MODULE 3
The Fundamental Theorem of Arithmetic
1. Sieve of Eratosthenes
Two is the only odd prime. Every even number > 2 is composite because it is divisible
by 2. Similarly, all multiples of 3 ≥ 6 are composite; all multiple of 5 ≥ 10 are composite;
and so on.
Eratosthenes (an old Greek guy) figured out a way to use this fact to make a list of
the primes. His method uses a technique called “Calculatus Eliminatus” (coined by Dr.
Seuss) which means “to discover where something is, discover where it’s not.” His idea is
very simple. Write down all the numbers from 1 to 100. Ignore 1 because it is a unit—it’s
neither prime nor composite. After 1 the first number we come to is 2, which must be
prime because there are no smaller numbers which could divide it. Using the fact that all
multiples of 2 are composite, mark an ‘X’ through every other number starting with 4. The
prime 3 has not been crossed out, so cross out every third number, starting with 6; these
are all composite because they are divisible by 3. Note that 6 was already crossed out as a
multiple of 2, but that’s okay. It’s like having a second opinion from another doctor. Four
has been crossed out, so we know it isn’t prime. The next number we come to which has
not been crossed out is 5. So we procced to cross out every multiple of 5, starting with 10.
(Actually we could have started with 25.) Six has been crossed out (twice), so we come to
7, the fourth prime on our list, and we cross out every seventh number starting with 14. At
this point the numbers on the list up to 100 which have not been crossed out are all primes.
The reason why we don’t need to cross out the multiples of 11, 13, etc, is given by the
following theorem:
Theorem #56. If n is composite then some prime number p which is ≤
√
n divides n.
Problem #57. Use the Sieve of Erostothenes to construct a list of the primes up to 200.
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30
3. THE FUNDAMENTAL THEOREM OF ARITHMETIC
2. Factoring into Primes
Every composite number n can be factored into two smaller number n = ab, where
2 ≤ a < n and 2 ≤ b < n. This factorization is not unique. For example, 30 factors in three
different way:
(1) 30 = 2 × 15
(2) 30 = 3 × 10
(3) 30 = 5 × 6
But if we continue factoring until we reach all primes
(1) 30 = 2 × 15 = 2 × (3 × 5)
(2) 30 = 3 × 10 = 3 × (2 × 5)
(3) 30 = 5 × 6 = 5 × (2 × 3)
We discover a fundamental fact: the list of primes is always the same, up to order. This
fact is called
The Fundamental Theorem of Arithmetic. Every integer n > 1 can be factored into
a product of primes, and the factorization is unique, up to the order in which we write the
primes.
Suppose you factor n as
n = p1 · p2 · p3 · · · pr
where the pi ’s are all primes. To simplify things, suppose your primes are ordered from
smallest to largest: p1 ≤ p2 ≤ · · · ≤ pr .
Now suppose your worst enemy claims to factor n in a superior way to yours, say as
n = q1 · q2 · q3 · · · qs
where the qj ’s are all prime and are in increasing order: q1 ≤ q2 ≤ · · · ≤ qs .
The Fundamental Theorem of Arithmetic says that you both have the same number of
primes in your two lists, that is, r = s, and the primes in both lists are the same: p1 = q1 ,
p2 = q2 , etc.
Maybe it seems unthinkable that there could possibly be any other outcome. But there
are many similar algebraic systems where factorization into “primes” is not unique.
3. SOME PRELIMINARIES
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Crazy Example. Let S be the set of all positive integers which are congruent to 1 mod
4, that is,
S = {1, 5, 9, 13, 17, 21, 25, 29, . . . }.
The set S acts likes the set of positive integers.
Problem #58. Show that is a and b are in S, then the product ab lies in S.
Definition. We define a number n > 1 in set S to be an S–prime if n cannot be written
as n = ab, where a and b lie S and 1 < a, b < n. Note that 9 and 21 are S-primes. Even
though 21 = 3 × 7, this is not a legitimate factorization of 21 by elements in S since 3 and
7 are not in S. (The integers have a similar restriction. The equation 7 = 3 × 37 does not
prove that 7 is a composite; we are not allowed to use the non-integer fraction 73 .)
Problem #59. Make a list of all the S primes less than 100.
Just like the positive integers, the elements of S can all be factored into S primes. For
example, 65 = 5 × 13. But is this factorization into S primes always unique? (For n = 65,
it is.)
Problem #60. Find a number n in S such that n can be factored in two entirely different
ways into a product of S-primes.
Moral: Unique factorization cannot be taken for granted.
We now turn our attention to proving unique factorization of the integers.
3. Some Preliminaries
Recall the Pails of Water Theorem, which asserts that you can solve the Pails of Water
Problem with buckets of capacity a and b whenever a and b are relatively prime.
POW Theorem. If a and b are positive integers with gcd(a,b) = 1, then there exist
integers m and n such that ma + nb = 1.
Theorem #61. If a | bc and gcd(a, b) = 1, then a | c.
Problem #62. Show that this theorem is false if we remove the restriction that a and b
are relatively prime.
Hint: Use the POW Theorem on a and b. Then multiply by c.
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3. THE FUNDAMENTAL THEOREM OF ARITHMETIC
Primes have a special property not shared by composites.
Proposition #63. If p is a prime and n is a positive integer, then either p | n or
gcd(p, n) = 1.
Theorem #64. If p | ab, then p | a or p | b.
Problem #64. Show that this theorem is false if we remove the restriction that p is a
prime.
4. Proof of Fundamental Theorem
Now we are ready to give a proof of the Fundamental Thoerem of Arithemetic. Let n
be a positive integer > 1 and suppose we have two factorizations of n into primes:
n = p 1 p 2 p 3 · · · pr = q 1 q 2 q 3 · · · q s
where the pi and qj are primes, in increasing order
p 1 ≤ p 2 ≤ p 3 ≤ · · · ≤ pr
and
q1 ≤ q 2 ≤ q 3 ≤ · · · ≤ qs .
Exercise #65. Give an argument that shows that p1 = q1 .
If p1 = q1 , then our previous equation becomes
n = p1 p2 p3 · · · pr = p1 q 2 q 3 · · · q s
Cancelling p1 gives
p 2 p 3 · · · pr = q 2 q 3 · · · q s
Exercise #66. Give an argument that shows that p2 = q2 .
Exercise #67. Explain how continuing this process gives p3 = q3 , p4 = q4 , and so on.
What can’t we run out of p’s before q’s, or the other way around, run out of q’s before p’s?
Conclude that r = s and the two lists of primes were identical.
5. Applications of The Fundamental Theorem
The equation
n = pe11 pe22 · · · perr
where the pi are distinct primes, is called the prime power factorization of n.
6. PRIMES OF SPECIAL TYPE
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Exercise #68. Find the prime power factorization of 20! = 1 · 2 · 3 · 4 · · · 19 · 20.
Exercise #69. If we write
100! = 2e 3f 5g ·
what are the values of the exponents e, f , and g?
The floor and ceiling functions are defined as follows: Given a real number x, the floor
of x, written ⌊x⌋ is the largest integer ≤ x; the ceiling of x, written ⌈x⌉ is the smallest
integer ≥ x. For example, ⌊π⌋ = 3, ⌈π⌉ =4.
Exercise #70. Can you think of a way to use the floor function to express the value of
g in the previous problem.
Exercise #71. Find gcd(314 · 722 · 115 · 173 , 52 · 114 · 138 · 17)
Exercise #72. Find LCM(314 · 722 · 115 · 173 , 52 · 114 · 138 · 17)
Exercise #73. Make a conjecture that generalizes the ideas you used to solve the
previous two problems.
The Fundamental Theorem can also be used to prove that certain equations do not have
integer solutions.
Proposition #74. There do not exist integers m and n such that m2 = 5n2 . Can you
conclude that the square root of 5 must be irrational?
Exercise #74. Show that the cube root
√
3
5 is irrational.
6. Primes of Special Type
All primes are odd—except 2.
These odd primes lie in two camps based on their remainders when we divide them by
4: (i) p = 4n + 1 and (ii) p = 4n + 3.
Table of primes:
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3. THE FUNDAMENTAL THEOREM OF ARITHMETIC
4k + 1 4k + 3
5
3
13
7
17
11
29
19
37
23
41
29
53
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Exercise #75. Extend this list using the primes up to 200.
One could ask whether both of these lists are infinite. That is, are there infinitely many
primes of the form 4n + 1? Are there infinitely many primes of the form 4n + 3? Also, it
looks like the primes congruent to 3 mod 4 are more numerous than those primes congruent
to 1 mod 4. Does this lead last out to infinity?
It is know that
• Yes, there are infinitely many primes of the form 4n + 1.
• Yes, there are infinitely many primes of the form 4n + 3.
• Roughly speaking (we need to use, ugh, limits to be precise here) there about as
many primes of the form 4n + 1 as primes of the form 4n + 3.
• Out to infinity, the race between the two types of primes is tied infinitely often.
Put another way, if C1 (x) counts the number of primes of the form 4n + 1 which
are ≤ x and C3 (x) counts the number of primes of the form 4n + 3 which are ≤ x,
then for infinitely many integers x, C1 (x) = C3 (x).
Exercise #76. For every odd prime p ≤ 100, determine whether p can be written as a
sum of two squares. For example, 13 = 22 + 32 , but 11 is not a sum of two squares; the
closest we come with 11 is 32 + 12 + 12 or 22 + 22 + 12 , which are both sums of three squares.
Exercise #77. Make a conjecture about which primes can be be written as a sum of
two squares.
Exercise #78. Make a list of the primes ≤ 200 which can be written in the form
p = n2 + 1. For example, 37 = 62 + 1, while 41 cannot be written as n2 + 1 for the simple
reason that 40 is not a square.
6. PRIMES OF SPECIAL TYPE
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One could ask if this list is infinite. The somewhat surprising answer is that “nobody
knows.” All sane mathematicians believe so, but so far, no one has been able to give a
proof.