Some Theorems on Prime-Generating Polynomials
© February 6, 2016 by Spencer W. Earnshaw
Presented to the Faculty of the Graduate School of Sonoma State College in partial
fulfillment of the requirements for the Degree of
MASTER OF ARTS
June 1969
Completely corrected, revised, edited, and expanded
by the author
since its original publication
© Feb-16, Micky Earnshaw
Master’s Thesis
Page 1
Preface
n
The purpose of this paper is to determine conditions on polynomials f  x    a i x i of degree
i 0
greater than 1 such that the sequence f 1, f 2, f 3,  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, a1 x  a 0 , with integral coefficient a 0 and a1 , assumes infinitely
many prime values for positive integral values of x if and only if a 0 and a1 are relatively prime.
In this paper, 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 polynomial 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 contained 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 definition 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 polynomials 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 integers, 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 expresses 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 primegenerating 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 major theorems (if I ever do!), the conditions in the statement of each theorem might need to be
revised or “fine-tuned”.
Master’s Thesis
© Feb-16, Micky Earnshaw
Page 2
Definition 1 I  denotes the positive integers.
n
Definition 2 The notation f  x    a i x i , sometimes denoted just by f x  , will denote a polynomial of
i 0
degree n with rational (and usually integral) coefficients a i , with a n  0 and n  1 .
n
 
Definition 3 D will denote the set of polynomials f  x    a i x i such that f I  contains infinitely
i 0
many prime numbers. See the first paragraph.
n
Theorem 1 Let n be a polynomial f  x    a i x i . If f x  is divisible by a fixed integer d  x  I  ,
i 0
then f x   D .
Proof If d divides every f  x  , then the only way that f x  could be prime is if d were itself a prime
number and there were no other divisors of f x  except for d. Since there can be at most n integral
solutions to f x   d , f x  must be composite for all but at most n values of x. Therefore, f x   D .
Definition 4 Let a 0 , a1 ,  , a n  denote the greatest common divisor of the integers a 0 , a1 ,  , a n .
n
Definition 5 I will call a polynomial f  x    a i x i with integral coefficients primitive if and only if
a0 , a1 , , a n   1 .
i 0
n
Theorem 2 If f  x    a i x i is not primitive, then f x   D .
i 0
Proof If f x  is not primitive, then d  a 0 , a1 ,  , a n  divides every coefficient. Therefore, d divides
f x  x  I . Thus, by Theorem 1, f x   D .
Discussion
With the above definitions, I’ll restate Dirichlet’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, f x   x n  x n 1  2  x n  2  xx  1  2 is a primitive polynomial of degree n n  2 , and
yet is divisible by 2 x  I  .
Master’s Thesis
© Feb-16, Micky Earnshaw
Page 3
Lemma 1 Given f x   a 0  a1 x  a 2 x 2   a m x m and g x   b0  b1 x  b2 x 2   bn x n , then the matrix
b 0 
b 
of coefficients of their product is given by Aij   1  , where Aij  is an m  n  1 by n  1 matrix such

 
bn 
a i  j if 0  i  j  m
that each element Aij  
.
0 otherwise
 
Proof The coefficient of each power of x in both f x  and g x has a subscript equal to its
corresponding power of x. The power of x associated with any product pair a j x j  bk x k is x j  k .
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 f x  and the other from the coefficients of g x – such that the subscripts
of each product pair add up to j. The above matrix product accomplishes this end. The ith row in the

n
resulting product matrix is
a
k 0


b . The sum of the subscripts of each product pair a i  k 1bk is i  1,
i  k 1 k
which is the power of x in the product which corresponds to that coefficient. The requirement that
0  i  j  m , zero otherwise, assures that each product pair in this sum is defined.
Discussion
To create the elements of matrix Aij of Lemma 1, consider the elements of each row of Aij to
 
 
be windows through which we view the following ordered array of coefficients of f x  :
Q  a m a m1  a1 a0 . Start with coefficient a 0 of Q in the upper left-hand corner window A11 .
When you move down one row, slide the array Q to the right one column. Continue this process to the
last row of Aij , where the coefficient a m will be in the lower right-hand corner window Am  n 1n 1 .
 
Every window not exhibiting an element of Q will have the value 0. By the m  1st row, a m is in
column 1 (followed by a m1 a m 2 etc.). After another n rows, a m has moved to the lower right-hand
corner of Aij , preceded by all zeros in this last row. This is a total of m  1  n  m  n  1 rows.
 
I’ll illustrate the above procedure with two general functions: a 0  a1 x  a 2 x 2  a 3 x 3  a 4 x 4 and
b0  b1 x  b2 x 2 . Refer to the matrices below. First, let f x  be the polynomial of degree 4 and g x be
the polynomial of degree 2. Then m  4 and n  2 , and the matrix of coefficients of the product is
shown in (1). Now reverse the roles of the two functions and let f x  be the polynomial of degree 2 and
g x be the polynomial of degree 4. Then m  2 and n  4 , and the matrix of coefficients of the
product is shown in (2). In either case, the resulting product matrix is the same.
a 0
a
 1
a 2

(1)  a 3
a 4

0
0

0
a0
a1
a2
a3
a4
0
0
a 0 b0





0
 a1b0  a 0 b1

a 0  b0   a 2 b0  a1b1  a 0 b2 



a1    b1    a 3 b0  a 2 b1  a1b2 
a 2  b2  a 4 b0  a 3 b1  a 2 b2 



a3 
 a 4 b1  a 3 b2




a4 
a 4 b2


b0
b
 1
b2

(2)  0
0

0
0

0
b0
b1
b2
0
0
0
0
0
b0
b1
b2
0
0
0
0
0
b0
b1
b2
0
0
a 0 b0





0  a 0   a1b0  a 0 b1

0   a1   a 2 b0  a1b1  a 0 b2 



0   a 2    a 3 b0  a 2 b1  a1b2 
 
b0   a 3  a 4 b0  a 3 b1  a 2 b2 



b1  a 4   a 4 b1  a 3 b2




b2 
a 4 b2


Master’s Thesis
© Feb-16, Micky Earnshaw
Page 4
Theorem 3 The product of two primitive polynomials is a primitive polynomial.
Proof Let the two polynomials be f x   a 0  a1 x  a 2 x 2   a m x m and g x   b0  b1 x  b2 x 2   bn x n .
Refer to the matrices of Lemma 1. If f x g x is not primitive, then some prime number p divides
every coefficient of f x g x , i.e., some prime number p divides every element of the product matrix
of Lemma 1. Let a M be the first coefficient of f x  , starting from a 0 , which is not divisible by p. There
must be such a coefficient, otherwise p would divide every coefficient of f x  , i.e., f x  would not be
primitive. Likewise, let b N be the first coefficient of g x , starting from b0 , which is not divisible by p.
In the product matrix, locate the coefficient c which contains this specific product pair, a M bN . Starting
from the term a M bN in C, the subscripts of the a i ' s decrease by 1 as you travel in one direction in the
sum of terms, and the subscripts of the b j ' s decrease by 1 as you travel in the opposite direction. By
our choice of a M , every coefficient a i with i  M is divisible by p, and, likewise, by our choice of b N ,
every coefficient b j with j  N is divisible by p. Thus, every product pair (except possibly for a M bN )
in c contains a factor which is divisible by p. Thus, every product pair (except possibly for a M bN ) is
divisible by p. Since we are assuming that c itself is divisible by p, and since every product pair (except
possibly for a M bN ) in c is divisible by p, it must be that a M bN is divisible by p1. But if so, then p –
being a prime number – must divide into either a M or into b N , but not into both. This contradicts our
choice of a M and b N . Therefore, our assumption that the product f x g x is not primitive must have
been false. Therefore, the product f x g x 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.
n
Theorem 5 Given a polynomial
f  x    a i x i , then f mx  b  mPx  f b . More specifically,
f mx  b  f mx  f b  a 0  mQx  .
i 0
Proof In the expansion of f mx  b  a0  a1 mx  b  a 2 mx  b  a3 mx  b   a n mx  b , the
2
3
n
last term in the expansion of each a k mx  b is a k b k . These constant terms, from k  0 to n, add up
to f b . 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, Px , times m. The second conclusion can be
k
drawn from observing that the first terms in the expansion of each a k mx  b for k  1 add up to
f mx  a 0 [the a 0 has already been accounted for in f b ]. Replacing mPx  by f mx  a 0 plus
another polynomial in x, Q(x), gives the second, more specific, result.
k
Definition 6 Let a, b  I  . The notation a | b will mean that a divides evenly into b, i.e., that there exists
a k  I  such that b  ak .
1
We can factor p out of every term of C (other than a M bN ) to express C as: C  pk1  a M b N . Since
p | C , C  k 2 p . So k 2 p  pk1  a M bN . Solving for a M bN , we have a M bN  pk 2  k1  , or
a b
k 2  k1  M N . Since k 2  k1 is an integer, p must divide evenly into a M bN .
p
© Feb-16, Micky Earnshaw
Master’s Thesis
Page 5
Definition 7 Let a, b  I  . The notation a  b (mod m) will mean that m | a  b , read, “a is congruent
to b modulo m”.
Definition 8 bm will denote the set of all integers i such that i  b mod m .
Definition 9 Given an integer m, let c m  r0 , r1 , , rm1  , where ri  i m and  no two elements of c m
are congruent modulo m. We call c m a complete residue system modulo m.
Theorem 6 Let c m  r0 , r1 , , rm1  be a complete residue system modulo m, m>1, where each subscript
n
of r represents the remainder when ri is divided by m, and let f  x    a i x i . If
i 0
m | a 0 , f r1 , f r2 , , f rm1  , then m | f x x and f x  D .
Proof By Theorem 5, f mx  ri   mQi x   f ri  i , where Qi  I . Thus, if m | f ri  , then
m | f mx  ri  x . Since any value of x can be expressed in the form x  md  ri for some
d  I , ri  c m , then m | f x x . By Theorem 1, f x   D . From the second part of Lemma 1,
f r0   f md  0  f md   f 0  a 0  mQx   f md   a 0  a 0  mQx   mQx   f md  
mQx   mRx   a 0  mS x   a 0 . That is, f r0   mS x   a 0 . Thus, m | f r0  if and only if m | a 0 , so
we can replace f r0  by a 0 in the statement of the theorem.
Definition 10 C kn will denote the number of different possible combinations of n things taken k at a time,
n!
specifically C kn 
, where n!  1 2  3 4  n .
k! n  k !
Lemma 2 Let f x   a 0  a1 x  a 2 x 2   a m x m and g x   b0  b1 x  b2 x 2   bn x m , where the
coefficients of the product f x g x are rational. Then the coefficients of f x  are rational if and only if
the coefficients of g x are rational.
Proof The proof of each part is an inductive argument on the subscript k of each a k or b k , referring to
Lemma 1 and using the fact that the rational numbers are closed under ,, and  . In each case, we’re
assuming that each element of the product matrix is rational.
Part 1 – If f x  has rational coefficients, then g x has rational coefficients
a b
Row 1: Since a 0 b0 is rational and b0 is rational, then a 0  0 0 , being a quotient of rational numbers,
b0
is rational.
a b  a0 b1   a0 b1
Row 2: Since a1b0  a 0 b1 is rational and a 0 , b0 and b1 are rational, then a1  1 0
is
b0
rational, since the rational numbers are closed under ,, and  .
Row3 through Row (m+1): As we continue down the product matrix row by row, only one new
coefficient a i of f x  is introduced at each step into a combination of rational numbers whose sum is
itself rational, requiring that new element a i to also be rational. Continuing this inductive process right
up through the (m+1)st row assures that each coefficient of f x  , a 0 through a m , inclusive, is rational.
Part 2 – If g x has rational coefficients, then f x  has rational coefficients The proof of this part
parallels the previous proof exactly, but with the roles of the a i s and the bi s reversed.
© Feb-16, Micky Earnshaw
Master’s Thesis
Page 6
Lemma 3 Let f x   a 0  a1 x  a 2 x 2   a m x m and g x   b0  b1 x  b2 x 2   bn x m , where the
coefficients of their product f x g x are integers. If either f x  or g x is primitive, then the
coefficients of the other polynomial are integers.
Proof Given the hypothesis, assume that f x  is primitive. Since the coefficients of f x g x are
integers, by Lemma 2, the coefficients of g x are rational. If they are not all integral, then at least one
P
coefficient of g x is a fraction. Out of every coefficient bi of g x factor , where P  1 is the
Q
greatest common divisor of all the numerators and Q  1 is the least common multiple of all the
P
denominators of the bi ’s. This leaves a primitive polynomial px , i.e., g  x    p x  . Then the
Q
P
P
product f x   g x   f  x    p x    f x   p x  . Since f x  px  is a product of primitive
Q
Q
polynomials, by Theorem 3, it is therefore equal to a primitive polynomial qx  . Thus
P
f  x g x   q x  . Distributing the denominator Q through each coefficient of qx  , there will be at
Q
least one coefficient of qx  into which Q will not divide (because qx  is a primitive polynomial). This
contradicts that fact that the coefficients of f x g x are all integers. This contradiction resulted from
assuming that the coefficients of g x were not all integers. Therefore, if f x  is primitive, the
coefficients of g x are all integers. The case where g x is primitive is proven identically, but with the
roles of the a i ’s and bi ’s reversed.
© Feb-16, Micky Earnshaw
Master’s Thesis
Page 7
n
Theorem 7 If f  x    a i x i has integral coefficients, n  2 , and f x  has one rational root, then
f x   D .
i 0
Proof If f x  has one rational root,
c
c

, then f x  can be expressed as f x   Px    x   
d
d

1
P  x   dx  c  , where the polynomial P(x), of degree n 1 , must have rational coefficients [let
d
c
f x   Px  and g  x   x  in Lemma 2]. Let L be the least common multiple of all the denominators
d
P'  x dx  c 
1 1
1
of the coefficients of P  x  . Factoring out , we have f x     P' x   dx  c  
. The
L d
dL
L
numerator of this fraction is a product of two polynomial factors, each with integral coefficients. Thus,
for any integral value x 0 of x , 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 dL  360 , then T  6
P '  x dx  c 
because 360  2  2  2  3 3 5 . In order for the quotient
to be a prime number for some
dL
integral x 0 , all of the prime factors in the denominator must cancel into the two factors in the
numerator, leaving a single prime factor – either P' x 0  or dx 0  c  – 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 factors of dL into dx  c 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 C 0T  C1T    CTT 
dx  c
 1 (where p is some combination of the factors of dL) having exactly
2 T , with each equation
P
one solution for x. Regardless of whether the remaining factors divide into P' x  to result in a prime is
dx  c
 1 as p assumes the
irrelevant… 2 T is an upper bound on the number of values of x for which
P
value of all possible combinations of factors of dL. Similarly, for the factor P' x  in the numerator,
P' x 
 1 (where p is some combination of the factors of dL) has at most n solutions for
each equation
P
x. Thus, since P' x  is of degree n 1 , n  1 2 T is an upper bound on the number of values of x for
P' x 
 1 . Summing up, the total of these two upper bounds on the number of values of x for
which
P
which either P' x  or dx  c is 1 is 2 T  n  1  2 T  n  2 T . This is the maximum number of values of x
P '  x   dx  c 
for which the quotient
could be a prime number. Therefore, for all but a finite (  n2 T )
dL
number of values of x, f x  is composite. Therefore, f x   D . Q.E.D.
Master’s Thesis
© Feb-16, Micky Earnshaw
Page 8
Discussion
Any polynomial with integral coefficients can be factored into linear factors of the form x  r  ,
where the r takes on the values of all of the real and complex roots of f x   0 . Since complex roots of
f x   0 occur in conjugate pairs, and the product of x  a  bi  and x  a  bi  is the quadratic


x 2  2ax  a 2  b 2 , we can conclude that any polynomial can be factored into linear and quadratic
factors.
n
Theorem 8 If f  x    a i x i has integral coefficients, n  3 , and f x  has one imaginary root c  di
i 0
with c a rational number, and d either a rational number or the square root of a rational number, then
f x   D .
Proof Since imaginary roots occur in conjugate pairs, and therefore both x  c  di  and x  c  di  are
factors of f x  , then g x   x  c  di  x  c  di   x 2  2ax  c 2  d 2 is a factor of f x  . Clearly,
given the conditions on coefficients c and d, the coefficients of g x are rational. Since n  3 ,
f x  Px  g x  , where Px must have rational coefficients [by Lemma 2 – let f x   Px  and let
g x   x 2  2ax  c 2  d 2 ].




If we factor out the lowest common denominator L1 out of all the denominators of Px , and
factor out the lowest common denominator L2 out of the denominators of g x , then we can write
P' x   g ' x 
f x   Px   g x  
, where both P' x  and g' x have integral coefficients.
L1 L 2
I’ll leave it to the reader to supply the details, parallel to the proof of the previous theorem, to
show that n2 T is an upper bound on the number of prime values which can be assumed by
P' x   g ' x 
f x  
, where T is the number of factors (each to the power 1) of the integer L1 L2 (as in
L1 L 2
the proof of the previous theorem). This quotient must therefore be composite for all but a finite number
of integral values of x. Therefore, f x   D .
Master’s Thesis
© Feb-16, Micky Earnshaw
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 factors, 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 f x  contained integral
coefficients, in which case replacing x by an integer would result in an integral factor of f x  . However,
the linear or quadratic factors of a polynomial with integral coefficients often contain irrational
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 n  3 .
f x   x 2  1  x  i x  i 
f x   2 x 2  3 

2x  3

2x  3

f x   x 3  4 x 2  2 x  6  x  4.1326374936...x  1.140435369...x  1.273072863...
f  x   x 3  3x 2  7 
x  1.27901878616x  2.13950939308  .946279541565i x  2.13950939308  .946279541565i  
x  1.27901878616x 2  4.27901878616 x  5.47294541386
Below are five sample polynomials, including the four above, which might belong to D. I let x  1,2,3,  ,
and wrote down prime functional values as they occurred – until I had ten such prime values.
f x   x 3  4 x 2  2 x  6
f x   x 3  3x 2  7
f x   x 5  4 x 4  x 2  1
prime values prime values
prime values
prime values
prime values
f 2   5
f 43  72,031
f 2   3
f 7   7,253
f x   x 2  1
f x   2 x 2  3
f 1  2
f 2   5
f 4   17
f 6   37
f 5  47
f 10   197
f 10  101
f 14  197
f 16  257
f 20   401
f 24   577
f 26   677
f 4   29
f 11  239
f 14   389
f 16   509
f 19   719
f 20   797
f 34   2,309
f 7   139
f 71  33,761
f 73  367,561
f 85  585,061
f 115  1,467,751
f 121  1,712,761
f 151  3,351,451
f 157   3,770,989
f 265  18,328,201
f 1  5
f 3  7
f 4   23
f 7   203
f 12   1,303
f 13  1,697
f 15  2,707
f 19   5,783
f 22   9,203
f 4   17
f 9  32,887
f 10  60,101
f 13  257,219
f 34  40,091,237
f 48  233,572,609
f 52  350,960,273
f 57   559,471,303
f 58  611,094,149
Master’s Thesis
© Feb-16, Micky Earnshaw
Page 10
n
Theorem 10 Let f  x    a i x i , and let g x be a subsequence of f 1, f 2, f 3,  obtained by
i 0
starting with the d
th
term of that sequence and choosing every c th term thereafter. Then the subsequence
n
n
polynomial is g  x   f cx  d    a i cx  d  
b x
i
i 0
i 0
i
i
, c  1, d  1, x  0,1,2,3,  . The relationship
between the coefficients a i of f x  and the coefficients bi of the subsequence polynomial g x is given
 1 C 01 d
b 0  
b   0
c
 1 
0
b2   0
   0
0
b3   
 
   
bn 
0
 0
by these two matrix equations:
C 02 d 2
C12 cd
c2
0


0
 C 0n 1 d n 1
 C1n 1 cd n  2
 C 2n 1 c 2 d n 3
 C 3n 1 c 3 d n  4



c n 1

0
C 0n d n  a
  0
C1n cd n 1   a 
 1
C 2n c 2 d n  2  a 
2

C 3n c 3 d n 3    a  ,

  3
  
nc n 1 d   
a n 
cn

2
n 1
n


i j
i j
1 d
2 d
n 1 d
n d




1

C
C


1
C

1
C
0
0
0
0

2
n 1
n 
c
c
c
c 

n2
1
d
d
d n 1
i j
i j
0
 C12 2   1 C1n 1 n 1  1 C1n n 
a 0  
c
c
c
c  b 0 
n 3
n2  
a  
1
i j
i j
b1 
n 1 d
n d




0


1
C

1
C

 1 0

2
2
c2
c n 1
c n  b2 
a 2  
n4
n 3
or, in reverse,    
  .
i j
i j
n 1 d
n d




0
0
0


1
C

1
C
a
b
3
3
3
n 1
n   3
  
c
c
 
  


  
  



 bn 
1
d
a n  



 n n 1


n 1
c
c


1
0

0
0

0


cn
In both of the above coefficient matrices, c  1, d  1 . For each ith-row jth-column element Aij of the first
n  1 n  1 coefficient matrix,
Aij  Ci j11c i 1 d j i ; for each ith-row jth-column element Aij of the second
d j i
. In both matrices, CQP  0 if P  Q .
c i 1
Proof Following is a derivation of the matrix of coefficients in the first matrix equation.
n
n
n
n
 i

 i

i
k
g x    bi x i   a i cx  d    a i  C ki cx  d i  k    a i  C ki c k d i  k x k  . To find the
i 0
i 0
i 0
i 0
 k 0

 k 0

i
coefficient bi of each x in g x , I’ll expand g x  f cx  d  with n  4 , then extrapolate and
n  1 n  1 coefficient matrix,
Aij   1
i j
C i j11
4
generalize. When n  4 , g x    bi x i  b0  b1 x1  b2 x 2  b3 x 3  b4 x 4 
i 0
f cx  d   a0  a1 cx  d   a 2 cx  d   a3 cx  d   a 4 cx  d  
2

3


4

a 0  a1 cx  d   a 2 c 2 x 2  2cdx  d 2  a 3 c 3 x 3  3c 2 dx 2  3cd 2 x  d 3 

a 4 c x  4c dx  6c d x  4cd x  d
4
4
3
3
2
2
2
3
4
=
Master’s Thesis
© Feb-16, Micky Earnshaw
a
0
Page 11
 
 

x  c a x . Equating corresponding coefficients of gx and f cx  d  …
 da 1  d 2 a 2  d 3 a 3  d 4 a 4  ca1  2cda 2  3cd 2 a 3  4cd 3 a 4 x  c 2 a 2  3c 2 da 3  6c 2 d 2 a 4 x 2 
c a
3
3
 4c 3 da 4
3
4
4
4
b0  a 0  da 1  d 2 a 2  d 3 a 3  d 4 a 4  C 00 a 0  C11 da1  C 22 d 2 a 2  C 33 d 3 a 3  C 44 d 4 a 4

 
b1  c a 1  2da 2  3d 2 a 3  4d 3 a 4  c C 01 a1  C12 da 2  C 23 d 2 a 3  C 34 d 3 a 4



b2  c 2 a 2  3da 3  6d 2 a 4  c 2 C 02 a 2  C13 da 3  C 24 d 2 a 4


. Expressing this set of equations in matrix form, we have…
b3  c 3 a 3  4da 4   c 3 C 03 a 3  C14 da 4

b4  c a 4   c C a 4
4
4
b0  1
b  
 1  0
b2   0
  
b3  0
b4  0
4
0
d d2
c 2cd
0 c2
0 0
0 0


d 4  a 0 
  
4cd 3   a1 
6c 2 d 2   a 2 
  
4c 3 d   a 3 
c 4  a 4 
d3
3cd 2
3c 2 d
c3
0

b0  C00
b   0
 1  C1
b2   C20
   0
b3  C3
b4  C40
C01d
C11c
C21
C31
C41
C02 d 2
C12cd
C22c 2
C32
C42
C03d 3
C13cd 2
C23c 2 d
C33c 3
C43
C04 d 4  a0 
  
C14cd 3   a1 
C24c 2 d 2   a2 
  
C34c 3d   a3 
C44c 4  a4 
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 bi ' s
in terms of the a i ' s ), because multiplying both sides of the first matrix equation by the inverse of the
coefficient matrix gives me the second matrix equation (the a i ' s in terms of the bi ' s ). 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 n  1, 2,3 & 4.
1 d 
n 1 

0 c 
1

0
n3 
0

0
1
d

1  c 

1 
0

c 

d
c
d2
2cd
0
0
c2
0
d3 

3cd 2 
3c 2 d 

c 3 
1 d d 2 


n  2 0 c 2cd 
0 0 c 2 


1

d
1 
c

1
0
c


0 0

0 0

d2
c2
d
2 2
c
1
c2
0
d3 
 3 
c 
d2 
3 3
c 
d
3 3 
c 
1 
c 3 
1

d
1 
c

1
 0

c

0 0

d2 

c2 
d
2 2 
c 
1 

c2 
© Feb-16, Micky Earnshaw
Master’s Thesis
Page 12

d
d2
d3
d4 
1




c
c2
c3
c4 

1

1 d d 2
3d 2
4d 3 
d3
d4 
0 1  2 d



2

c
c2
c3
c4 
4cd 3 
0 c 2cd 3cd

1
d
d2 
n  4 0 0 c 2 3c 2 d 6c 2 d 2   0 0
3 3 6 4 


c2
c
c 
3
3

0
c
4c d 
0 0
1
4d 

0
 4 
0 0
0
0
c 4 
0 0

c3
c

1 
0
0
0 0

c4 

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 g x  f x will mean that there exist integers c and d, c  1, d  1 , such that
the relationship between the coefficients given in Theorem 10 holds. The values of g I 
 
are a subset of the values of f I  , i.e., g 1, g 2, g 3,  is a subsequence of
f 1, f 2, f 3,  .
 
Definition 12 Let g x  f x. Then I’ll say that g x generates f x  , or that f x  is generated by
g x , and I’ll refer to g x as the generator of f x  .
Master’s Thesis
© Feb-16, Micky Earnshaw
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 f x  and want to find a generator
polynomial g x , I’ll choose f x   3x 2  x  7 . For the generator polynomial g x , I’ll specify starting
on the 2 nd term of the sequence f 1, f 2, f 3,  , and choosing every 3 rd term thereafter, i,e.,
c  3, d  2 . Since a 0  7, a1  1, a 2  3 , the first matrix equation of Theorem 10 becomes
b0  1 d
 b   0 c
 1 
b2  0 0
d 2  a 0  b0  1 2 4   7  17 

2cd    a1    b1   0 3 12   1  33
c 2  a 2  b2  0 0 9   3  27 
Thus, g x   27 x 2  33x  17 . Here’s the sequence of values.
x
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
f x   3x 2  x  7
9
17
31
51
77
109
147
191
241
297
359
427
501
581
667
759
857
961
1071
1187
g x   27 x 2  33x  17  f 3x  2
17  g 0  f 3  0  2  f 2
77  g 1  f 3 1  2  f 5
191  g 2  f 3  2  2  f 8
359  g 3  f 3  3  2  f 11
581  g 4  f 3  4  2  f 14
857  g 5  f 3  5  2  f 17
1187  g 6  f 3  6  2  f 20
Master’s Thesis
© Feb-16, Micky Earnshaw
Page 14
Discussion (continued)
For the second matrix equation, where we’re given a generator polynomial g x and we want to
generate a polynomial f x  , I’ll choose g x   x 2  x  3 , resulting from starting with the 3 rd term of the
sequence f 1, f 2, f 3,  , and choosing every 4 th term thereafter, i,e., c  4, d  3 . Since
b0  3, b1  1 , and b2  1 , the second matrix equation of Theorem 10 becomes

d
1 
c
a 0  
 a   0 1
 1 
c
a 2  
0 0

Thus, f x  
x
0
1
2
d2 
3

1 

2

c  b0  a 0 
4
d     
1
 2 2   b1    a1   0
4
c 

1  b2  a 2  

0 0
c2 


1 2 5
69 1 2
x  x

x  10 x  69 . Here’s the sequence of values.
16
8
16 16
f x  

1 2
x  10 x  69
16

g x   x 2  x  3  f 4 x  3
3 34
3 165
3
3
4
2
5
2
6
2
3
7
8
9 
 69 

16  3   16 
 5
3
    1   
8
 8
1   1   1 
 16 
16 
3
13
16
3
4
13
16
3  g 0  f 4  0  3  f 3
3  g 1  f 4 1  3  f 7
5
16
3
4
9
3
10
4 165
5
5  g 2  f 4  2  3  f 11
11
12
5
13
6
14
7
9
9  g 3  f 4  3  3  f 15
10 165
3
11
4
13 165
15
15  g 4  f 4  4  3  f 19
15
16
17
18
19
13
16
3
4
13
16
© Feb-16, Micky Earnshaw
Master’s Thesis
Page 15
Theorem 11 Let g x  f x. If g x D , then f x   D . If c  1 , then g x D if and only if
f x   D .
Proof The first part is clear, since every value that g x assumes appears in the sequence
f 1, f 2, f 3,  . If c=1, then the sequence f d  1, f d  2, f d  3,  is identical to the
sequence g 1, g 2, g 3,  , so if one polynomial  D , then the other one is.
Discussion
The second example of applying Theorem 10 generated a polynomial f x  


1 2
x  10 x  69
16
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 polynomials f x  and g x with g x  f x, 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 generator and the generated polynomial are integers.
Theorem 12 Given nth degree polynomials f x  and g x with g x  f x. If the coefficients a i of
the generated polynomial f x  are integers, then the coefficients bi of the generator polynomial g x are
integers. If the coefficients bi of the generator polynomial g x are integers, with the added requirement
that c i | bi for i  1,2,  , n , then the coefficients a i of the generated polynomial f x  are integers.
Proof Assume integral coefficients a i for f x  . From the relationship between the coefficients a i of
f x  and the coefficients bi of g x given in the first equation in the proof of Theorem 10, namely,
n
that bi  c i  C jji d j i a j with c and d integers, then each coefficient bi of g x is an integral linear
j i
combination of the integral a i ' s , and is therefore itself an integer. For the second conclusion, assume
integral coefficients bi for g x , where c i | bi , i  1,2,  , n . Referring to the second matrix of
coefficients in Theorem 10, which converts coefficients bi of g x into coefficients a i of f x  , in
order to end up with integers for a 0 , a1 ,  , a n from integral coefficients b0 , b1 ,  , bn , we must have
1
c i | bi i . This is clear if you look at the diagonal elements Aii  i , each of which is multiplied by bi
c
i
in the product matrix. Since c is in the denominator with bi the only integer in the numerator, and we
b
require that each a i of f x  be an integer, we must have c i | bi i . Since c i | bi i , each quotient ii
c
is an integer i . Since each a i is a linear combination of terms, each having the same form: a i  {the
b
integer C QP }  {a power of the integer d}  {the integer ii }, we can conclude that each a i must itself be
c
an integer.
Master’s Thesis
© Feb-16, Micky Earnshaw
Page 16
Theorem 13 Given nth degree polynomials f x  and g x . Then g x  f x if and only if the first
matrix equation of Theorem 10 is satisfied for all of the coefficients a i of f x  and bi of g x , for
b
1
i  0,1,2,  , n , with integers c  n n and d 
na n
an

a
1 
 bn 1

bn 1  n
 n 1  a n 1  
c
 na n 
 bn




n 1
n

 a n 1  .


Proof From the last product in the first matrix equation of Theorem 10, bn  c n a n , from which c  n
From the next-to-last product in the same matrix equation, bn 1  c n 1 a n 1  na n d  , from which
d
1  bn 1

 n 1  a n 1  . The second version for d in the theorem results from replacing c by
na n  c

n
bn
.
an
bn
.
an
Discussion
A question suggested by Theorem 13 is whether a polynomial g x which is not primitive could
still generate a polynomial f x  which is primitive, even if c, d   1. The answer is, YES. Here are such
a g x , f x  , c and d: g x   36 x 2  12 x  18, c  6, d  8, f x   x 2  14 x  66 . Note that 36,12,18  6 ,
which makes g x non-primitive, and c, d   2 . Here’s the matrix relationship between the coefficients
a i of f x  and the coefficients bi of g x .
General...

d
1 
c
a 0  
 a   0 1
 1 
c
a 2  
0 0

Specific...
d 

c 2  b0 
d
 2 2    b1  
c 
1  b2 

c2 
2
4

1 

3
 66  
 14  0 1

 
6
 1  
0 0
16 
9  18 
.
4
   12 
9
1  36
36 
As you can see, to guarantee that a polynomial f x  which is generated from another polynomial
g x of the same degree is primitive, all that is required is that the coefficient bn of x n in g x be equal
to c n . The example above provides a counterexample to the converse of Theorem 13: Given g x  f x ,
if f x  is primitive then g x is primitive. The fact is, however, that non-primitive generators g x don’t
do us much good. This is because the point here is to find primitive polynomials g x D which generate
infinitely many other polynomials f x  , depending upon the values of c and d. Each such generated
polynomial f x  is a member of D because every value of g x for x  0,1,2,  ,  is also a value of
f x  for x  1,2,  ,  . Therefore, if the sequence g 1, g 2, g 3,  contains infinitely many primes,
then so does the sequence f 1, f 2, f 3,  , because g 1, g 2, g 3,  is a subsequence of
f 1, f 2, f 3,  .
© Feb-16, Micky Earnshaw
Master’s Thesis
Page 17
Theorem 14 Given nth degree polynomials f x  and g x with g x  f x. If g x is primitive, and
c i divides each integral coefficient bi of x i in g x , then f x  is primitive.
Proof Referring to the first matrix of coefficients of Theorem 12, if f x  were not primitive, then the
greatest common divisor of the a i ' s , r  a 0 , a1 , , a n  , would divide each coefficient bi of g x , and
therefore g x could not be primitive. From Theorem 12, the requirement that c i | bi i guarantees that
the coefficients of f x  are integers.
?Theorem If f x  is of degree 2 or greater, is primitive, and has no real or complex roots with rational
coefficients, then f x   D .
?Theorem If f x  is of degree 2 or greater, is primitive, and assumes a prime value for two distinct
integral values of x, then f x   D .