PRIME FACTORIZATION
WOUTER KAGER
Suppose that a and b are two positive integer numbers. We have learned
in school how to divide a by b, e.g. by long division. This yields two unique
integer numbers q and r, the quotient and remainder of the division, such
that
a = qb + r, q ≥ 0, and 0 ≤ r < b.
If the remainder r is 0, then we we call b a divisor of a, and say that a is
divisible by b, or also that b divides a.
A prime number is an integer p ≥ 2 whose only divisors are 1 and p
itself. The first ten prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. A
positive integer n which is not prime is called composite. By definition, it
must have a divisor b between 1 and n, which implies that it can be written
as a product ab of integers a and b satisfying 1 < a < n and 1 < b < n.
Let p1 , p2 , . . . , pk be prime numbers satisfying p1 ≤ p2 ≤ · · · ≤ pk . Then
their product is called a prime factorization of the number n = p1 p2 · · · pk ,
and the primes p1 , . . . , pk are called prime factors of n. Note that it is part
of the definition of a prime factorization that the prime factors are written
down in order from smallest to largest when read from left to right (where
the same prime number may appear multiple times in the product).
The goal of this short note is to prove the following theorem:
Theorem 1 (Prime factorization theorem, or the Fundamental theorem of
arithmetic). Every integer n > 1 has a unique prime factorization.
The proof requires a number of lemmas, the first of which establishes that
every integer larger than 1 admits at least one prime factorization.
Lemma 2. Every integer number n > 1 is equal to a product of (possibly
just one) prime numbers.
Proof. Suppose that the claim is false. Then there is at least one integer
larger than 1 which is not equal to a product of prime numbers. Let m be
the smallest such integer. Since m is not a product of prime numbers, m
cannot be prime and must be composite. So there are integers a and b such
that m = ab, 1 < a < m and 1 < b < m. However, m was the smallest
integer larger than 1 which was not equal to a product of primes, so both a
and b must be equal to products of primes, implying that m = ab has to be
equal to a product of primes after all. Contradiction.
Date: November 20, 2013.
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WOUTER KAGER
Lemma 2 says that every integer larger than 1 is equal to a product of
primes, and hence has a prime factorization (just write down the factors
in increasing order). To prove Theorem 1, it remains to establish that this
prime factorization is unique. For this we need two fundamental lemma’s.
Lemma 3 (Bézout’s lemma). The greatest common divisor of two positive
integer numbers a and b is the smallest positive integer of the form ax + by,
where x and y are integer (but not necessarily positive) numbers.
Proof. Let A be the set of all integer numbers of the form ax + by, where x
and y are integer (possibly 0 or negative) numbers, and let d be the smallest
positive number in the set A. By definition of A, there are integer numbers
u and v such that d = au + bv.
Let q and r be the quotient and remainder of the division of a by d, so
that a = qd + r with 0 ≤ r < d. Then
r = a − qd = a − q(au + bv) = a(1 − qu) + b(−qv),
which shows that r is an element of A. However, d was the smallest positive
element of A, so r must be 0. Hence, d divides a. A similar argument shows
that d divides b, so d is a common divisor of a and b.
Now suppose that e is also a common divisor of a and b. Then there are
integer quotients q1 and q2 such that a = q1 e and b = q2 e. But then
d = au + bv = q1 eu + q2 ev = (q1 u + q2 v)e,
and since d and e are positive, it must be the case that (q1 u + q2 v) ≥ 1, and
hence d ≥ e. It follows that d is the greatest common divisor of a and b. As an aside, we remark that the above proof shows that every common
divisor of two positive integer numbers a and b is in fact also a divisor of
the greatest common divisor of a and b.
Lemma 4 (Euclid’s lemma). Let p be a prime number, and let a1 , a2 , . . . ak
be positive integer numbers (k ≥ 2). If p does not divide any of the numbers
a1 , a2 , . . . , ak , then p does not divide the product a1 a2 · · · ak .
Proof. Let p be a prime which does not divide any of the positive integer
numbers a1 , a2 , . . . , ak . Assume that p does divide the product a1 a2 · · · ak .
Then, since p does not divide a1 , there must be an integer i satisfying
2 ≤ i ≤ k, such that p divides a1 a2 · · · ai , but does not divide a1 a2 · · · ai−1 .
Since p divides a1 a2 · · · ai , we have a1 a2 · · · ai = qp for some quotient q.
Moreover, because p is prime and does not divide ai , the greatest common
divisor of p and ai must be 1, so that by Bézout’s lemma, there are integer
numbers x and y such that ai x + py = 1. Therefore,
a1 a2 · · · ai−1 = a1 a2 · · · ai−1 (ai x + py) = (qx + a1 · · · ai−1 y)p,
which shows that a1 a2 · · · ai−1 is divisible by p. But this contradicts our
definition of i, so we conclude that p cannot divide a1 a2 · · · ak .
PRIME FACTORIZATION
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Proof of Theorem 1. Lemma 2 establishes that every integer larger than 1
has at least one prime factorization, but we still have to prove that this
prime factorization is unique. To prove this, suppose that there are integer
numbers larger than 1 that have more than 1 prime factorization. Then
there is a smallest such integer. Call it n, and let p1 p2 · · · pk and q1 q2 · · · ql
be two different prime factorizations of n. Clearly, if k = 1 or l = 1, then
the number n would itself be prime and the two prime factorizations would
have to be equal, so it must be the case that k ≥ 2 and l ≥ 2.
Now let r be the largest of the two primes pk and ql . Regardless which
of the two is larger, from the the two prime factorizations of n it follows
that the number n is divisible by r. In particular, the product p1 p2 · · · pk is
divisible by r, so by Euclid’s lemma, at least one of the factors p1 , p2 , . . . , pk
must be divisible by r. Since p1 ≤ p2 ≤ · · · ≤ pk ≤ r, it follows that r = pk .
By a similar argument, r = ql as well. But this is impossible, because
then the number n/r would be an integer number smaller than n with two
different prime factorizations p1 p2 · · · pk−1 and q1 q2 · · · ql−1 .
Remark. The unique prime factorization of an integer number n > 1 is
also commonly written in the form
n = pe11 pe22 · · · pekk ,
where p1 , p2 , . . . , pk are distinct prime numbers such that p1 < p2 < · · · < pk ,
and e1 , e2 , . . . , ek are the respective powers to which each of these prime
factors is raised, where each of these powers is an integer larger than 0.
If we now choose integer numbers d1 , d2 , . . . , dk such that 0 ≤ d1 ≤ e1 ,
0 ≤ d2 ≤ e2 , . . . , 0 ≤ dk ≤ ek , then the integer number whose prime
factorization is
pd11 pd22 · · · pdkk
is clearly a divisor of n. In fact, by the prime factorization theorem, every
divisor of n must have a prime factorization of this form. It follows that n
has exactly (e1 + 1)(e2 + 1) · · · (ek + 1) divisors.
VU University, Department of Mathematics, De Boelelaan 1081a, 1081 HV
Amsterdam, The Netherlands