44
Chapter 1. The Integers
Theorem 1.4.7. Let m, a, b be nonzero integers. If m | (ab) and gcd(m, a) = 1, then m | b.
Proof. See Exercise 3.
Lemma 1.4.8. Let n, m 2 Z and suppose that m 6= 0 and that n = qm + r. Then gcd(n, m) =
gcd(m, r).
Proof. See Exercise 4.
Exercises 1.4
1. For each pair of integers n and d, find q and r such that n = dq + r where 0  r < d.
(a)
(b)
(c)
(d)
n = 335 and d = 17.
n = 335 and d = 17.
n = 121 and d = 13.
n = 121 and d = 13.
2. Use the division algorithm with d = 3 to prove that the square of every natural number n has
the form 3k or 3k + 1 for some natural number k.
3. Let m, a, b 2 Z where gcd(m, a) = 1. By Theorem 1.4.3 there are integers s and t such that
1 = sa + tm. Using this equation, prove Theorem 1.4.7.
4. Prove Lemma 1.4.8.
5. Let m and n be nonzero integers that are relatively prime. Suppose that m | ` and n | ` for
some integer `. Prove that mn | `.
6. Prove the following theorem.
Theorem. Every natural number n
n = ai 3i + ai
0 can be expressed as
13
i 1
+ · · · + a 2 32 + a 1 3 + a 0
where i is a natural number and 0  aj  2 for all j = 0, 1, 2, . . . , i.
For Exercise 3, multiply both sides of the equation 1 = sa + tm by b. For Exercise 4, let d =
gcd(n, m) and e = gcd(m, r). Using the equation n = qm + r and Definition 1.4.2, show that d | e
and e | d. Thus, d = e. For Exercise 5, since gcd(m, n) = 1, there are integers s and t such that
1 = sm + tn. Now muliply both sides of this equation by `. For Exercise 6, use the strong induction
strategy 1.3.3. By Theorem 1.4.1 every natural number n can be written as n = 3q + r where q
and r are natural numbers and 0  r  2.
1.5
The Fundamental Theorem of Arithmetic
In number theory, Euclid’s fundamental theorem of arithmetic (or unique factorization theorem)
states that every natural number greater than 1 can be written as a unique product of prime
numbers. For example; 23,456,700 = 22 · 33 · 52 · 67 · 389 and there is no other possible factorization
of 23,456,700 into prime numbers. Since multiplication is commutative and associative, the order
of the prime factors is usually written in ascending order, that is, from least to greatest.
We shall prove the fundamental theorem of arithmetic in two steps. First we prove Theorem
1.5.1 which states that every natural number n 2 can be written as a product of primes. In the
second step we prove that there is only one such prime factorization (see Theorem 1.5.6). Both of
these proofs use the well-ordering principle.
1.5 The Fundamental Theorem of Arithmetic
Theorem 1.5.1 (Existence of Prime Factorization). Every natural number n
as a product of primes.
45
2 can be expressed
Proof. We prove that every n 2 is a product of primes. Suppose, for a contradiction, that there
is an n 2 which is not a product of primes. Let N 2 be the least such natural number. Thus,
N is the smallest integer n
2 that is not the product of prime numbers.
(1.5)
Observe that n = 2 is clearly a ‘product’ of one prime; namely, itself. Thus, N > 2. Clearly, either
N is a prime or N is not a prime. Thus, there are two cases to consider:
Case 1: N is a prime p. Since N = p is a prime, we conclude that N is a ‘product’ of one
prime; namely, itself. This contradicts our assumption (1.5).
Case 2: N is not a prime. Since N is not a prime, then N can be expressed as a product
n = ij where 2  i < n and 2  j < n. By (1.5), i and j can expressed as a product of primes, say
i = p1 p2 · · · pk and j = q1 q2 · · · ql . Thus,
N = ij = (p1 p2 · · · pk )(q1 q2 · · · ql ) = p1 p2 · · · pk q1 q2 · · · ql .
Therefore, N can be expressed as a product of primes. This contradiction completes the proof.
Euclid was the first to prove the following lemma which shows that if a prime number p evenly
divides the product of two whole numbers ab, then either p divides a or p divides b. Euclid’s lemma
is used to prove the uniqueness of a prime factorization.
Lemma 1.5.2 (Euclid’s Lemma). Let a and b be natural numbers, and let p be a prime. If p | (ab),
then p | a or p | b.
Proof. Let a and b be natural numbers and let p be a prime so that p | (ab). Suppose p - a. Since
p is a prime and p - a, it follows that gcd(a, p) = 1. Theorem 1.4.7 now implies that p | b.
We can now derive the following corollary.5
Corollary 1.5.3. Let a be a natural number and p be a prime. If p | a2 , then p | a.
One can prove our next theorem by induction on n, using Lemma 1.5.2 in the inductive step.
Theorem 1.5.4. Let a1 , a2 , . . . , an be natural numbers and let p be a prime. If p | (a1 a2 · · · an ),
then p | ai for some i where 1  i  n.
Definition 1.5.5. Let n > 1 be a natural number. We shall say that a prime factorization
n = p1 p2 · · · pk is in ascending order if pi  pj when 1  i  j  k. We shall refer to such a
factorization as an ascending prime factorization.
Example 1. Ascending prime factorizations: 10 = 2 · 5, 20 = 2 · 2 · 5, 13 = 13, 84 = 2 · 3 · 3 · 7.
Theorem 1.5.6 (Uniqueness of Prime Factorization). Let n
2 be a natural number. Suppose
that n = p1 p2 · · · pr is an ascending prime factorization and that n = q1 q2 · · · qs is also an ascending
prime factorization. Then r = s and p1 = q1 , p2 = q2 , . . . , pr = qs .
5
A corollary is a statement that follows from a previously proven theorem.
46
Chapter 1. The Integers
Proof. We prove, using the well-ordering principle, that for all natural numbers n
2, there is
only one ascending prime factorization of n. Suppose, for a contradiction, that there is a natural
number n
2 that has two di↵erent ascending prime factorizations. Let N be the least such
natural number.
N is the smallest integer n
2 that has two di↵erent ascending prime factorizations.
(1.6)
Since n = 2 is the only prime factorization of n, we see that N > 2. Clearly, either N is a prime
or N is not a prime. Thus, there are two cases to consider:
Case 1: N is a prime p. Since N = p is a prime, we conclude that N is the unique ‘product’
of one prime; namely, itself. This contradicts our assumption (1.6).
Case 2: N is not a prime. Let
N = p1 p2 · · · pr and N = q1 q2 · · · qs .
(1.7)
be two di↵erent ascending prime factorizations of N where r > 1 and s > 1. We now prove that
pr = qs . Suppose, for a contradiction, that pr 6= qs . Thus, either pr < qs or pr > qs . Suppose
that (?) pr < qs . Since N = q1 q2 · · · qs , it follows that qs | n. Moreover, because N = p1 p2 · · · pr , it
follows that qs | (p1 p2 · · · pr ). By Theorem 1.5.4, there is an i with 1  i  r such that qs | pi . Since
qs and pi are both primes, it follows that qs = pi . Since pi  pr , we conclude that qs  pr . Since
the inequality qs  pr contradicts (?), we infer that pr < qs is impossible. A similar argument
(see Exercise 7) shows that pr > qs is also impossible. Therefore, pr = qs . Now since pr = qs , let
p = pr = qs be this common value. From (1.7) we see that
N = p 1 p 2 · · · pr
1p
= q1 q2 · · · qs
1 p.
By canceling p, we obtain k = p1 p2 · · · pr 1 = q1 q2 · · · qs 1 with 2  k < N . Since 2  k < N ,
it follows from (1.6) that r 1 = s 1 and p1 = q1 , p2 = q2 , . . . , pr 1 = qs 1 . Thus, r = s and
because pr = qs , we conclude that the ascending prime factorizations in (1.7) are exactly the same.
This contradiction completes the proof of the theorem.
It often happens that certain primes occur more than once in an ascending prime factorization.
We can simplify such ascending prime factorizations; for example, 100 = 2 · 2 · 5 · 5 = 22 · 52 ,
56 = 2 · 2 · 2 · 7 = 23 · 7, 882 = 2 · 3 · 3 · 7 · 7 = 2 · 32 · 72 , 6936 = 23 · 3 · 172 , 1200 = 24 · 3 · 52 . Thus,
we can simplify a prime factorization of any natural number n > 1 by using exponents and write
n = pa11 pa22 · · · pakk where p1 , p2 , . . . , pk are distinct ascending primes and a1 1, a2 1, . . . , ak 1.
Theorem 1.5.7 (Fundamental Theorem of Arithmetic). Let n > 1 be a natural number. There
exists distinct primes p1 , p2 , . . . , pk in ascending order and exponents a1
1, a2
1, . . . , ak
1
such that n = pa11 pa22 · · · pakk . Furthermore, if n = q1b1 q2b2 · · · q`b` is any ascending prime factorization
into distinct primes, then ` = k, p1 = q1 , p2 = q2 , . . . , pk = q` and a1 = b1 , a2 = b2 , . . . , ak = b` .
Proof. This follows immediately for Theorems 1.5.1 and 1.5.6.
Exercises 1.5
1. Prove Theorem 1.5.4.
2. Let a and b be integers, and let p be a prime. Prove that if p | a and p | (a2 + b2 ), then p | b.
p
3. Let p be a prime. Prove that p is irrational.
p
4. Let p be a prime. Let m 1 be a natural number. Prove that if p - m then mp is irrational.
1.6 Congruence Modulo m
5. Let a, b 2 N and p be a prime. Prove for all natural numbers n
pn | b.
47
1, if pn | (ab) and p - a, then
6. In the proof of Theorem 1.5.6 it was stated that if n is a prime number p, then n = p is the
only prime factorization of n. Explain why this is true.
7. In our proof of Theorem 1.5.6 we stated that “A similar argument shows that pr > qs is also
impossible.” Under the assumptions used in our proof of the theorem, present this similar
argument.
Exercise Notes: For Exercise 5, use induction on n.
1.6
Congruence Modulo m
Karl Friedrich Gauss (1777-1855) has been called the “Prince of Mathematicians” for his many
contributions to pure and applied mathematics. One of Gauss’s most important contributions to
number theory was the introduction of an equivalence relation on the integers called congruence
modulo m, where m 1 is an integer. We will investigate Gauss’s congruence relation. We will
also show that the operations of addition, subtraction, and multiplication preserve Gauss’s relation
(see Theorem 1.6.2).
Definition 1.6.1 (Congruence modulo m). Let m 1 be an integer. For any integers a and b, we
define
a ⌘ b (mod m) if and only if m | (a b).
For example, 10 ⌘ 2 (mod 4) because 4 | (10
24 ⌘ 0 (mod 4) because 4 | (24 0).
2),
5 ⌘ 3 (mod 4) since 4 | ( 5
3), and
Remark. When a ⌘ b (mod m) we say that a is congruent to b modulo m. We also write
a 6⌘ b (mod m), when we wish to say that a is not congruent to b modulo m.
The notation a ⌘ b (mod m) given in Definition 1.6.1 is just a statement about divisibility and
is used mainly to simplify reasoning about the divisibility concept. When m 1 is an integer and
a, b are integers, one can easily verify that the following are all equivalent:
1. a ⌘ b (mod m),
2. m | (a
3. a
b),
b = km for some k 2 Z,
4. a = b + km for some k 2 Z.
Example 1. Let n be any integer. The division algorithm (see Theorem 1.4.1) implies that there
are integers k and r such that n = 4k + r and 0  r < 4. Thus, we have n = 4k + 0, n = 4k + 1,
n = 4k+2 or n = 4k+3. We conclude for every integer n that either n ⌘ 0 (mod 4) or n ⌘ 1 (mod 4)
or n ⌘ 2 (mod 4) or n ⌘ 3 (mod 4).
1.6.1
Fundamental Properties
In this section we shall establish a series of theorems that will allow us to develop a so-called
congruence algebra. The following theorem is a fundamental result showing that the congruence
modulo m relation is preserved under the operations of addition, subtraction, and multiplication.