A FEW EXCURSIONS THROUGH PRIMES, POLYNOMIALS, AND POLYGONS
Take a few minutes and see how many prime numbers you can find that are one less than
some base raised to a power (that is, of the form an-1).
First, note that a6-1 = (a-1)(a5+a4+a3+a2+a+1) [Check this out algebraically.]
This indicates that there are no primes of the form a6-1 unless a=2 since a-1 divides a6-1
regardless of what the base a is, and unless a=2 (in which case a-1=1), this means that a6-1 is
composite.
This result generalizes to other powers. an-1=(a-1)(an-1+an-2+an-3+ … +a2+a+1), so a-1 divides
an-1 which means an-1 is composite unless a=2 (again in which case a-1=1, so this does not
give a factorization of an-1 that involves two smaller factors).
Now unless the exponent n is prime, even 2n-1 cannot be prime. For if n=jk, then we have
2jk-1=(2j)k-1=(2j-1)[(2j)k-1+(2j)k-2+(2j)k-3+…+ (2j)2+(2j)+1]
Primes of the form 2p-1 (where p is prime) are called “Mersenne Primes.” A few of these are:
22-1 = 3
23-1 = 7
25-1 = 31
27-1 = 127
However, 211-1= 89·23 so is not prime.
213-1 is prime
217-1 is prime
219-1 is prime
However neither 223-1 nor 229-1 is prime.
There are about 44 known Mersenne primes at this time.
Mersenne primes are useful in the search for perfect numbers. A perfect number is one that is
the sum of its proper divisors such as 6=1+2+3 and 28=1+2+4+7+14. Note that
6=2٠3=21٠(22-1) and 28=4٠7=22٠(23-1). The next perfect number 496=16٠31=24٠(25-1).
All even perfect numbers are products of the form 2p-1٠(2p-1) where 2p-1 is a Mersenne prime,
and all Mersenne primes yield a perfect number of the form 2p-1٠(2p-1). As you can imagine,
it is easier to search for Mersenne primes than it is to search for perfect numbers directly.
Incidentally, it is not known if there are any odd perfect numbers.
Now let’s turn our attention to a different type of prime: an+1. See if you can find some
primes of this form.
You may have discovered some primes of the form an+1 such as 37=62+1 and 17=24+1. Let’s
explore the form an+1 a bit and see if we can discover any rules for when such a form might
be prime. First, note that a10+1=(a2+1)(a8-a6+a4-a2+1) [Check this out by multiplying the
binomial times the polynomial.] This implies that no number that is one more than a tenth
power of some base can be prime, since such a number is divisible by a2+1. For instance,
610+1 is divisible by 37=62+1. We may also note that a15+1=(a3+1)(a12-a9+a6-a3+1), so no
number that is one more than a fifteenth power of some base can be prime either. However, if
we try to apply this pattern to a12+1 using a2+1 as a divisor, we run into a problem:
(a2+1)(a10-a8+a6-a4+a2-1) is not the same as a12+1. We may conclude that if the signs of the
polynomial are going to alternate and if the sign on the 1 at the end is going to be positive,
then that polynomial needs an odd number of terms. So how many terms are in the
polynomial when we write an+1=(am+1)(an-m-an-2m+an-3m-an-4m+…+a2m-am+1)? Noting that the
n
final 1 is a0=an-km for some k, we conclude that k must be . For instance when we factored
m
10
a10+1, the polynomial had 5 terms, which is . Turning the logic around a bit, we can factor
2
n
a +1 provided n has an odd factor. If m is odd and divides n, then an/m+1 divides an+1. Thus
the only primes of the form an+1 must be those where n has no odd divisors. These are
precisely the powers of 2, so an+1 can only be a prime if n=2k for some k. If the base a is 2,
then these kind of primes are known as “Fermat primes.” Numbers of this form that are not
prime are referred to as “Fermat numbers.” If a is some base other than a power of 2, then a
prime of this form is known as a “generalized Fermat prime.” So 37=62+1 is a generalized
Fermat prime. There are quite a list of generalized Fermat primes; the last to be discovered
was found in 1998. Let’s examine some Fermat primes:
2
F0= 2  1  2  1  3
0
F1= 2
21
 1  22  1  5
 1  2 4  1  17
23
8
F3= 2  1  2  1  257
24
16
F4= 2  1  2  1  65,537
F2= 2
22
However, the Fermat number F5= 22  1  4, 294,967, 297 was shown to be composite by
Euler in the eighteenth century. Since then all Fermat numbers Fn for 5≤n≤32 have been
shown to be composite. It is not known whether there are more Fermat primes than the five
that are already known. Some mathematicians believe there are not, but it has still not been
proven that there may not be an infinite number of Fermat primes. In fact, it is possible that
there are only a finite number of Fermat numbers that are composite (that is, beyond some
point all Fermat numbers may be prime). Incidentally, if F33 is prime, it would require over
two billion digits to express this number in decimal form (that would require more pages than
the Encyclopedia Britannica!).
5
So do Fermat primes have any applications such as using Mersenne primes to find perfect
numbers? It turns out that they do. This application is in the field of geometry rather than
number theory. As you know, it is easy to use a compass and straightedge to construct a
regular triangle (equilateral triangle) and a regular quadrilateral (square). It is not too difficult
to construct a regular pentagon, regular hexagon, regular octagon, regular decagon (10 sides),
or regular dodecagon (12 sides) using a compass and straightedge. A question that remained
open until 1796 was, “For which values of n is it possible to construct a regular polygon with
n sides with a compass and straightedge?”
The Greeks realized that if it was possible to construct a regular polygon with n sides, it was
also possible to construct a regular polygon with 2n sides. Note that in the figure, a regular
pentagon has been inscribed in a circle. (This can be done with any regular polygon; just find
the intersection of the perpendicular bisectors of 2 of the sides, and that’s the center of the
circumscribed circle.) Now notice that when the vertices of the polygon are connected to the
360o
center of the circle, this gives an angle of
. Thus if it is possible to construct a regular
n
360o
polygon with n sides, it is possible to construct an angle of
. Conversely, if it is possible
n
360o
to construct an angle of
, it is possible to construct a regular polygon with n sides. Since
n
bisecting an angle is one of the basic constructions that is possible with a compass and
straightedge, this leads to the conclusion that if we can construct a regular polygon with n
360o
sides, we can then construct an angle of
which we can then bisect to get an angle of
n
360o
from which we can construct a regular polygon with 2n sides.
2n
Another fairly easy result is that if we can construct a regular polygon with m sides and one
with n sides, and if m and n are relatively prime (that is, they have no common divisors), then
we can construct a regular polygon with m·n sides. One of the basic theorems of number
theory is that if m and n have no common divisors, there are integers a and b such that
am+bn=1 (of course either a or b will be negative). For example, 3 and 7 are relatively prime,
c d cn  dm
and 5·3-2·7=1. Now note that  
. If we choose c and d so that cn+dm=1, this
m n
mn
360o
360o
360o
gives a way to construct an angle of
from an angle of
and an angle of
. If
n
mn
m
360o
360o
we can construct an angle of
, we can construct an angle of c·
by simply adding on
m
m
360o
360o
360o
360o
c-1 such angles to the original angle
. Then either |c|·
-|d|·
or |d|·
n
n
m
m
360o
360o
|c|·
is
(we can easily construct the difference of two angles). This leaves only the
m
mn
question of what prime and prime power values of n allow us to construct a regular polygon
with n sides. Carl Frederick Gauss proved in 1796 that if n is a Fermat prime, then a regular
polygon with n sides can be constructed with a compass and a straightedge (four decades
later, another mathematician proved that this requirement is not only sufficient, but also
necessary for odd primes). Gauss’ proof involved an interesting switch from a geometric
question to an algebraic question.
First, we need to discuss something called “constructible” or “Euclidean” numbers. A
number is called “constructible” if a line segment of that length can be constructed using a
straightedge and compass given a line segment that is one unit long (Descartes seems to have
been the first person to give this idea much thought). Given a line segment a units long and a
segment b units long, it is easy to construct line segments a+b units long and a-b units long:
a
1
units long,
units long, and a
b
a
long provided we are also given a segment that is 1 unit long:
We can also construct segments that are a·b units long,
To construct a·b, form an angle with sides of length a and b. Mark off a segment of unit
length on the side of length b (see the figure on the left). Draw a line from the end of the unit
length segment to the end of the segment of length a. Construct a line at the end of the
segment of length b that is parallel to the segment just drawn. Extend the segment of length a
x a
to intersect that line. Call the intersection x. Now by similar triangles,  , so x=a·b. Try
b 1
this when b<1 to see that it also works. Looking at the middle figure, you should be able to
a
show that in this case, x  . If you let a=1 in the middle figure, you can construct the
b
reciprocal of b. The square root of a is shown on the right. Here, AB=a, and a line segment
of length 1 is added to the end of AB (ending at C). Find the midpoint of this segment of
length a+1, and draw a circle through A and C. Then construct a perpendicular at B, and
where this intersects the circle, identify the point D. Now an angle inscribed in a semicircle is
a right angle so angle ADC is a right angle. But then triangle ABD is similar to triangle DBC,
BD AB

so
. This says BD2=AB, so BD is the square root of AB which was of length a.
1
BD
Thus we can construct segments of any integer length from a segment of unit length. Then
since we can divide, we can construct segments of any rational length. We can then construct
the square root of any previously constructible number. This means that any number, which
consists of rational numbers, or square roots of rational numbers, or square roots of those
types of expressions, is constructible. It is significantly harder to prove that these are the only
constructible numbers, but they are! Now, what does this have to do with regular polygons?
In the above triangle, suppose AB=1. Then cos A = AC. If we can construct angle A with a
straightedge and compass, we can construct the length of AC. Conversely, if we can
construct a segment of length AC<1, then we can construct angle A whose cosine is AC.
Thus our question about constructible polygons can be changed into the question of which
360o
angles
have a cosine that is a constructible number. To understand what Gauss did
n
next, we need to explore complex numbers a little.
1
3
1
1
i and find z3.

i and find w8 [Hint: w8=(w2)4]. Now let z=  
2 2
2
2
8
w is known as an eighth root of unity since w =1. z is known as a third root of unity. Now let’s
1
3
 1
i . Like any complex number, it has two parts, a real part    and
look again at z=  
 2
2 2
 3
an imaginary part   . We can graph this on a coordinate axis system where we call the
 2
horizontal axis the real axis, and the vertical axis the imaginary axis. So a complex number a+bi
would look like this on a graph:
First, let w=
Now note that in the diagram, r  a 2  b 2 and   tan  1
(If a<0,   tan  1
b
a
b
 180 o to get our angle into the proper quadrant)
a
Now note from the diagram that a = r cos θ and b = r sin θ. Then a+bi can be written as
r(cos θ + i sin θ). For example, 1+i = 2 (cos 45o + i sin 45o) and
1
3
 
i = cos 120o + i sin 120o. This is known as the polar form for a complex number, and
2 2
it has some distinct advantages over the rectangular form. For instance, if we wish to multiply
r1(cos θ1 + i sin θ1) and r2(cos θ2 + i sin θ2), we can just multiply the moduli (the r’s) and add
the arguments (the angles) to get r1r2[cos(θ1+ θ2) + i sin(θ1+ θ2)]. (You can easily prove this
by expanding that last expression using the trigonometric formulas for the sum of the sine and
cosine functions.) This also makes it easy to raise a complex number to a power since
[r(cos θ + i sin θ)]2 = r2(cos 2θ + i sin 2θ) and extending this idea, [r(cos θ + i sin θ)]n =
3
 1
3 
r (cos nθ + i sin nθ). So   
i = (cos 120o + i sin 120o)3 = cos 3600 + i sin 3600 = 1.
 2 2 
n
1
3
i is a solution to the equation
Now note that this means  
2 2
1
1
x3 – 1 = 0 and from earlier work,

i = cos 450 + i sin 45o is a solution to x8 – 1 = 0.
2
2
Turning this around a bit, we can deduce that one of the solutions of the equation x n – 1 = 0 must
be
360 0
3600
k  360 0
k  360 0
. The other roots are then cos
for k=2, 3, . . .n
cos
 i  sin
 i  sin
n
n
n
n
and are spaced evenly around the unit circle centered at the origin:
In the figures above, the figure on the left illustrates that w is an eighth root of unity (and the other
7 eighth roots of unity are w2, w3, … w8=1), z is a cube root of unity (and z2 is a second cube root
of unity).
Now if y is the first nth root of unity (see the figure on the right), and if the circle is a unit circle,
360o
then y = cos θ + i sin θ where θ=
. Thus the real part of this nth root of unity is just
n
o
360
cos
. Thus if the real part of y is a constructible number, then the angle θ can be constructed
n
with a compass and straightedge.
We have now changed a geometry problem into an algebra problem! For which values of n, do
complex roots of xn-1=0 have real parts that are constructible numbers? In abstract algebra, there
is a topic called Galois theory that allows the discovery of when the real part of the nth root of
unity is a constructible number. It turns out that Galois theory implies that the real part of an n th
root of unity is a constructible number if and only if the number of positive integers less than n
that are relatively prime to n is a power of 2. The number of positive integers less than n that are
relatively prime to n is denoted by φ(n), and is called the Euler phi function or the totient function.
k
k
k
It can be shown that if n  p1 1 p2 2    pm m then
k 1
φ(n)= ( p1  1) p1 1 ( p2  1) p2
k2 1
   ( pm  1) pm
k 1
km 1
. Now pi
ki 1
is certainly not a power of 2
unless pi =2, or unless ki=1 in which case pi i = pi  1. So none of the primes that make up the
factorization of n can be raised to a power higher than 1 except for possibly the prime 2. Then we
have φ(n)= 2 k1 1 ( p 2  1)( p3  1)    ( p m  1) and p i 1 must be a power of 2 for each of the odd
primes. But that means each of the odd primes must be one more than a power of 2, which means
it must be a Fermat, prime. So finally we have the following result. If n=2k for some k greater
than 1, the polygon is constructible. And if n is a non-negative power of 2 times a product of
distinct Fermat primes, the polygon is constructible. For n≤40, this gives 3, 4, 5, 6, 8, 10, 12, 15,
16, 17, 20, 24, 30, 32, 34, and 40 as the constructible regular polygons.
0
Gauss did not actually show the construction of the 17-gon. This was done a few years later. In
1832, a description of the construction of a 257-gon was published; the description took over 200
pages! It would have to be a big 257-gon, or it would look like a circle, and I can’t imagine how
much accuracy could be maintained with a straightedge and compass throughout a construction
that takes 200 pages to describe!
There is an interesting story regarding the construction of a regular polygon with 65,537 sides. In
the early part of the 20th century, a professor at the University of Gottingen had a pesky Ph.D.
student. In desperation one day, he suggested that the student write up the construction of a
regular polygon with 65,537 sides. The student left, and did not return the next day, or the next
week, or the next month, or the next year. Everyone thought he had gotten discouraged and left
town to pursue other interests. Ten years later, he appeared at his advisor’s office, placed a huge
stack of papers on the advisor’s desk and declared, “Here it is.” If you go to the library at the
University of Gottingen today and ask to see the construction of the 65,537-gon, you will be led
down a dark and dank hallway, and a huge stack of papers will be retrieved from a dusty shelf and
you will be allowed to look at it. One researcher that did this reported that the final statement in
this publication is, “I have done it,” but in the language this mathematician used (German), it
could also mean, “I give up,” so it is not even certain that the student himself thought he had
succeeded. A prominent contemporary mathematician, John Horton Conway, has said he doubts
such a project could have been completed successfully in the early part of the 20 th century (an
ambiguous statement in itself: does that mean it could be don now, or not?). There have been
reports that the story about the graduate student is apocryphal, and that in fact it was a Gottingen
professor that spent 10 years on the problem. In any case, the report that such a document exists
at the University of Gottingen seems fairly well documented.
A little more detail for the gung-ho!
Let’s justify the statement that the numbers we determined to be constructible really are the
only ones that we can do with a straightedge and compass. Let S={x|x is constructible or –x
is constructible or x=0} Now consider the coordinate plane that we’ve all come to know and
love, except let’s only consider points for which both coordinates are taken from S. We can,
given an origin, two perpendicular axes, and a line segment of unit length, construct any of
these points with a straightedge and compass. What other kinds of things can we do with a
straightedge and compass on this set of points? Well, we can certainly draw the line through
two such points. Now we know that the equation of a line through two points (x1, y1) and

y  y1
y  y1
y  y1 
(x2, y2) is given by y  y1  2
( x  x1 )  2
x   y1  2
x1  Since each of
x2  x1
x2  x1
x2  x1 

these values is in S, the coefficients of the equation of this line are also in S. Now if we
construct a second line that is not parallel to the first, they must intersect, and the point of
intersection is a point we can construct with a straightedge and compass. But if we have two
lines y=m1x+b1 and y=m2x+b2, the coordinates of their point of intersection are
 b1  b2 b1m2  b2 m1 

 , and both of these are in S.
,
 m2  m1 m2  m1 
Another thing we can do is set our compass to the distance between two of the points with
coordinates in S. But the distance between them is ( x1  x 2 ) 2  ( y1  y 2 ) 2 and all of the
operations involved here can be done with a straightedge and compass, so this length is
constructible and therefore in S. We can then draw one or more circles using one of our
points as center and a value from S as radius. The equation of such a circle is (x-h)2+(y-k)2=r2
which is x2+y2+ax+by+c=0 for some value of a, b, and c, and h, k, r, a, b, and c are all in S.
We can then find intersections between a line and a circle or between two circles. In the first
case, consider the intersection of y=mx+b and x2+y2+dx+ey+f=0 (the equation of a line and
the equation of a circle in general) where all coefficients are from S. To do this, we substitute
the value for y into the equation of the circle: x2+(mx+b)2+dx+e(mx+b)+f=0 This gives us a
quadratic equation in x: (m2+1)x2+(2mb+d+em)x+(b2+be+f) with all coefficients in S. But
since the quadratic formula involves nothing more serious than taking a square root (along
with the other 4 basic operations), this value for x is again a in S, and substituting this value
back into the solved equation for y gives a number from S for this coordinate of the point as
well.
So how do we find the intersection of two circles? Subtract one equation from the other to
give a linear equation in x and y, solve for y, substitute back into one of the equations for the
circles, and use the quadratic formula to solve. Again, we get numbers from S.
Thus we can’t use our straightedge and compass to construct a point that doesn’t have
coordinates that are in S, and with a little bit of trimming, this constitutes a proof that no other
numbers are constructible. Note in particular, that no cube roots of non-perfect cubes are
constructible. In fact the only roots that are constructible are roots with an index that is a
power of 2.
In case you were wondering how in the world the totient function gets into this problem, we
can touch on that topic a bit more. First, let’s define a primitive nth root of unity. If ω is the
nth root of unity that makes the smallest angle with the positive x-axis, then the other roots of
unity can be expressed as ω2, ω3, … ωn=1 as we said earlier. However, it is possible that a
smaller power than the nth power of some of these roots might be equal to 1. For instance, if
we consider the sixth roots of unity, then the sixth power of any of these roots is equal to 1,
but notice that (ω2)3=ω6=1, (ω3)2=ω6=1, and (ω4)3=ω12=(ω6)2=12=1, so only ω and ω5 do not
have smaller powers than the sixth power that are equal to 1. These are called the primitive
roots of unity. Notice also that
x6-1=(x-1)(x+1)(x2+x+1)(x2-x+1) [you can check this out by multiplying these polynomials
together[. Now obviously neither ω or ω5 are roots of the first two factors here, and if you
notice that x3-1=(x-1)(x2+x+1), then the third factor contains ω2 and ω4, so the primitive sixth
roots of unity are solutions of x2-x+1. This is called the sixth cyclotomic polynomial. If you
think about it awhile, you should be able to convince yourself that the primitive nth roots of
unity are the powers of ω that are relatively prime to n. This implies that the degree of the nth
cyclotomic polynomial will be φ(n), and according to Galois theory, the real part of the roots
of the nth cyclotomic polynomial are constructible if the degree of the cyclotomic polynomial
is a power of 2.
Ok, one final thing.. I can’t resist demonstrating how to prove that the pentagon is
constructible using just algebraic theory, not geometry. We have determined that a regular
polygon with n sides in constructible if the real part of an nth root of unity is constructible.
2
2
2
 i  sin
The primitive root of x5-1=0 is cos
so we need to show that cos
is
5
5
5
constructible. (I’m using radian measure here.) The complex conjugate of a+bi is a-bi. Note
that (a+bi)+(a-bi)=2a. Also note that (a+bi)(a-bi)=a2+b2 and if a+bi is a root of unity,
a2+b2=1, so for roots of unity, (a+bi)-1=a-bi. Now let ω be the primitive fifth root of unity, so
2
2
 i  sin
ω= cos
. Since ω satisfies x5-1=0, it satisfies (x-1)(x4+x3+x2+x+1)=0 [check to
5
5
see that this is true], and since certainly ω-1 is not 0, ω must satisfy x4+x3+x2+x+1=0 so we
have ω4+ω3+ω2+ω+1=0. Now multiply both sides of this equation by ω-2 to get
2
ω2+ω+1+ω-1+ω-2=0. But ω+ω-1 = 2 cos
and since the argument (angle) for ω2 is twice the
5
argument for ω, we have
4
2
2
ω2+ω-2 = 2 cos
= 2(2 cos 2
- 1) = 4 cos 2
- 2 since cos(2θ) = 2 cos2 θ – 1 for
5
5
5
any angle θ.
Then ω4+ω3+ω2+ω+1 = 4 cos 2
2
2
+ 2 cos
- 1 = 0. Using the quadratic formula we get
5
5
2  2  4  4(4)( 1)
1 1
=
 
5 where we took the positive square root since we
5
8
4 4
2
know cos
>0. Now would you like to try this on the polynomial x16+x15+…+1 to find
5
2
cos
? The method can be found in Galois Theory by Ian Stewart (Chapman and Hall,
17
1973) as well as some websites listed two pages below.
cos
Several years before the formula we discussed above was proven both necessary and
sufficient using Galois theory, Gauss did prove that this condition on the number of sides was
a sufficient condition. How did Gauss derive this formula several years before Galois theory
was discovered? We can “hint” at that with a little work that doesn’t require much
2
background beyond what we’ve been discussing. We will start by finding cos
which is
5
the angle needed to construct a regular pentagon using a method more like what Gauss used to
2
derive cos
.
17
Let ς, ς2, ς3, and ς4 be the 4 complex fifth roots of unity (solutions to x5-1=0).
2
2
+ sin
5
5
8
8
2
2
 2 
 2 
and ς4= cos
+ sin
= cos 
- sin
 + sin  
 = cos
 5
 5
5
5
5
5
2
So ς+ς4=2 cos
5
Note that ς= cos
Now let’s find the first four powers of 3 modulo 5:
1, 3, 4, 2 (30=1≡1 mod 5, 32=9≡4 mod 5, 33=27≡2 mod 5)
Now define x1=ς+ ς4 and x2=ς3+ ς2 (use every other number in the list of powers of 3
modulo 5 to define x1 and x2. This step is not very well motivated here, but we will come
back to that later.)
Now x5-1=(x-1)(x4+x3+x2+x+1) so ς+ ς2+ ς3+ ς4+1=0 and ς+ ς2+ ς3+ ς4=-1
So x1+x2=-1
Now the hard part (Note that ςj+ςk=ςj+k=ςj+k mod 5. E.G.: ς3∙ς4=ς7=ς2):
x1∙x2=(ς+ς4) (ς3+ς2)= ςς3+ςς2+ς4ς3+ς4ς2=ς4+ς3+ς2+ς=-1
So x2=-1-x1 and then x1∙x2=x1(-1-x1)=-x1-x12=-1 which gives
1 5
x12+x1-1=0 so x1 
2
2
2  1  5
but since x1=ς+ς4, we have x1 = 2 cos
so cos
=
5
5
4
We know that x1 is the positive root of the quadratic equation above since we know that
2
2 cos
must be positive.0.
5
2
in order to construct the heptadecagon, we will use Gauss’ method
17
which is available at several websites as well as the Stewart book mentioned earlier:
Now moving on to cos
http://bdillon41.tripod.com/bob/heptadecagon.doc
http://www.fermatstheorem.com/2008/01/gauss-seventeenth-root-of-unity.html
http://www.sanjosemathcircle.org/archive/Heptadecagon.pdf
The first link is on my page. Please refer to it now. It’s a very tedious procedure, but notice
the first few steps:
1) Gauss finds the powers of 3 modulo 17 (this is not clearly done in the second website, but
that’s how the order of the roots is found in step 2)
2) Gauss uses this list to define x1 and x2
3) Gauss finds the sum and product of x1 and x2 (the product is very tedious since 64 terms
must be evaluated).
4) Gauss then solves a quadratic equation to find x1 and x2
2
These are the same four steps we used to find cos
5
Then Gauss divides each of x1 and x2 into two sums of four roots of unity using subscripted
values of y. Eventually these get divided into sums of just two roots of unity using
subscripted values of z. Now let’s make a couple observations.
1) If we want the last division of terms to give just two terms, and we want that to be half the
number of terms in the previous sum, the previous sum must have four terms. If we want
that to be half the number of terms in the previous sum, that previous sum must have eight
terms. The sum before that must have sixteen terms. Thus the exponent in xn-1=0 must be
one more than a power of 2. That’s half the battle!
2) Now why does n need to be prime? Well, we need to be able to find some number for
which the first n-1 powers modulo n are distinct. If you’ve had a course in Abstract
Algebra, you should have no trouble realizing that that requires n to be a prime. (If n is not
a prime, the products of 1 to n-1 modulo n don’t form a group.)
So why do powers of something modulo n give us a method to divide the roots of unity in half
so that this process actually works? That has to do with a subject called Gaussian sums which
is beyond what I want to explain here. The fact that it does work, but only for Fermat primes,
no doubt convinced Gauss that there was no way to construct a polygon with n sides with n on
odd prime unless n was a Fermat prime, and although his method does prove that this
condition is sufficient, it doesn’t rigorously prove that it is necessary.
A final note: During the Greek age, when the city of Delphi was plagued by a plague, the city
Father’s asked the oracle at Delphi how to end the disaster. The oracle’s altar was a cube, and
the oracle replied that if the city would build her a cube with twice the volume, the plague
would end. They understood that this meant they were to draw a line segment equal to the
length of a side of the existing cube, and use a straightedge and compass to construct a line
segment that would give twice the volume. They tried a line segment twice as long as the
original, but quickly discovered that gave a cube that had eight times the volume. To solve
the problem, it is necessary to construct a line segment that is 3 2 units long, and from our
work above, that can’t be done!