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Prime

Definition. An n is said to be a prime if and only if n > 1 and the only positive divisors of n are 1 and n.

A positive integer n is said to be composite if and only if n > 1 and n is not a prime. Thus, n > 1 is composite if and only if there exist a and b with 1 < a, b < n such that n = a b. ·

As an exercise using the algorithm, we prove the following:

Theorem. If p and p2 + 2 are both primes, then p2 2 is also a prime. −

Proof. Suppose that p is a prime, when the division algorithm is used to divide p by 3,

p = 3q + r

where 0 r 2, so the only possible remainders are r = 0, r = 1, and r = 2. ≤ ≤

case 1 : If r = 0, then p = 3q for some positive integer q, and since p is prime, we must have q = 1, so that p = 3. In this case, p2 + 2 = 32 + 2 = 9 + 2 = 11 is a prime, and p2 2 = 9 2 = 7 is also a prime. Thus, the − − implication is true for p = 3. The implication is also true if p = 3 and 3 p, since in this case the hypothesis is false. 6

case 2 : If r = 1, then p = 3q + 1 for some positive integer q, and

p2 + 2 = (3q + 1)2 + 2 = 9q2 + 6q + 3 = 3(3q2 + 2q + 1)

2 2 and 3 p + 2. Since the second factor is clearly greater than 1, then p + 2 is composite in this case, and again the implication is true since the hypothesis is false.

case 3 : If r = 2, then p = 3q + 2 for some positive integer q, and

p2 + 2 = (3q + 2)2 + 2 = 9q2 + 12q + 6 = 3(3q2 + 4q + 2)

2 2 and again 3 p + 2. Since the second factor is clearly greater than 1, then p + 2 is also composite in this case, and again the implication is true since the hypothesis is false.

Therefore, if p and p2 + 2 are both prime, then p2 2 is also prime. − We will show that there are infinitely many primes, in fact, the proof we give is Euclid’s original proof.

Lemma. Every positive integer n > 1 has a prime divisor.

Proof. Let S = n Z n > 1 and n has no prime divisors . If S = , since S is bounded below, by the { ∈ | } 6 ∅ well ordering property S has a smallest element, say n0 S. ∈

Since n0 > 1 and n0 has no prime divisors, then n0 is composite, and there exist integers a0, b0 Z such that ∈ n0 = a0 b0 · where 1 < a0 < n0 and 1 < b0 < n0.

However, since 1 < a0 < n0 and n0 is the smallest element in S, then a0 S, which implies that a0 has a 6∈ prime divisor, say p a0, but then p n0 also, which is a contradiction.

Therefore, the assumption that S = leads to a contradiction, and we must have S = , so that every positive integer n > 1 has a prime divisor.6 ∅ ∅

Theorem. There are infinitely many primes.

Proof. Suppose not, suppose that p1, p2, . . . , pN are the only primes. Now consider the integer

M = p1 p2 pN + 1, · · · ·

from the previous lemma, M has a prime divisor, and it must therefore be one of the primes p1, p2, . . . , pN . This is a contradiction, since none of these primes divides M.

The Fundamental Theorem of

The Fundamental Theorem of Arithmetic states that if n > 1 is a positive integer, then n can be written as a product of primes in only one way, apart from the of the factors.

Recall that an integer n is said to be a prime if and only if n > 1 and the only positive divisors of n are 1 and n.

In order to prove the fundamental theorem of arithmetic, we need the following lemmas.

Lemma 1. Every integer n > 1 is either a prime or a product of prime numbers.

proof. We will prove this by induction on n. The lemma is clearly true for n = 2. Assume now that it is true for every positive integer k with 2 k < n. If n is not a prime, then it has a positive divisor d with d = 1 and d = n. Therefore, n = m d, ≤where m = n. However, both m and d are less than n and greater than6 1, so by6 the induction hypothesis· each of m and6 d is a product of primes, therefore n is also a product of primes. This completes the induction. Lemma 2. If a prime p does not divide a, then gcd(p, a) = 1. proof. Let d = gcd(p, a), then d p and p is prime, so that d = 1 or d = p. However, d a, so we must have

d = p, since p a. Therefore, d = 1. 6 6

Lemma 3. If a prime p divides ab, then p a or p b. More generally, if a prime p divides a product a1a2 an, · · · the p divides at least one of the integers a i, for 1 i n. ≤ ≤

proof. Suppose that p ab and that p a. We will prove that p b. From Lemma 2 we have gcd(p, a) = 1, and by the there exist6 integers x and y such that 1 = xa + yp, and therefore,

b = x ab + yb p, · · so that p b. For the more general statement, use induction on n.

Theorem (Fundamental Theorem of Arithmetic). Every integer n > 1 can be represented as a product of prime factors in only one way, apart from the order of the factors.

proof. The proof is by induction on n. The theorem is true for n = 2. Assume, then, that the theorem is true for all integers k with 1 < k < n. We will show that this implies that it is also true for n. If n is prime, then there is nothing more to prove. Assume, then, that n is composite and that n has two , say

n = p1p2 ps = q1q2 qt. ( ) · · · · · · ∗

We want to show that s = t and that each equals some qj .

Since p1 q1q2 qt and p1 is prime, then by Lemma 3, p1 must divide some qj . We may assume then (relabel) · · · that p1 q1, and therefore p1 = q1 since they are both primes. In ( ) we can cancel p1 on both sides to get ∗ n = p2 ps = q2 . . . qt. p1 · · · n n If s > 1 or t > 1, then 1 < < n, and by the induction hypothesis the two factorizations of must be p1 p1 identical, apart from the order of the factors. Therefore s = t and the factorizations in ( ) are also identical, apart from the order of the factors. The induction is complete. ∗ Note: In the of an integer n, a particular prime p may occur more than once. If the distinct prime factors of n are p1, p2, . . . , pk, and if pi occurs as a prime factor αi times, for 1 i k then we can write ≤ ≤ α1 α2 αk n = p1 p2 p , · · · k that is, k αi n = Y pi , i=1 and this is called the factorization of n into prime powers. We can also express 1 in this form by taking each exponent αi = 0.

Corollary. If k αi n = Y pi , i=1 then the set of positive divisors of n is the set of all numbers d of the form

k βi d = Y pi , i=1 where 0 βi αi for i = 1, 2, . . . , k, and the number of positive divisors of n, denoted by τ(n), is given by ≤ ≤

τ(n) = (1 + α1)(1 + α2) (1 + αk). · · ·

As an example of the use of the prime factorization of an integer, we have the following result.

Theorem. If n > 1 is a positive integer, then n is a perfect square if and only if n has an odd number of divisiors.

Proof. Let the prime factorization of n be given by

α1 α2 αk n = p1 p2 p · · · k where p1 < p2 < < pk are distinct primes and for 1 i k, each of the integers αi 1. · · · ≤ ≤ ≥

Now note that n is a perfect square if and only if each αi is an even integer, that is, if and only if there exist positive integers βi such that αi = 2βi for 1 i k. ≤ ≤

From the previous corollary, n is a perfect square if and only if the number of divisors of n is

τ(n) = (2β1 + 1)(2β2 + 1) (2βk + 1), · · · that is, if and only if τ(n) is an odd integer. Example. The Locker Problem

A certain locker room contains n lockers numbered 1, 2, . . . , n and they are all originally locked. An attendant performs a of operations T1, T2, . . . , Tn whereby with the operation Tk, 1 k n, the condition of being locked or unlocked is changed for all those lockers and only those lockers whose≤ n≤umbers are multiples of k. Show that after all the n operations have been performed, all those lockers whose numbers are perfect squares (and only those lockers) are now open or unlocked.

proof. Locker number m, for 1 m n, will be unlocked after the n operations have been performed if and only if it has changed state ≤an odd≤ number of times, that is, if and only if the integer m has an odd number of positive divisors. Therefore, locker number m is unlocked after all n operations are performed if and only if m is a perfect square.

We will show that the prime factorization of an integer leads to a method to find the and the of two positive integers. First the definition.

Definition. If a and b are positive integers, then the least common multiple of a and b is the smallest positive integer m such that a m and b m.

Note: The least common multiple of a and b is denoted by [a, b] or lcm(a, b), and its existence is guaranteed by the well ordering property of the positive integers.

Once the decomposition of the positive integers a and b are known, it is trivial to find both gcd(a, b) and lcm(a, b), in fact, if

α1 α2 αk β1 β2 βk a = p1 p2 p and b = p1 p2 p · · · k · · · k

where p1 < p2 < < pk are distinct primes, 0 αi and 0 βi, for 1 i k, (zero exponents are allowed so that we may use· · ·the same primes in the factorization≤ of b≤oth a and≤b), then≤

min{α1,β1} min{α2,β2} min{αk,βk} gcd(a, b) = p1 p2 p · · · k and max{α1,β1} max{α2,β2} max{αk ,βk} lcm(a, b) = p1 p2 p . · · · k The hard part (how hard is an open question) is finding the prime power decomosition of a positive integer.

From the above, we have an easy proof of the following theorem.

Theorem. If a and b are positive integers, then

gcd(a, b) lcm(a, b) = a b. · ·

Proof. Note that for any two real numbers x and y we have

max x, y + min x, y = x + y { } { } since on the left, one is x and the other is y. Now multiply the prime power decompositions of a and b together to get gcd(a, b) lcm(a, b) = a b. · · As an application of the fundmental theorem of arithmetic, we prove a special case of Dirichlet’s theorem, which says that if a and b are relatively prime positive integers, then the arithmetic progression

a n + b, n 0 · ≥ contains infinitely many primes.

Theorem. There are infinitely many primes of the form 4n 1, where n is a positive integer. −

Proof. Note that any odd integer is of the form 4n 1 or 4n + 1, and the product of any two odd integers a = 4k 1 and b = 4` 1, or the product of an two o−dd integers a = 4k + 1 and 4` + 1, is of the form 4n + 1. − −

Suppose that there are only finitely many primes of the from 4n 1, say p1, p2, . . . , pk, where p1 = 3, and let − Q = 4p2 pk 1, · · · − then Q is not divisible by any of the primes p1, p2, . . . , pk. However, this implies that every prime divisor of Q is of the form 4n + 1, and as noted above, this implies that Q itself if of the form 4n + 1, which is a contradiction. Therefore there are an infinite number of primes of the form 4n 1. − The

Now we give a theorem to determine whether a positive integer is a prime, that is, a simple .

Theorem. If n is a composite integer, then n has a prime divisor p such that p √n. ≤ Proof. If n is composite, then n = a b, where 1 < a b < n. If a > √n, then b a > √n, and this implies that · ≤ ≥ n = a b > √n √n = n · · which is a contradiction. Therefore, a √n and since a > 1, then a has a prime divisor p such that ≤ p a √n, and since p a and a n, then p n also. ≤ ≤

Before the advent of the computer, one of the most efficient methods of constructing tables of primes was the sieving process, invented by the Greek mathematician Eratosthenes (276-194 B.C.). The method is called the Sieve of Eratosthenes.

We illustrate the method by constructing a table containing all primes less that 130. We begin by listing all the integers from 2 to 129 (since 1 is not a prime it is not listed). The work involved in the process is simplified by the previous lemma.

The first number in our list, 2, must be a prime and no multiple of 2 except 2 itself can be prime. We remove all multiples of 2 (except 2 1) from our list. The next remaining number, 3, must be a prime, so we delete all the multiples of 3. We no· w delete the multiples of 5, the multiples of 7, and the multiples of 11. Because the largest prime less than √130 is 11, then from the lemma, all the remaining numbers must be primes. In the table below, rather than delete multiples of 2, 3, 5, 7, 11, we have underlined the integers that were not deleted, thus, the primes less than 130 are all underlined.

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129

The sieve of Eratosthenes can be used to derive a formula for the number of primes less than or equal to n if the primes less than or equal to √n are known. Although this formula is very awkward to use, it is the only formula for the exact value of π(n).