18 Divisibility. and 0 r < d. Lemma Let n,d Z with d 0. If n = qd+r = q d+r with 0 r,r < d, then q = q and r = r.

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1 DIVISIBILITY 18 Divisibility Chapter V Theory of the Integers One of the oldest surviving mathematical texts is Euclid s Elements, a collection of 13 books. This book, dating back to several hundred years BC, is one of the earliest examples of logical reasoning in mathematics still available for us to read. Although most of the books are devoted to theorems concerning geometry (many of which you may have seen in some form in a high school geometry class), books seven and nine deal with the arithmetic of the integers. In this chapter, we will deal with some of the material found in these two books. In particular, Euclid dealt with the topics of divisibility and greatest common divisors. It is remarkable that, thousands of years before the advent of electronic computers, Euclid wrote down a very efficient algorithm for computing GCDs, which is still used essentially without change in modern computer systems. In addition, Euclid defines prime numbers, proves that there are infinitely many primes, and proves the Fundamental theorem of arithmetic, which states that every natural number greater than 1 has a unique factorization into prime numbers. The results that we present in this chapter have thus stood the test of time and been studied by mathematicians over millennia. Besides being important and useful results on their own, they form a significant part of the common heritage of mathematics. 18.A The division algorithm A fundamental property of the integers that relates addition and multiplication is the division algorithm. The fact we can divide integers and get a unique quotient and remainder is the key to understanding divisibility, congruence, and modular arithmetic. We begin with the statement of the division algorithm. In order to simplify the organization of the proof, we then prove uniqueness of the quotient and remainder as a lemma, before proceeding with the remainder of the proof. A portion of the proof is left for the reader, in Exercise Theorem 18.1 (The Division Algorithm). Let n,d Z with d 0. Then there are unique integers q,r such that and 0 r < d. n = qd+r Lemma Let n,d Z with d 0. If n = qd+r = q d+r with 0 r,r < d, then q = q and r = r. Proof of Lemma. Suppose that n = qd+r = q d+r with 0 r < d and 0 r < d. Then we have (q q )d = r r with d < r r < d and r r is a multiple of d. Dividing by d, and using the fact that (r r)/d = q q, we see that 1 < q q < 1. Since 0 is the only integer between 1 and 1, we have q q = 0 so q = q. It then follows that r = r. Wewillprovetheexistenceofq andr forn 0andd > 0. Theremainder of the proof will be left to the exercises (see Exercise 18.1).

2 CHAPTER V. THEORY OF THE INTEGERS DIVISIBILITY Proof of Theorem 18.1 for n 0 and d > 0. Fix d > 0, and let P(n) be the statement P(n): There are integers q n,r n such that n = q n d+r n and 0 r n < d. We note that taking q 0 = r 0 = 0, we have 0 = q 0 d+r 0, and 0 r 0 < d. Hence, P(0) is true. Assume, then that P(k) is true; in other words, there are integers q k and r k such that k = q k d + r k, with 0 r k < d. Then we have that k+1 = q k d+r k +1. Note that 0 r k < r k +1 d. If r k +1 < d, then we see that P(k +1) is true (with q k+1 = q k and r k+1 = r k +1), and if r k +1 = d, then k+1 = q k d+d = (q k +1)d+0, so P(k+1) is true (with q k+1 = q k +1 and r k+1 = 0). Hence, by induction, P(n) is true for all n 0 and d > B Divisibility and common divisors We now prove several facts about divisibility. Theorem Let a and b be nonzero integers. If a b, then a b. Proof. Assume that a b. Then b = ak for some k Z. Hence, b = a k. Note that k 0, so k 1. Dividing by k, we find that 1/ k 1. Then 0 < a b = 1 k 1. Multiplying by b, we see that a b. Theorem Let b Z with b 0. Then there are finitely many integers that divide b. Proof. If a Z divides b, then a < b. Hence, a { b,..., b }. This set is finite, so there are only finitely many possibilities for a. Example If b = 12, the divisors of b are { 12, 6, 4, 3, 2, 1,1,2,3,4,6,12}. Definition Acommon divisor oftwointegersaand bisanonzero integer c such that c a and c b. Example If a = 12 and b = 18, then the common divisors of a and b are ±1, ±2, ±3, and ±6. Theorem Let a and b be integers, not both 0. Then the set of common divisors of a and b has a largest element. Proof. Without loss of generality, let a 0. The set of divisors of a is finite, and contains the set of common divisors of a and b, so the set of common divisors is finite. Since it is finite and nonempty (note that it contains 1), this set has a largest element. Definition The greatest common divisor (or GCD) of two integers a and b (not both zero) is the largest common divisor of a and b. We denote the greatest common divisor of a and b by GCD(a,b), or, more commonly, by (a, b). Example If a = 12 and b = 18, the list of divisors of a is and the list of divisors of b is { 12, 6, 4, 3, 2, 1,1,2,3,4,6,12} { 18, 9, 6, 3, 2, 1,1,2,3,6,9,18}. The numbers common to both of these sets (their intersection) are { 6, 3, 2, 1,1,2,3,6}. Hence, the greatest common divisor of 12 and 18 is 6. Example Let a Z be nonzero. Then every divisor of a is smaller than a, and, in fact a is a divisor of a. In addition, a is a divisor of 0. Hence, GCD(a,0) = a.

3 CHAPTER V. THEORY OF THE INTEGERS DIVISIBILITY 18.C Computing the GCD Note that listing out all the divisors of a and b is a very inefficient way of computing the GCD. We will now give a very efficient algorithm to compute GCD(a,b). It is based on the following theorem. Theorem Let a,b Z with b 0, and assume that a = qb+r. Then GCD(a,b) = GCD(b,r). Proof. Let S be the set of common divisors of a and b. Let T be the set of common divisors of b and r. We will show that S = T. Once this is proven, the largest element of S must be the same as the largest element of T, and the theorem will be proven. Suppose, then, that d S. Then d a and d b. Now r = a qb, so we must have d r. Hence, d is a common divisor of b and r, so d T. Thus, S T. Now assume that d T. Then d b and d r. Since a = qb+r, it must be the case that d a. Hence, d is a common divisor of a and b, so d S. Therefore, S T, and we see that S = T. Remark Note that the theorem does not require that r < b. The division algorithm guarantees that for b 0, we can find q,r with a = qb+r and 0 r < b. Algorithm Given two integers a, b not both 0, assume that a 0 and that a > b (if either of these does not hold, swap a and b so that both hold). If b = 0, then the GCD(a,b) = a, and we are finished. Otherwise, apply the division algorithm multiple times, as follows. Divide a by b a = q 1 b+r 1 with 0 r 1 < b. Divide b by r 1 b = q 2 r 1 +r 2 with 0 r 2 < r 1 Divide r 1 by r 2 r 1 = q 3 r 2 +r 3 with 0 r 3 < r 2... Divide r n 2 by r n 1 r n 1 = q n r n 1 +r n with 0 r n < r n 1. Continue to divide until we get a remainder r n = 0 (we can t go any farther, since we can t divide by 0). If r 1 = 0, then GCD(a,b) = b, and we are finished. If r n = 0 for n > 1, then GCD(a,b) = r n 1, and we are finished. To show that an algorithm is correct, there are two things that need to be demonstrated. First, the answer that the algorithm computes must be correct. Second, the algorithm must terminate after finitely many steps; it does us no good if the algorithm takes forever to compute an answer. We will demonstrate both of these. Notice that if b = 0, then the algorithm completes by saying GCD(a,b) = a. By Example 18.11, this is the correct answer. Choose n N so that r n is 0. Then we have the following sequence of equalities, from Theorem GCD(a,b) = GCD(b,r 1 ) = GCD(r 1,r 2 ) = = GCD(r n 1,r n ) 18.D The Euclidean Algorithm We now describe an algorithm that very efficiently compute GCD(a, b). This algorithm is recursive, meaning that in the process of computing the GCD of a and b, it will require computing the GCD of other pairs of numbers. After describing the algorithm, we will prove that it gives the correct answer. If r 1 = 0, then GCD(a,b) = GCD(b,0) = b by Example 18.11, and the algorithm gives a correct answer. If r n = 0, with n > 1, then we have GCD(a,b) = GCD(b,r 1 ) = GCD(r 1,r 2 ) = = GCD(r n 1,r n ) = GCD(r n 1,0) = r n 1, and the algorithm gives the correct answer.

4 CHAPTER V. THEORY OF THE INTEGERS DIVISIBILITY Hence, the output of the algorithm is always correct, if it ever gets to the point where r n = 0. If it does not, then we will be forever dividing, and never getting an answer. Note however, that b > r 1 > r 2 > r 3 > r 4 > 0. It is easy to see that 0 r 1 b 1, and in general that 0 r i b i. Hence, after at most b divisions, we must get a remainder of 0. Therefore the algorithm completes after a finite number of steps. For experienced computer programmers, you may recognize that the algorithm can be written recursively; we give a recursive version here. Algorithm Given two integers a,b, assume that a 0. We will also assume that a b (if not, switch a and b). Perform the following steps. (1) If b = 0, then GCD(a,b) = a, and we are done. (2) Use the division algorithm to find a = qb + r with q,r Z and 0 r < b. (3) At this point, we know that GCD(a,b) = gcd(b,r). Use the algorithm to compute GCD(b, r). Example Suppose that we wish to compute the GCD of 39 and 57. We perform our divisions as follows 57 = = = The last nonzero remainder is 3, so the GCD of 39 and 57 is 3. Example Suppose that we wish to compute the GCD of 1037 and We perform our divisions as follows = The last nonzero remainder is 29, so the GCD of 1073 and 1537 is 29. Remark Notice that the number of divisions is actually significantly less that b. In fact, it can be shown (we will not do it) that the number of divisions required is always less than 2log 2 b, which is typically (except when b is small) much smaller than b. 18.E Exercises Exercise Complete the proof of Theorem 18.1 as follows. (a) Using the fact that the theorem is true for nonnegative d and n, prove the theorem for arbitrary n and positive d. (Hint: if n < 0, then n > 0. Use the proven case of the division algorithm for n and d.) (b) Usingthefactthatthetheoremistrueforpositived, provethetheorem for negative d. Exercise Write out all the divisors of 60 and all the divisors of 42. Circle the common divisors, and find the GCD. Exercise Use the Euclidean algorithm to compute the following GCD s. (a) GCD(60,42). (b) GCD(667,851). (c) GCD(1855,2345). (d) GCD(589,437). Exercise Recall that the Fibonacci numbers are defined by the relations F 1 = 1, F 2 = 1, and for n > 2, F n = F n 1 +F n 2. Prove by induction that for every n N, GCD(F n,f n+1 ) = 1. Exercise Let n Z. Prove that GCD(2n+1,4n+3) = 1. Exercise Let n Z. Prove that GCD(6n+2,12n+6) = = = =

5 CHAPTER V. THEORY OF THE INTEGERS THE EXTENDED EUCLIDEAN ALGORITHM 19 The Extended Euclidean Algorithm 19.A The GCD as a linear combination We now recall the result of Exercise Theorem Every nonempty subset of the natural numbers has a smallest element. We will use this theorem to prove an important statement about GCD(a, b); that it can be written as an integer linear combination of a and b. Definition An integer linear combination of two integers a and b is a number of the form ax+by where x,y Z. Theorem Let a and b be integers, not both equal to 0. Define S = {ax+by : x,y Z,ax+by > 0}. Then S is a nonempty subset of N, and the smallest element of S is equal to GCD(a, b). Remark Note that the elements of S are just the integer linear combinations of a and b that are positive. Proof. Let S = {ax+by : x,y Z,ax+by > 0}. It is clear that S is a subset of the natural numbers, since its elements are positive integers. In addition, S is nonempty, since it contains at least one of 1a+0b = a, ( 1)a+0b = a, b or b. Hence, S has a smallest element, which we will call s. We note that for some x,y Z, we have s = ax+by. Note that s > 0. Let d = GCD(a,b). Then d a and d b, so by Theorem 8.15, d ax+by, and we have that d s. Hence, d s. Now we use the division algorithm to write a = qs+r with 0 r < s. Then r = a qs = a q(ax+by) = a(1 qx)+b( qy) is an integer linear combination of a and b. If r were positive, then r would be an element of S that is smaller than s. Hence, r must be 0. Therefore a = qs and we see that s a. A similar argument shows that s b. Since s is a common divisor of a and b, it cannot be larger than the greatest common divisor d. Hence, s d. Combining the facts that d s and s d, we see that d = s, as desired. Example Leta = 6andb = 9. ThenthesetS contains6( 1)+9(1) = 3. It cannot contain 1 or 2 (since, for example if we had 2 = 6x+9y, then since 3 6 and 3 9, we would have 3 2). Therefore 3 is the smallest element of S. It is also the GCD of 6 and B Calculating the GCD as a linear combination Now that we know that GCD(a,b) can be written as an integer linear combination of a and b, the natural questions is how to compute x and y so that GCD(a,b) = ax+by. We begin by performing the Euclidean Algorithm for a and b, and solving each equation for the remainder. a = q 1 b+r 1 r 1 = a q 1 b b = q 2 r 1 +r 2 r 2 = b q 2 r 1. r n 3 = q n 1 r n 2 +r n 1 r n 2 = q n r n 1 +r n. r n 1 = r n 3 q n 1 r n 2 r n = r n 2 q n r n 1 The bottom right equation then expresses r n as an integer linear combination of the previous two remainders, r n 1 and r n 2. We replace r 1 in this equation by the integer linear combination expressed in the next equation up, so r n = r n 2 q n r n 1 = r n 2 q n (r n 3 q n 1 r n 2 ) = (1 q n 1 )r n 2 +( q n )r n 3. We now perform a similar replacement of r n 2 by an integer linear combination of r n 3 and r n 4. Although this sounds tricky, it is easy to understand if we do a couple of examples.

6 CHAPTER V. THEORY OF THE INTEGERS THE EXTENDED EUCLIDEAN ALGORITHM Example Find the GCD of 493 and 391, and write it as 391x+493y. We perform the Euclidean algorithm, and solve each of the resulting equations for the remainder. 493 = = = = = = = 5 17 Now we see that 17 = , from the bottom right equation. Looking up one equation higher, we see an expression for 85 that we plug into this equation, so 17 = ( ) = (1+3) = Going one equation higher, we see an expression for 102; namely, 102 = We plug this in to our expression for = = 4 ( ) = , and we see that we have expressed GCD(493,391) = 17 = ( 5) C Relative Primality The fact that GCD(a,b) can be written as an integer linear combination of a and b has many important consequences. In particular, we saw in the proof of Theorem 19.3 that GCD(a,b) is in fact the smallest positive integer linear combination of a and b. This yields the following theorem. Theorem Let a and b be integers, not both 0. Then GCD(a,b) = 1 if and only if 1 = ax+by for some x,y Z. Proof. If GCD(a,b) = 1, Theorem 19.3 tells us that 1 = ax + by for some x,y Z. Conversely, if 1 = ax+by with x,y Z, we note that 1 is the smallest positive number that can be written as an integer linear combination of a and b (since there are no positive integers smaller than 1). Hence, 1 = GCD(a,b). We give a special name to a pair of numbers that have GCD equal to 1. Definition Let a,b Z. If GCD(a,b) = 1, then we say that a and b are relatively prime. Example Since GCD(5,7) = 1, we say that 5 and 7 are relatively prime. Relatively prime integers have many useful properties. We prove some of them here. These theorems go back to the days of ancient Greek mathematicians; they were known to Euclid several hundred years BC. In fact the first of them is named Euclid s Lemma.. Theorem (Euclid s Lemma). Let a,b,c Z with a 0. If a bc and GCD(a,b) = 1, then a c. Proof. Assume that a bc and GCD(a,b) = 1. Since a bc, we see that bc = ak for some k Z. Also, for some x,y Z, we have 1 = ax + by. Multiplying this last equation by c, we obtain c = acx+bcy = acx+(bc)y = acx+(ak)y = a(cx+ky). Hence, since cx+ky Z, we see that a c. Example If 2 3x, then, since GCD(2,3) = 1, we see that 2 x. Theorem Let a,b,c Z, with a 0 and b 0. If a c and b c and GCD(a,b) = 1, then ab c. Proof. Assume that a c and b c and GCD(a,b) = 1. Then for some k,l,x,y Z, we have c = ak, c = bl, and 1 = ax+by. Multiplying this last equation by c, we get c = cax+cby = (bl)ax+(ak)by = ab(xl+ky).

7 CHAPTER V. THEORY OF THE INTEGERS THE EXTENDED EUCLIDEAN ALGORITHM Since xl+ky Z, we see that ab c. Example If 2 x and 3 x, we see that 6 x, since GCD(2,3) = 1. Warning Note that neither of these theorems is true if we just assume that a b instead of GCD(a,b) 1. For instance, taking a = 4, b = 6, and c = 2, we have that 4 6 2, and 4 6, but it is not the case that 4 2. Taking a = 12 and b = 18, we see that both a and b divide 36, but ab D Other properties of the GCD We now discuss a different way to describe the GCD of two integers. This description can be used to define the GCD, and defining the GCD this way allows the GCD to be defined in number systems where the idea of a greatest elementofasetmaynotmakesense. Ifyoucontinueontoanabstractalgebra course, you may see such number systems. Theorem Let a,b Z, not both 0. If d is any common divisor of a and b, then d GCD(a,b). Proof. Since GCD(a,b) = ax+by for some x,y Z, this follows from Theorem Theorem Let a,b Z, not both 0. Suppose that some integer d satisfies (a) d > 0, (b) d is a common divisor of a and b, (c) every common divisor of a and b divides d. Then d = GCD(a,b). Proof. Suppose that d satisfies (a), (b), and (c). Then, since GCD(a, b) is a common divisor of a and b, we see that (c) implies that GCD(a,b) d. However, GCD(a,b) = ax+by is divisible by any common divisor of a and b, so d GCD(a,b). This implies that d = ±GCD(a,b). However, both d and GCD(a,b) are positive, so d = GCD(a,b). Remark We note that of course GCD(a, b) satisfies conditions (a), (b), and (c) of the theorem. Using this characterization of the GCD, we can calculate GCD(ac, bc). Theorem Let a,b,c Z, with a and b not both 0, and c > 0. Then GCD(ac,bc) = cgcd(a,b). Proof. Let d = GCD(a,b). Then we can write d = ax+by for some x,y Z. Multiplying this equation by c, we see that dc = acx + bcy. Notice that dc > 0, since both d and c are greater than 0. In addition, since d a and d b, we see that dc ac and dc bc. Finally, suppose that e ac and e bc. Since dc = acx + bcy, we see that e dc. Hence, dc is a positive common divisor of ac and bc, that is divisible by every common divisor of ac and bc. Therefore, by Theorem 19.16, dc = GCD(ac,bc). 19.E Exercises Exercise Foreachpairofnumbersaandbbelow, calculategcd(a,b), and find x,y Z such that GCD(a,b) = ax+by. (a) Let a = 15 and b = 27. (b) Let a = 29 and b = 23. (c) Let a = 91 and b = 133. (d) Let a = 221 and b = 377. Exercise Let n N and a Z. Assume that GCD(a,n) = 1. Prove that there is some b Z such that ab 1 (mod n). Exercise Let a,b Z, not both zero, and let d = GCD(a,b). Prove that ( a GCD d, b ) = 1. d

8 CHAPTER V. THEORY OF THE INTEGERS PRIME NUMBERS Exercise Let a,b,c N. Assume that GCD(a,b) = 1. Let d = GCD(a,c) and e = GCD(b,c). (a) Prove that d and e are relatively prime. (b) Prove that de c. (c) Prove that de is an integer linear combination of ab and c. (d) Prove that GCD(a,c) GCD(b,c) = GCD(ab,c). Exercise Let a,b,c N. Assume that GCD(b,c) = 1. Prove that GCD(ab,c) = GCD(a,c). Exercise A common multiple of a and b is an integer n such that a n and b n. The least common multiple of a and b, written LCM(a,b) is the smallest positive common multiple of a and b. (a) Determine the LCM of 12 and 18. (b) Determine the LCM of 21 and 35. (c) Let a,b be positive integers. Prove that LCM(a,b) = ab/d, where d = GCD(a,b). (Hint: Show that ab/d is a common multiple of a and b. Then show that it divides (and is thus smaller than) every other positive common multiple of a and b. You may wish to factor a = a d and b = b d and use the fact that GCD(a,b ) = 1.) 20 Prime numbers Now that we have a good understanding of divisibility in the integers, we are prepared to define and study the multiplicative building blocks of the integers. As far as multiplication is concerned, prime numbers are the atoms from which other integers are formed. 20.A Definition of prime numbers Definition A prime number is an integer p > 1 such that the only positive divisors of p are 1 and p. An integer n > 1 that is not prime is said to be composite. Example We know that any positive divisors of an integer n are between 1 and n, so we may easily check whether a given integer is prime. The integer 2 is prime, since there are no integers between 1 and 2. The integer 3 is prime, since it is not divisible by 2. Similarly, 5 is prime, since it is not divisible by 2, 3, or 4. Note that 4 is not prime, since it is divisible by 2. The first few prime numbers are 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89. If a number is composite, then it has a positive factor besides itself and 1. We expand a bit on this fact in the following theorem. Theorem Let a Z, with a > 1. If a is composite, then there are positive integers b and c, with 1 < b,c < a such that a = bc. Proof. Since a is not prime, it has a positive divisor b with 1 b and b a. Since b a, there is some integer c such that a = bc. Since b is positive and smaller than a, we have 1 < b < a. Now c = a/b. Since b > 1, we see that a/b < a, and since b < a and positive, we see that a/b > 1. Hence, 1 < c < a. We note the following useful fact about prime numbers.

9 CHAPTER V. THEORY OF THE INTEGERS PRIME NUMBERS Theorem Let p be a prime number, and let a Z. Then { p if p a GCD(p,a) = 1 if p a. Proof. We know that GCD(p,a) must be a positive divisor of p, so it must be 1 or p. If p a then p is clearly the largest common divisor; similarly if p a, then 1 is the largest common divisor. When we combine this with Euclid s lemma(theorem 19.10), we obtain the following important description of prime numbers. Theorem Let a Z with a > 1. The following are equivalent. (a) a is a prime number. (b) For b,c Z, if a bc and a b, then a c. Proof. Assume that a is a prime number, and that a bc and a b. Then GCD(a,b) = 1, so by Euclid s Lemma (Theorem 19.10), we see that a c. Now suppose that a is not prime. Then a is composite, so a = bc for some integers b,c with 1 < b,c < a. Now a bc and a b and a c (since b and c are positive and smaller than a. Hence, (b) is false. 20.B Divisibility by primes We know that if a number is composite, then it has a factorization into smaller numbers. One might wonder if it is possible to guarantee that these smaller numbers must also be composite. In other words, is is possible to find a number that is so composite that all of its factors are composite? The following theorem answers that question. Theorem Every integer larger than 1 is divisible by a prime number. Proof. Let a Z be larger than 1. If a is prime, then it is divisible by a prime number, namely itself. Otherwise the set S = {b Z : b a,1 < b < a} is finite and nonempty, so it has a least element p. Suppose, by way of contradiction, that p is not prime. Then there is some c with 1 < c < p and c p. Since c p and p a, we have that c a, so c S. This is a contradiction, since p was the smallest element of S. Hence, p must be a prime divisor of a. Theorem Let a Z be greater than 1. Then a is either prime or is a product of primes. Proof. We proceed by strong induction. Let P(n) be the statement P(n) : n is prime or a product of primes. We wish to prove that P(n) is true for all n 2. For our base case, we note that P(2) is true, since 2 is prime. Suppose that, for some k 2, P(2) P(k) is true. In other words, every integer from 2 to k is either prime or a product of primes. We now break the proof into cases, depending on whether k+1 is prime or composite. Case 1. Suppose k +1 is prime. Then P(k +1) is true. Case 2. Suppose k +1 is composite. Then we may factor k +1 = bc, with 1 < b,c < k + 1. Hence, by our induction hypothesis, P(b) and P(c) are both true; both b and c are prime or products of primes. Hence, the product bc is a product of primes, so P(k +1) is true. Therefore, by induction the theorem is true. This theorem can be greatly strengthened; we will now prove that every integer not only has a prime factorization, but that that prime factorization is unique. The importance of this theorem is evident in the fact that it has a name: The Fundamental Theorem of Arithmetic. Theorem 20.8 (The Fundamental Theorem of Arithmetic). Let n Z be greater than 1. Then n has a factorization into primes n = p 1 p 2 p k with p 1 p 2 p r, and this factorization is unique.

10 CHAPTER V. THEORY OF THE INTEGERS PRIME NUMBERS Proof. We have already seen that n can be written as a product of primes. It is easy to see that we can reorder the primes so that they are in increasing order. All that remains to be proven is the uniqueness statement, which we will prove by strong induction. Let P(n) be the statement P(n): n has a unique factorization into prime numbers. Clearly, P(2) is true; 2 is prime, so the only way to factor it into 2 = p 1 p r is to have r = 1 and p 1 = 2. A similar argument works for any prime p; P(p) is true. Assume that k 2 and that P(2)... P(k) is true. In other words, every integer from 2 to k has a unique prime factorization. We wish to prove that P(k +1) is true. We divide the proof into two cases. Case 1: If k +1 is prime, then P(k +1) is true, as explained above. Case 2: If k +1 is not prime, Assume that k +1 = p 1 p k = q 1 q l has two prime factorizations, with p 1,...,p k,q q,...,q l all prime, and p 1 p k and q 1 q l. Note that p 1 k +1, so p 1 q 1 q k. Hence, we have that p 1 q i for some i. Since q i is prime and p 1 1, we must have p 1 = q i. Now we know that q 1 q i = p 1. In addition, in a similar fashion to what we did above, q 1 = p j p 1 for some j, so in fact, q 1 = p 1. Now, (k+1)/p 1 = p 2 p k = q 2 q l. Since 2 (k+1)/p 1 k, we see that (k + 1)/p 1 has a unique factorization into prime. Hence, k = l and each p i = q i for i from 2 to k. Since p 1 = q 1, we see that the two factorization that we had for k +1 were identical. Hence, k +1 has a unique factorization into primes, and P(k +1) is true. Thus, by induction, every integer greater than 1 has a unique factorization into primes. Remark We note that typically, the factorization of a number into primes will be simplified by combining copies of the same prime together. For instance, if we wish to factor 120, rather than writing 720 = , we might write 720 = This provides for a more compact representation of the factorization. Using this convention, we can restate Theorem 20.8 as saying that every integer n > 1 can be uniquely written as n = k i=1 p ai i, with k N, each p i prime, p 1 < p 2 < p k, and each a i > C The infinitude of primes Thousands of years ago, the ancient Greeks knew that there were infinitely many primes. We give here an adaptation of the proof given by Euclid of this fact. What we will prove is actually that no finite set of primes can contain all the primes. This clearly implies that the set of all primes must be infinite. Theorem No finite set of prime numbers contains all the prime numbers. Proof. Let S be any finite nonempty set of prime numbers. Let M = n Sn. Then M 2 since S is nonempty, and any prime is greater than 1. Let N = 1+M. We note that N > 1, so N is divisible by some prime number p. Suppose, by way of contradiction, that p S. Then clearly p M. Since p N and p M, we see that p (N M), so p 1. This is a contradiction, since p > 1. Hence, p is a prime that is not in S. Remark We can try this proof out, by selecting any finite set of primes that we wish, and constructing a prime not in that set. For instance, let S = {2,3,5,7,11,13}. Then M = 2730 and N = In this case, N is prime, and is not contained in S. Now suppose that S = {2,3,5,7,11,13,17,19}. Then M = and N = = Both 593 and 1487 are primes that are not in S.

11 CHAPTER V. THEORY OF THE INTEGERS PRIME NUMBERS 20.D Computing GCD in terms of prime factorization We now demonstrate a method of computing the GCD of two integers if the prime factorization of each integer is known. Definition Let n 1 be an integer and let p be a prime. If p n, define v p (n) to be the exponent of p in the factorization of n ; otherwise, if p n, define v p (n) = 0. We call v p (n) the valuation of n at p. Example Ifn = 720 = , thenv 2 (n) = 4, v 3 (n) = 2, v 5 (n) = 2, and v p (n) = 0 for all p > 5. You will prove the following theorem in the exercises. Theorem Let m,n Z be nonzero. (a) For any prime p, we have v p (mn) = v p (m)+v p (n). (b) If m n then for any prime p, we have v p (m) v p (n). (c) If v p (m) v p (n) for all primes p, then m n. Using this theorem, we may give the following description of the GCD. Theorem Let a,b Z be nonzero. For each prime p, define c p = min(v p (a),v p (b)). Then GCD(a,b) = p abp cp. Example Let a = and let b = Then GCD(a,b) = E Exercises Exercise For each of the following integers n, write the prime factorization of n. (a) n = 27. (b) n = (c) n = 60. Exercise An integer n is said to be squarefree if, for m N, m 2 n m = 1. In other words, the only square of an integer dividing n is 1. (a) Prove that n is squarefree if and only if v p (n) 1 for all primes p. (b) Suppose that n is squarefree. Prove that if n = ab with a,b Z, then GCD(a,b) = 1. (c) Prove or disprove the converse of (b). Exercise Let m,n Z be nonzero. Prove the following. (a) For any prime p, we have v p (mn) = v p (m)+v p (n). (b) If m n then for any prime p, we have v p (m) v p (n). (c) If v p (m) v p (n) for all primes p, then m n. Exercise Determine the GCD of 1081 and Is this easier using the Euclidean algorithm, or using prime factorization? Exercise Let a and b be nonzero integers. Prove that LCM(a,b) = p abp max(vp(a),vp(b)). Proof. Let d = p abp cp. Note that d > 0 and that v p (d) = c p for any prime p. Since c p v p (a) for all p, we see that d a. Similarly d b. So d is a common divisor of a,b. Let e be any common divisor of a and b. Then v p (e) v p (a) and v p (e) v p (b), so v p (e) c p. Hence, e d. Therefore, by Theorem 19.16, d = GCD(a, b).