 2^n, (since steps not using the compass have degree one). It follows that a
point (x,y) cannot be constructed unless its coefficients are solutions of an equation of
some degree 2^k with rational coefficients.
Now this applies also to complex numbers x+iy corresponding to our points. I.e, if (x,y)
is a constructible point, then the complex number z = x+iy must satisfy an equation of
degree 2^n for some n. But the first vertex on the unit circle of a regular p - gon, after
the point (1,0), is exactly “1/p^ th” of the way around the circle. Since multiplying
complex numbers adds their angles and multiplies their lengths, it is a complex solution
of the equation z^p – 1 = 0.
Now this equation factors as z^p – 1 = (z-1)(z^(p-1) +….+z + 1) = 0, and since z=1 is
the only solution of the first factor, the complex number we want is a solution of the
second factor, which has degree p-1. For that point, which gives the first vertex of the
regular p-gon, to be constructible, we must have p-1 = 2^n, i.e. p = 2^n+1. Then of
course since p is prime, we have seen it must have form 2^(2^k) + 1, i.e. it must be a
Fermat prime.
These ideas are usually taught in a college abstract algebra course as an application of
linear algebra. One possible source is the book Abstract algebra, a geometric
approach, by Theodore Shifrin, or my math 4000 notes #4f, the last couple lectures, on
my web page at UGA.
http://www.math.uga.edu/~roy/
