Some Theorems on Prime

Generating Polynomials
©
October 10, 2013
by Spencer W. Earnshaw
Presented to the Faculty of the Graduate School of Sonoma State College in partial
fulfillment of the requirements for t
he Degree of
MASTER OF ARTS
June 1969
Completely
corrected, revised, edited,
and
expanded
by the author
since its original publication
©
Oct

13
, Micky Earnshaw
Master’s Thesis
Page
1
Preface
The purpose of this paper is to determine conditions on polynomials
of degree
greater
than 1 such that the sequence
will contain infinitely many prime numbers.
The determination of such conditions on linear polynomials has been solved by Dirichlet (1805

1859),
who proved that a linear polynomial,
, with integral coefficient
and
, assumes infinitely
many prime values for positive integral values of
x
if and only if
and
are relatively prime.
In this pap
er, although I have not succeeded in determining and proving sufficient criteria for a
polynomial of degree greater than one to assume infinitely many prime values for positive integral values
of x, I have established some conditions sufficient for a polyn
omial
not
to be a member of such a set D. I
also determine conditions under which the values of one polynomial for positive integral values of
x
are a
subset of those of another polynomial of the same degree.
Overview of definitions, lemmas, and theorems c
ontained in this paper
Definition 3, defining the set D of polynomials which have the property given in the first sentence
of the preface above, saves me much verbiage throughout this paper. Theorem 1 is an intuitively obvious
and trivial result. The defi
nition of a
primitive polynomial
, one in which the coefficients have no common
divisor, is an extremely useful definition
–
I apply it throughout this paper. Once the definition of a
primitive polynomial is understood, Theorem 2 is as intuitively obvious (
and trivial) as is Theorem 1.
Lemma 1, expressing the coefficients of the product of two polynomials in matrix form, is a
very
important and useful result which I invoke in later proofs. Theorems 3 and 4, which state that a product of
primitive polynomia
ls is also a primitive polynomial, are very significant results. Theorem 5 is relevant
only as a support for Theorem 6, which gives a condition under which a polynomial is
not
a member of D.
Lemmas 2 and 3, reflecting the closure of the rationals and the i
ntegers, respectively, are likewise only
important in as much as they support subsequent theorems. Theorems 7 and 8 each give a simple
condition which insures that a polynomial does
not
belong to D.
Theorem 10 is a major result. Using matrices, it express
es the relationship between the coefficients
of two polynomials, one of whose consecutive integral functional values are a subsequence of those of the
other polynomial (I use subset notation to indicate this relationship
–
see definition 11). Theorems 11,
12
and 13 express the interrelationship between and conditions on two such polynomials, one of which is a
member of D. Theorem 14 expresses conditions under which two such polynomials are both primitive.
Throughout this paper, I intersperse ‘Discussions’
and examples of various aspects of prime

generating polynomials which the Theorems and Lemmas suggest. At the end of this paper, which is still
“in process”, are two theorems which I intuit are true, but have yet to prove. In the process of proving
these m
ajor theorems (if I ever do!), the conditions in the statement of each theorem might need to be
revised or “fine

tuned”.
©
Oct

13
, Micky Earnshaw
Master’s Thesis
Page
2
Definition 1
denotes the positive integers.
Definition 2
The notation
, sometimes denot
ed just by
, will denote a polynomial of
degree
n
with rational (and usually integral) coefficients
, with
and
.
Definition 3
D will denote the set of polynomials
such that
contains infinitely
many prime numbers. See the first paragraph.
Theorem 1
Let
n
be a polynomial
. If
is divisible by a fixed integer
d
,
then
.
Proof
If
d
divides every
, then t
he only way
that
could be prime is if
d
were itself a prime
number and there were no other divisors of
except for
d
. Since
there can be at most
n
integral
solutions to
,
must be composite for all but at most
n
values of
x
. Therefore,
.
Definition 4
Let
denote the
greatest common divisor
of
the integers
.
Definition 5
I will call a polynomial
with integral coefficients
primitive
if and only if
.
Theorem 2
If
is not primitive, then
.
Proof
If
is not primitive, then
divides every coefficient. Therefore,
d
divides
. Thus, by Theorem 1,
.
Discussion
With the above definitions, I’ll restate Dirich
let’s Theorem, referred to in the first paragraph, in
terms of the above terminology:
Every primitive linear polynomial is a member of D
.
The extension of Dirichlet’s Theorem to degrees greater than 1, the converse of Theorem 2, does
not
follow. That is,
it is
not
true that every primitive polynomial (regardless of degree) is a member of D.
For example,
is a primitive polynomial of degree
, and
yet is divisible by 2
.
©
Oct

13
, Micky Earnshaw
Master’s Thesis
Page
3
Lemma 1
Given
and
, then the matrix
of coefficients of their product is given by
, where
is an
by
matrix such
that each element
.
Proof
The coefficient of each power of
x
in both
and
has a subscript equal to its
corresponding power of
x
. The power of
x
associated with any product pair
is
.
Thus, in the product, the coefficient of each power
j
of
x
will be the sum of all possible product pairs
–
one from the coefficients of
and the other from the coefficients of
–
such that the subscripts
of each product pair add up to
j
. The above matrix product accomplishes this end. The
i
th row in the
resulting product matrix is
. The sum of the subscripts of each product pair
is
,
which is the power of
x
in the product which corresponds to that coefficient. The requirement that
, zero otherwise, assures that each product pair in this sum is defined.
Discussion
To create th
e elements of matrix
of Lemma 1, consider the elements of each row of
to
be windows through which we view the following ordered array of coefficients of
:
. Start
with coefficient
of Q in the upper left

hand corner window
.
When you move down one row, slide the array Q to the right one column. Continue this process to the
last row of
, where the
coefficient
will be in the lower right

hand corner window
.
Every window not exhibiting an element of Q will have the value 0. By the
row,
is in
column 1 (follow
ed by
etc.). After another
n
rows,
has moved to the lower right

hand
corner of
, preceded by all zeros in this last row. This is a total of
rows.
I’ll illustra
te the above procedure with two general functions:
and
. Refer to the matrices below. First, let
be the polynomial of degree 4 and
be
the polynomial of degree 2.
Then
and
, and the matrix of coefficients of the product is
shown in (1). Now reverse the roles of the two functions and let
be the polynomial of degree 2 and
b
e the polynomial of degree 4. Then
and
, and the matrix of coefficients of the
product is shown in (2). In either case, the resulting product matrix is the same.
(1)
(2)
©
Oct

13
, Micky Earnshaw
Master’s Thesis
Page
4
Theorem 3
The product of two primitive polynomials is a primitive polynomial.
Proof
Let the two polynomials be
and
.
Refer to the matrices of Lemma 1. If
is not primitive, t
hen some prime number
p
divides
every coefficient of
, i.e., some prime number
p
divides every element of the product matrix
of Lemma 1. Let
be the first coefficient of
, starting from
, which is
not
divisible by
p
. There
must be such a coefficient, otherwise
p
would divide
every
coefficient of
, i.e.,
would not be
primitive. Likewise, let
be the
first coefficient of
, starting from
, which is
not
divisible by
p
.
In the product matrix, locate the coefficient
c
which contains this specific product pair,
. Starting
from the term
in C, the subscripts of the
decrease by 1 as you travel in one direction in the
sum of terms, and the subscripts of the
decrease by 1 as you travel in the opposite direction. By
our choice
of
, every coefficient
with
is divisible by
p
, and, likewise, by our choice of
,
every coefficient
with
is divisi
ble by p. Thus, every product pair (except possibly for
)
in
c
contains a factor which is divisible by
p
. Thus, every product pair (except possibly for
) is
divisible by
p
. Since we are assuming that
c
itself i
s divisible by
p
, and since every product pair (except
possibly for
) in
c
is divisible by
p
, it must be that
is divisible by
p
1
. But if so, then
p
–
being a prime number
–
must divide into either
or into
, but not into both. This contradicts our
choice of
and
. Therefore, our assumption that the product
is not primitive must have
been false. Th
erefore, the product
is primitive.
Theorem 4
The product of any number of primitive polynomials is a primitive polynomial.
Proof
This follows from the previous theorem by simple induction.
Theorem 5
Given a polynomial
, then
. More specifically,
.
Proof
In the expansion of
, the
last term in the expansion of each
is
. These constant terms, from
to
n
, add up
to
. Each remaining term is a multiple of
m
times some positive power of
x
. Thus, the sum of the
remaining terms can be expressed as a polynomial in
x
,
, times
m
. The seco
nd conclusion can be
drawn from observing that the first terms in the expansion of each
for
add up to
[the
has already been accounted for in
]. Replacing
by
plus
another polynomial in
x
, Q(x), gives the second, more specific, result.
Definition 6
Let
. The notation
will mean that
a
divides evenly
into
b
, i.e., that there exists
a
such that
.
1
We can factor p out of every term of C (othe
r than
) to express C as:
. Since
,
. So
. Solving for
, we have
, or
.
Since
is an integer,
p
must divide evenly into
.
©
Oct

13
, Micky Earnshaw
Master’s Thesis
Page
5
Definition 7
Let
. The notation
will mean that
, read, “
a
is
congruent
to
b
modulo
m
”.
Defin
ition 8
will denote the set of all integers
such that
.
Definition 9
Given an integer
m
, let
, where
and
no two el
ements of
are congruent modulo
m
. We call
a
complete residue system modulo
m
.
Theorem 6
Let
be a complete residue system modulo m, m>1, where each subscript
of r represents the remaind
er when
is divided by m, and let
. If
, then
.
Proof
By Theorem 5,
, where
. Thus, if
, then
. Since
any
value of
x
can be expressed in the form
for some
, then
. By Theorem 1,
. From the second part of Lemma 1,
. That is,
. Thus,
if and only if
, so
we can replace
by
in the statement of the theorem.
Definition 10
will denote the number of different possible combinations of
n
things taken k at a time,
specifically
, where
.
Lemma 2
Let
and
, where the
coefficien
ts of the product
are rational. Then the coefficients of
are rational if and only if
the coefficients of
are rational.
Proof
The proof of each part is an inductive argument on the sub
script k of each
or
, referring to
Lemma 1 and using the fact that the rational numbers are closed under
. In each case, we’re
assuming that each element of the product matrix is rationa
l.
Part 1
–
If
has rational coefficients, then
has rational coefficients
Row 1
: Since
is rational and
is rational, then
, being a quoti
ent of rational numbers,
is rational.
Row 2
: Since
is rational and
and
are rational, then
is
rational, since the rational numbers are closed under
.
Row3 through Row (m+1)
: As we continue down the product matrix row by row, only one new
coefficient
of
is introduced at each step into a combination of rational numbers whose sum is
itself ratio
nal, requiring that new element
to also be rational. Continuing this inductive process right
up through the (m+1)st row assures that each coefficient of
,
through
, inclusive, is rational.
Part 2
–
If
has rational coefficients, then
has rational coefficients
The proof of this part
parallels the previous proof exactly, but with the roles of the
s and the
s reversed.
©
Oct

13
, Micky Earnshaw
Master’s Thesis
Page
6
Lemma 3
Let
and
, where the
coefficients of their product
are integers. If either
or
is pri
mitive, then the
coefficients of the other polynomial are integers.
Proof
Given the hypothesis, assume that
is primitive. Since the coefficients of
are
integers, by Lemma 2, the coefficients of
are rational. If they are not all integral, then at least one
coefficient of
is a fraction. Out of every coefficient
of
factor
, where
is the
greatest common divisor of all the numerators and
is the least common multiple of all the
denominators of the
’
s. This leaves a primitive polynomial
, i.e.,
. Then the
product
. Since
is a product of primitive
polynomials, by Theorem 3, it is therefore equal to a primitive polynomial
. Thus
. Distributing the
denominator
Q
through each coefficient of
, there will be at
least one coefficient of
into which
Q
will
not
divide (because
is a primitive polynomial). This
contradicts that fact that t
he coefficients of
are all integers. This contradiction resulted from
assuming that the coefficients of
were not all integers. Therefore, if
is primitive, the
coefficients of
are all integers. The case where
is primitive is proven identically, but with the
roles of the
’
s and
’
s reversed.
©
Oct

13
, Micky Earnshaw
Master’s Thesis
Page
7
Theorem 7
If
has integral coeffi
cients,
, and
has one rational root, then
.
Proof
If
has one rational root,
, then
can be expressed as
, where the polynomial P(x), of degree
, must have rational coefficients [let
and
in Lemma 2]. Let
L
be the least common multiple of all the denominat
ors
of the coefficients of
. Factoring out
, we have
. The
numerator of this fraction is a product of two polynomial factors, each with integral coefficients. Thus,
for any integral val
ue
of
, each factor in the numerator is an integer. Let
T
be the number of factors
(each to the power 1) of the integer
dL
in the denominator. For example, if
, then
because
. In order for the quotient
to be a prime number for some
integral
, all of the prime factors in the denominator must cancel into the two factors in the
numerator, le
aving a single prime factor
–
either
or
–
with the other factor having the
value of 1.
I’ll take each factor in the numerator separately. For how many different values of
x
can we divide
some of the
T
facto
rs of
dL
into
so that the resulting quotient is 1? We could divide them either
not at all, or one

at

a

time, or two

at

a

time, …, or T

at

a

time. This is a total of
, with each equation
(where
p
is some combination of the factors of
dL
) having exactly
one solution for x. Regardless of whether the remaining factors divide into
to result in a prime is
irrelevant…
is an
upper bound on the number of values of
x
for which
as
p
assumes the
value of all possible combinations of factors of
dL
. Similarly, for the factor
in the numerator,
each equation
(wher
e
p
is some combination of the factors of
dL
) has at most
n
solutions for
x. Thus, since
is of degree
,
is an upper bound on the number of values of
x
for
which
.
Summing up, the total of these two upper bounds on the number of values of
x
for
which either
or
is 1 is
. This is the
maximum
number of values of
x
for which the quotient
could be a prime number. Therefore, for all but a finite (
)
number of values of x,
is composite. Therefore,
. Q.E.D.
©
Oct

13
, Micky Earnshaw
Master’s Thesis
Page
8
Discussion
Any polynomial with integral coefficients can
be factored into linear factors of the form
,
where the r takes on the values of all of the real and complex roots of
. Since complex roots of
occur in conjugate pairs, and the product
of
and
is the quadratic
, we can conclude that any polynomial can be factored into linear and quadratic
factors.
Theorem 8
If
has integral coefficients,
, and
has one imaginary root
with
c
a rational number, and d either a rational number or the square root of a rational number, then
.
Proof
Since imaginary roots occur in c
onjugate pairs, and therefore both
and
are
factors of
, then
is a factor of
. Clearly,
given the conditions on coefficients
c
and d, the c
oefficients of
are rational. Since
,
, where
must have rational coefficients [by Lemma 2
–
let
and let
].
If we f
actor out the lowest common denominator
out of all the denominators of
, and
factor out the lowest common denominator
out of the denominators of
, then we can wri
te
, where both
and
have integral coefficients.
I’ll leave it to the reader to supply the details, parallel to the proof of the previous theorem, to
show that
i
s an upper bound on the number of prime values which can be assumed by
, where T is the number of factors (each to the power 1) of the integer
(as in
the proof of the previous theorem). This quotient must ther
efore be composite for all but a finite number
of integral values of x. Therefore,
.
©
Oct

13
, Micky Earnshaw
Master’s Thesis
Page
9
Discussion
Parallel to the reasoning in the proof of the previous theorem, and since any polynomial can be
factored into linear and quadratic fact
ors, we might be tempted to erroneously conclude that there are only
a finite number of values of
x
which could result in a prime value for any polynomial of degree greater
than 1. This
would
be the case if one of the linear or quadratic factors of
contained integral
coefficients, in which case replacing
x
by an integer would result in an integral factor of
. However,
the linear or quadratic factors of a polynomial with integral coefficients often contain irrat
ional
coefficients and irrational terms. Here are four examples, each of which is probably a member of D, as the
table of values which follows would imply. The first example illustrates why the preceding Theorem
requires
.
Below are five sample polynomials, including the four above, which might belong to D. I let
,
and wrote down prime functional values as they occurr
ed
–
until I had ten such prime values.
©
Oct

13
, Micky Earnshaw
Master’s Thesis
Page
10
Theorem 10
Let
, and let
be a subsequence
of
obtained by
starting with the
term of that sequence and choosing every
term thereafter. Then the subsequence
polynomial is
,
. The relationship
between the coefficients
of
and the coefficients
of the subsequence polynomial
is given
by these two matrix equations:
,
or, in reverse,
.
In both of the above coefficient matrices,
. For each i
th

row j
th

column element
of the first
coefficient matrix,
; for each i
th

row j
th

column element
of the second
coefficient matrix,
. In both matrices,
.
Proof
Following is a derivation of the matrix of coefficie
nts in the first matrix equation.
. To find the
coefficient
of each
in
, I’ll expand
with
, then extrapolate and
generalize. When
,
=
©
Oct

13
, Micky Earnshaw
Master’s Thesis
Page
11
. Equating corresponding coefficients of
and
…
. Expressing this set of equations in matrix form, we have…
Generalizing and extrapolating from this example results in the generalized (first) matrix shown
and described in the statement of the theorem.
For the second matrix equation, I want to find the inverse of the above coefficient matrix (the
in terms of the
), because multiplying both sides of the first matrix equation by the inverse of the
coefficient matrix gives me the second matrix equation (the
in terms of the
). Again, en route
to generalizing and extrapolating from specific cases, I’ve evaluated the inverse of the matrix of
coefficients for the first matrix equation with
&
4.
©
Oct

13
, Micky Earnshaw
Master’s Thesis
Page
12
Generalizing and extrapolating from this example results in the generalized (second) matrix shown
and described in the statement of the theorem.
Definition 11
The notation
wil
l mean that there exist integers
c
and d,
, such that
the relationship between the coefficients given in Theorem 10 holds. The values of
are a subset of the values of
, i.e.,
is a
subsequence
of
.
Definition 12
Let
. Then I’ll say that
generates
, or that
is generated by
, and I’ll ref
er to
as the
generator
of
.
©
Oct

13
, Micky Earnshaw
Master’s Thesis
Page
13
Discussion
I’ll give examples of applying each of the two matrix equations in Theorem 10. For the first
matrix equation, where we’re given a generated polynomial
and want to find a generator
polynomial
, I’ll choose
. For the generator polynomial
, I’ll specify starting
on the
term of the sequence
, and choosing every
term thereafter, i,e.,
. Since
, the first matrix equation of Theorem 10 becomes
Thus,
. Here’s the sequence of values.
0
1
9
2
17
3
31
4
51
5
77
6
109
7
147
8
191
9
241
10
297
11
359
12
427
13
501
14
581
15
667
16
759
17
857
18
961
19
1071
20
1187
©
Oct

13
, Micky Earnshaw
Master’s Thesis
Page
14
Discussion (continued)
For the second matrix equation, where we’re given a generator polynomial
and we want to
generate a polynomial
, I’ll choose
, resulting from starting with the
term of the
sequence
, and choosing every
term
thereafter, i,e.,
. Since
, and
, the second matrix equation of Theorem 10 becomes
Thus,
. Here’s the sequence of values.
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
©
Oct

13
, Micky Earnshaw
Master’s Thesis
Page
15
Theorem 11
Let
. If
, then
. If
, then
if and only if
.
Proof
The first part is clear, since every value that
assumes appears in the sequence
.
If c=1, then the sequence
is identical to the
sequence
, so if one polynomial
, then the other one is.
Discussion
The second example of applying Theorem 10 generated a polynomial
which had rational coefficients. The concept of a primitive polynomial, in relation to polynomials which
assume infinitely many prime values for integral values of x, has meaning only if its coefficients are
integers. Thus, given polynomial
s
and
with
, both polynomials are primitive only
if the coefficients of
both
polynomials are integers. The following theorem gives conditions under which
the coefficients of both the gener
ator
and
the generated polynomial are integers.
Theorem 12
Given nth degree polynomials
and
with
. If the coefficients
of
the generated polynomial
are integers, then the coefficients
of the generator polynomial
are
integers. If the coefficients
of the generator polynomial
are integers, with the added requirement
that
, then the coefficients
of the generated polynomial
are integers.
Proof
Assume integral coefficients
for
. From the relationship between the coefficient
s
of
and the coefficients
of
given in the first equation in the proof of Theorem 10, namely,
that
with
c
and
d
integers, then each coefficien
t
of
is an integral linear
combination of the integral
, and is therefore
itself
an integer. For the second conclusion, assume
integral coefficients
for
, where
. Referring to the second matrix of
coefficients in Theorem 10, which converts coefficients
of
into coefficients
of
, in
orde
r to end up with integers for
from integral coefficients
, we must have
. This is clear if you look at the diagonal elements
, each of which is multiplied by
in the product matrix. Since
is in the denominator with
the only integer in the numerator, and we
require that each
of
be an integer, we must have
. Since
, each quotient
is an integer
. Since each
is a linear combination of terms, each having the same form:
{the
integer
}
{a power of the integer
d
}
{the integer
}, we can conclude that each
must itself be
an integer.
©
Oct

13
, Micky Earnshaw
Master’s Thesis
Page
16
Theorem 13
Given nth degree polynomials
and
. Then
if and only if the first
matrix equation of Theorem 10 is satisfied for all of the coefficients
of
and
of
, for
, with integers
and
.
Proof
From the last product in the first matrix equation of Theorem 10,
, from which
.
From the
next

to

last product in the same matrix equation,
, from which
. The second version for
d
in the theorem results from replacing
c
by
.
Discussion
A question suggested by Theorem 13 is wh
ether a polynomial
which is
not
primitive could
still generate a polynomial
which is primitive, even if
. The answer is, YES. Here are such
a
,
,
c
and d:
. Note that
,
which makes
non

primitive, and
. Here’s the matrix relationship between the coefficients
of
a
nd the coefficients
of
.
.
As you can see, to guarantee that a polynomial
which is generated from another polynomial
of the same degree is primitive, all that is required is that the coefficient
of
in
be equal
to
. The example above provides a counterexample to the converse of
Theorem 13: Given
,
if
is primitive then
is primitive. The fact is, however, that
non

primitive generators
don’t
do us much good. This is because the point here is
to find
primitive
polynomials
which generate
infinitely many other polynomials
, depending upon the values of
c
and
d
. Each such generated
polynomial
is a member of D because every value o
f
for
is also a value of
for
. Therefore, if the sequence
contains infinitely many primes,
then so does the sequence
, b
ecause
is a subsequence of
.
©
Oct

13
, Micky Earnshaw
Master’s Thesis
Page
17
Theorem 14
Given
n
th degree polynomials
and
with
. If
is primitive, and
divides each integral coefficient
of
in
, then
is primitive.
Proof
Referring to the first matrix of coefficients of Theorem 12, if
wer
e not primitive, then the
greatest common divisor of the
,
, would divide each coefficient
of
, and
therefore
could not be primitive. From Theor
em 12, the requirement that
guarantees that
the coefficients of
are integers.
?Theorem
If
is of degree 2 or greater, is primitive, and has no real or complex root
s
with rational
coeff
icients, then
.
?Theorem
If
is of degree 2 or greater, is primitive, and assumes a prime value for two distinct
integral values of x, then
.
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