Chapter 2: Primes and their Factorization
17
SECTION C Properties of Prime Numbers
By the end of this section you will be able to
 understand some unsolved problems regarding primes
 prove properties of primes in an arithmetic progression
There are many unanswered questions in Number Theory. In this section we examine some
of these.
C1 Unsolved Problems of Primes
A conjecture is a statement which has no proof. Generally for a conjecture our intuition
tells us that the statement is true but we cannot find a proof.
1.
Twin Prime Conjecture (Euclid around 300BC.)
Primes like 11 and 13 are examples of twin primes.
Definition (2.12).
Twin primes are of the form p and p  2 where p is prime.
Other examples are
(i) 17 and 19
(ii) 41 and 43
(iii) 1 000 000 000 061 and 1 000 000 000 063
The still outstanding conjectures of Number theory is:
There infinitely many primes of the form p and p  2 .
No one has come up with a proof of this result although most mathematicians do seem to
believe it is true. The largest known twin primes are:
(3 756 801 695 685·2666669 )  1
These numbers were discovered in 2011.
2.
Lagrange’s Conjecture (1775)
Every odd integer greater than 5 can be written as a sum p  2q , where p and q are both
primes.
For example
7  3  2  2  , 9  5  2  2  , 11  5  2  3 , 13  3  2  5  , 15  5  2  5  , 
3.
Goldbach’s Conjecture (1642)
Another unproved conjecture in Number Theory is Goldbach’s Conjecture:
Every even number greater than 2 can be written as a sum of two primes.
For example
4  22
6  33
8  53
10  5  5

100  57  43

1000  509  491

Chapter 2: Primes and their Factorization
18
Again this result has not been proven. Mathematicians have been trying to prove this result
for over 270 years. Computers have verified this result for all the even integers up to
4  1018 . The nearest proposition to Goldbach’s conjecture that has been proven is:
‘Every even integer > 2 is the sum of six or fewer primes.’
There also exists another Goldbach Conjecture known as Goldbach’s Weak Conjecture
which claims:
Every integer greater than 5 can be written as the sum of three primes.
Christian Goldbach was born in Königsberg, Russia in 1690
and attended the local university. He studied law as well as
mathematics. At the young age of 20 he went on a
education tour of Europe. In 1712 whilst in Oxford he met
another famous mathematician Bernoulli, where the two of
them discussed infinite series. However this topic did not
appeal to Goldbach. 5 years later in 1717 he read an article
by Leibniz on evaluating the area of a circle using infinite
series. This time he was fascinated.
Figure 6 Christian Goldbach 1690 to 1764
In 1719 Goldbach began a correspondence with Euler which lasted for 35 years. In one such
correspondence in 1742 he made his famous conjecture.
4. Landau’s Conjecture (1912)
Landau gave the following conjecture in 1912.
There are infinitely many primes of the form n 2  1 .
22  1  5, 42  1  17, 62  1  37, 
All 4 of these problems have remained unproved for hundreds of years, or in the case of
twin prime conjecture for thousands of years.
C2 Distribution of Primes
We have already proven that there are an infinite number of primes. How are these primes
distributed?
This is one of the most interesting questions in the whole of mathematics.
While we see our number system as an ordered, predictable set, its primes are
unpredictable. There is no formula which locates all the primes so we cannot say where the
next prime will sprout out from.
Mathematicians keep on searching for larger and larger primes. In 1876, the French
mathematician Édourd Lucas proved 2127  1 is prime and this remained the largest known
prime for 75 years. It is only with the advent of the digital computer that we keep on
finding larger and larger primes on a regular basis. The largest prime number at present (6th
February 2013) is 257 885 161  1 which is 17 million digits long.
Electronic Frontier Foundation are offering a prize of $150 000 to anyone who finds the
first 100 million digit prime.
The table below shows the number of primes   N  below a certain number N and the
distribution of primes below N:
Chapter 2: Primes and their Factorization
19
 N / N Proportion of primes
  N  - Number of
N - Number
Percentage of
primes below N
primes  N
100
25
0.25
25
1 000
168
0.168
16.8
10 000
1 229
0.1229
12.29
100 000
9 592
0.095 92
9.592
1 000 000
78 498
0.078 498
7.8498
10 000 000
664 579
0.066 4579
6.64579
100 000 000
5 761 455
0.05 761 455
5.761455
1 000 000 000
50 847 534
0.050 847 534
5.0847534
Note that the proportion of the numbers which are primes below 100 is 25%, the proportion
of primes below 1 000 is 16.8%. The proportion is getting smaller for larger N as you
observe from the last column of the above table. What does this mean?
The primes are thinning out for larger and larger values of N. Will they eventually vanish?
No because we have already proven that there are an infinite number of primes. Below is
the graph of the prime counting function   x  against x:
Figure 7
As the slope of the curve is getting smaller for larger N so this suggests that the primes are
thinning out. So for large numbers we expect to see large gaps between consecutive primes
which also implies that we should have a large number of consecutive composite numbers.
How do we know there are a large number of consecutive composite numbers?
As we’ve already seen, there is no distribution of the primes which would allow us to tell
where the next prime is located. The twin prime conjecture has shown us that there are
situations where we only have 1 composite number between two primes. The difference
between two successive prime numbers is called the prime gap, the table below gives the
first 12 prime numbers and the corresponding prime gap:
2
Gap
3
1
5
2
7
2
11
4
13
2
17
4
19
2
23
4
29
6
31
2
37
6
Chapter 2: Primes and their Factorization
20
If we subtract one from the prime gap, then we get the number of consecutive composite
integers between the two primes. We can see that between 23 and 29 there are 6  1  5
composite integers.
Generalising this:
Proposition (2.13).
For any positive integer n there exists a consecutive sequence of n composite integers.
What does this mean?
This means that between some pair of successive primes on the number line there will be n
composite integers. It does not mean that the occurrence is unique, there may be several
pairs of successive primes which sit either side of n consecutive composite integers.
Proof.
In order to prove this we need to write out the all integers in a way that separates the
composite numbers from the primes.
Consider the list of integers
I1   n  1 ! 2, I 2   n  1 ! 3, I 3   n  1 ! 4,  ,
I n 1   n  1 ! n, I n   n  1 !  n  1
Consider the first number in this list, I1   n  1 ! 2 . We know that 2 is a factor of this
number. How?
Because  n  1 !  1 2  3  4     n  1 so 2 is factor of  n  1 ! therefore
2
 *
 n  1! 2
Since 2 is a factor of I1   n  1 ! 2 so it is a composite integer.
Similarly the second number in the list, I 2   n  1 ! 3 , is divisible by 3 and so is
composite. The last number I n   n  1 !  n  1 is divisible n  1 . Hence the n numbers in
this list are all composite because each of these has a factor other than 1 and number itself.
We have produced n consecutive composite integers.
■
What does this mean?
It means that we can choose any positive integer n, 5 for example, and somewhere else on
the real number line we will find 5 composite numbers sat between 2 primes.
In the case of 5 we find that further along the number line, 5 composite numbers 24, 25, 26,
27 and 28 sit between the prime numbers 23 and 29. Further along the number line,
between the primes 31 and 37 we encounter 5 consecutive composite integers.
This works for all n, no matter how large, so if we choose the positive integer to be 1
million, further up the number line we will find 1 million composite integers sat side by
side, framed by 2 primes.
The numbers in the list (*) can be used to generate n consecutive composite numbers. For
example if we choose n  6 then the following 6 integers are consecutive composite
integers:
Chapter 2: Primes and their Factorization
 6  1! 2  5042
 6  1! 3  5043
 6  1! 4  5044
 6  1! 5  5045
 6  1! 6  5046
 6  1! 7  5047
21
 2 divides 5042
3 divides 5043
 4 divides 5044
5 divides 5045
6 divides 5046
7 divides 5047 
C3 Primes of the form 4n  1
We can write every integer in one of the following forms:
4n, 4n  1, 4n  2 or 4n  3
How do we know this?
Well we can prove this.
Proposition (2.14).
Every integer has one of the following forms:
4n, 4n  1, 4n  2 or 4n  3
How do we prove this?
By applying the Division Algorithm (1.7):
Given integers a and b  1 , there exist unique integers q and r such that
a  bq  r
0r b
Proof.
Let a be any integer and b  4 then applying the Division Algorithm with q  n we have
a  4n  r
0r 4
The remainder r can only take values of 0, 1, 2 or 3 because 0  r  4 . Substituting these
values of r into a  4n  r gives
a  4n, 4n  1, 4n  2 or 4n  3
This means that any integer a can be represented by one of these numbers.
■
We are interested in primes of this format; 4n, 4n  1, 4n  2 or 4n  3 . However primes
cannot be of the form 4n or 4n  2 . Why not?
Because these are even numbers so are divisible by 2
Can we write all primes in the form 4n  1 or 4n  3 ?
These numbers 4n  1 or 4n  3 cover all the odd numbers. Hence all the odd primes (that
is all the primes other than 2) must be captured by either of these numbers.
Next we prove there are infinitely many primes of the form 4n  3 . For example
4 1  3  7, 4  2   3  11, 4  4   3  19, 4  5   3  23, 4  7   3  31, 
In order to prove this we need a lemma:
Lemma (2.15).
The product of two or more integers of the form 4n  1 can be written in the same form,
4n  1 .
Proof.
It is sufficient to prove this result for two arbitrary integers a and b of the form
4  integer   1
Let a  4n1  1 and b  4n2  1 then
Chapter 2: Primes and their Factorization
22
a  b   4n1  1 4n2  1  16n1n2  4n1  4n2  1
 4  4n1n2  n1  n2   1
 4n  1 where n  4n1n2  n1  n2
Hence the product a  b is of the form 4n  1 .
We can repeat this process with another multiple of 4  integer   1 and we will get an
answer of the form 4  integer   1 .
■
Proposition (2.16).
There are infinitely many primes of the form 4n  3 .
How do we prove this result?
By contradiction
Proof.
Suppose there are a finite number of primes of the form 4n  3 . Let these be
p1 , p2 , p3 ,  , P
 †
where P is the largest such prime. Consider the number
N  4  p1  p2  p3    P   1
We can rewrite N as
N  4  p1  p2  p3    P   1
 4  p1  p2  p3    P   1  3
Note that N is an odd number and of the form 4n  3 . Either N is composite or prime.
If N is prime then we are done because we have a larger prime of the form 4n  3 which is
greater than P.
If N is composite then it has a prime factor of the form 4n  3 . Why?
N is odd and all odd numbers have the form 4n  1 or 4n  3 . If all the factors of N were of
the form 4n  1 then by the previous Lemma (2.15):
The product of two or more integers of the form 4n  1 is of the same form, 4n  1 .
N would also be of this form 4n  1 but it is not. Hence N must have at least one prime
factor of the form 4n  3 . Call this prime factor Q.
However this prime factor Q of N cannot be one of the numbers in the list  †  . Why not?
Suppose Q is amongst the list  †  then Q
 p1  p2  p3    P 
and because Q is a prime
factor of N so Q N . Therefore
Q
N  4  p1  p2  p3    P   Q
This result Q
 1
 1
because N  4  p1  p2  p3    P   1
is impossible because Q is prime.
Our supposition that there are a finite number of primes of the form 4n  3 must be wrong.
This completes our proof.
■
There are also infinite number of primes of the form 4n  1 . (See Exercise 2(c).)
C4 Primes in an Arithmetic Progression
Chapter 2: Primes and their Factorization
23
Dirichlet proved that the number of primes in an arithmetic progression is infinite. Who
was Dirichlet?
Dirichlet was born in Germany in 1805 and by the age of
12 he had developed a love for mathematics; spending all
his pocket money on purchasing mathematics textbooks.
After leaving school he decided to study in Paris rather
than Germany because at that time the standard of
mathematics teaching and research was a lot higher in
France.
In 1828 Dirichlet manage to get an academic post at the
Military College in Berlin which meant that he could also
teach at the University of Berlin. He became Professor
of Mathematics at the University of Berlin where he
remained until 1855.
In 1855 when Gauss, one of the greatest mathematicians
of all time, passed away, Dirichlet took up his chair in
Göttingen. He stayed there until his death in 1859.
Figure 7
Johann Dirichlet 1805 to 1859
In 1837 Dirichlet proved that there are infinitely many primes in an arithmetic progression.
With this he became one of the founders of ‘Analytic Number Theory’ a branch of number
theory which uses mathematical analysis to prove and solve problems in this field.
Next we state Dirichlet’s Theorem about primes in an arithmetic progression. What is an
arithmetic progression?
Consider a sequence of numbers 1, 8, 15, 22, 29,  where each term is 7 more than the
previous term, then this sequence is an example of an arithmetic progression. We can write
a general arithmetic progression as follows:
a, a  b, a  2b, a  3b, 
where a is the first term and b is the common difference.
Dirichlet’s Theorem (2.17).
Let a and b be relatively prime positive integers, then the arithmetic progression
a, a  b, a  2b, a  3b, 
contains infinitely many primes.
Proof.
To be given later.
What does this mean?
Let a  2 and b  3 then there are an infinite number of primes in the list:
2, 2  3  5, 2  2  3  8, 2  3  3  11, 2  4  3  14, 2  5  3  17, 
Note that Dirichlet’s Theorem does not claim that the arithmetic progression:
a, a  b, a  2b, a  3b, 
consists solely of primes. In the above example, 8 and 14 are clearly not prime.
Example 3
Show that there are infinitely many primes that end with 999.
Solution
Chapter 2: Primes and their Factorization
24
Applying Dirichlet’s Theorem (2.17) with a  999, b  1000 and gcd  999, 1000   1 .We
have:
N  999, 999  1000, 999  2 1000  , 999  3 1000  , 
N  999, 1999, 2999, 3999, 
†
Dirichlet’s Theorem claims that there are infinitely many primes of this form  †  .
Note that not all these numbers are prime.
C4 Generating Primes
Is there a simple formula for generating all primes?
No. However there is a simple formula which generates some primes. For example the
following quadratic
f  n   n 2  n  41
produces primes for n  1, 2, 3, 4,  , 39 . But when n  40 we have
f  40   402  40  41
 40  40  1  41  41 40  1  412
When n  41 we have
f  41  412  41  41
 41 41  1  1  41 43
Hence f  40  and f  41 are composite, not prime.
SUMMARY
There are still many problems in number theory which remain unresolved such as
Goldbach’s conjecture and the twin prime conjecture.
There are an infinite number of primes but there is no particular pattern apart from them
thinning out for larger and larger values.