Some Theorems on Prime-Generating Polynomials

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10 Οκτ 2013 (πριν από 3 χρόνια και 10 μήνες)

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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
.