Math 290 Lecture #32
§11.6,11.7: More Number Theory
§11.6: The Fundamental Theorem of Arithmetic. The prime numbers occupy
a special place in the divisibility of integers.
We can express the number 420 as the product 2 · 2 · 3 · 5 · 7.
That we can write every integer greater than or equal to 2 as a product of primes is
amazing, the proof of which uses the Strong Principle of Mathematical Induction.
Theorem 11.17 (The Fundamental Theorem of Arithmetic). Every integer
n ≥ 2 is either prime or can be expressed as a product of primes,
n = p1 p2 · · · pm
for primes p1 , p2 , ·, pm . Furthermore this factorization of n is unique except possibly for
the order of the factors.
Proof. The base case of n = 2 has such a factorization, namely 2 = 2.
Now suppose for an integer k ≥ 2 that every integer i such that 2 ≤ i ≤ k is either prime
or can be expressed as a product of primes.
The goal is to show that k + 1 is either prime or can be expressed as a product of primes.
If k + 1 is prime, then we are done.
We thus assume that k + 1 is composite, so that by Lemma 11.1 there are integers a and
b such that k + 1 = ab with 2 ≤ a ≤ k and 2 ≤ b ≤ k.
By the Strong induction hypothesis, each of a and b is prime or can be expressed as a
product of primes.
In either case, we have that k + 1 can be expressed as a product of primes.
To prove that the factorization in unique, we argue by contradiction.
Suppose there is an integer n ≥ 2 such that
p 1 p 2 · · · p s = n = q1 q2 · · · qt ,
where in each factorization, the primes are in nondecreasing order, i.e., p1 ≤ p2 ≤ ··· ≤ ps
and q1 ≤ q2 ≤ · · · ≤ qt .
With the factorizations being different, there is a smallest positive integer r such that
pr 6= qr .
Then for each integer i with 2 ≤ i < r we have pi = qi , and canceling the common factors
gives
pr pr+1 · · · ps = qr qr+1 · · · qt .
Consider the prime pr .
Either r = s so that pr pr+1 · · · ps is exactly pr , or r < s so that the integer pr+1 pr+2 · · · ps
is the product of s − r primes.
In either case, we have pr | qr qr+1 · · · qt .
By Corollary 11.15, because pr is prime, we have that pr | qj for some j with r ≤ j ≤ t.
Because qj is prime we have that pr = qj .
Since qr ≤ qj it follows that qr ≤ pr .
Considering instead the prime qr , we get pr ≤ qr .
Thus we arrive at pr = qr , a contradiction.
An immediate consequence of the Fundamental Theorem of Arithmetic is the following.
Corollary 11.18. Every integer exceeding 1 has a prime factor.
For the factorization n = p1 p2 · · · pm we can group the equal primes and write
n = pα1 1 pα2 2 · · · pαk k
where p1 , p2 , . . . , pk are primes satisfying p1 < p2 < · · · < pk .
This we call the canonical factorization of n.
How do we find the canonical factorization of an composite integer n > 3 into prime
factors?
The following result gives a place to start.
Lemma
11.19. If n is a composite number, then n has a prime factor p such that
√
p≤
n.
Proof. Suppose n is composite.
Then n = ab for integers a and b that satisfy 1 < a < n and 1 < b < n.
WLOG assume that a ≤ b.
Then we have that a2 ≤ ab = n, so that a ≤
√
n.
Since a > 1 we have by Corollary 11.18 that a has a prime factor p.
√
Since p | a and a | n, then p | n and p ≤ a ≤ n.
.
Examples. (a) Find the smallest prime factor of the composite n = 2695.
By Lemma 11.19, we know that 2695 has a prime factor no bigger than
√
n ≈ 51.9.
For p = 2, we apply the Division Algorithm to write 2695 = 2(1347) + 1, so that 2 - n.
For p = 3 we apply the Division Algorithm to write 2695 = 3(898) + 1, so that 3 - n.
For p = 5 we apply the Division Algorithm to write 2695 = 5(539), and we have 5 is the
smallest prime factor of 2695.
(b) Express the integer n = 54587 as a product of primes.
By Lemma 11.19 we know that n has a prime factor no bigger than
The primes 2, 3, 5, 7, 11 all fail to be factors of 54587.
But 13 is a prime factor since 54587 = 13(4199).
√
n ≈ 233.6.
At this point√we now seek the smallest prime factor of 4199, which prime factor is no
bigger than 4199 ≈ 64.7.
None of the primes 2, 3, 5, 7, 11 are factors of 4199, but 13 is because 4199 = 13(323).
√
What is the smallest prime factor of 323? It is no bigger than 323 ≈ 17.9.
The primes 2 through 13 are not prime factors, but 17 is a prime factor of 323 because
323 = 17(19).
Since 19 is prime, we have determined all of the prime factors and can now express
54587 = 13 · 13 · 17 · 19 = 132 · 17 · 19.
Expressing a composite integer n ≥ 2 in the canonical factorization seems to presuppose
that there are infinite many primes, but we have not as yet proved this.
Theorem 11.22. The number of primes is infinite (and hence denumerable).
Proof. Suppose to the contrary that the number of primes is finite and list these finitely
many primes as P = {p1 , p2 , . . . , pn }.
We will get a contradiction by factoring the integer m = p1 p2 · · · pn + 1.
The integer m satisfies m ≥ 2 and so it has a prime factor.
Since every prime belongs to P , there is an integer i with 1 ≤ i ≤ n such pi | m.
Then m = pk for some integer k.
The integer l = p1 · · · pi−1 pi+1 · · · pn then satisfies
1 = m − p1 · · · pn = pi k − pi l = pi (k − l).
Since k − l ∈ Z, this says that pi | 1, a contradiction.
§11.7: Concepts Involving Sums of Divisors. For an integer n ≥ 2, a positive
integer a is called a proper divisor of n if a | n and a < n.
The proper divisors of 6 are 1, 2, and 3, the proper divisors of 10 are 1, 2 and 5, and the
proper divisors of 28 are 1, 2, 4, 7, and 14.
A positive integer n ≥ 2 is called perfect is the sum of its proper divisors is n.
The integer 6 is perfect because 1 + 2 + 3 = 6, the integer 10 is not perfect because
1 + 2 + 5 = 8 6= 10, while the integer 28 is perfect because 1 + 2 + 4 + 7 + 14 = 28.