is square-free. Suppose it were not and had a square factor x 2 p r 1. This is a contradiction, as the only possible primes dividing p r 1

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Hilary Simpson
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1 1. Section Problem 8 Let n be a positive integer. By the fundamental theorem of arithmetic we can factor n as n = p e 1 1 p e where the p i are distinct primes and e i 0 for each i. For each e i, by the division algorithm write e i = 2f i + r i with r i {0, 1}. Then we have n = (p f 1 1 p f )2 p r 1 1 p r We have (p f 1 1 p f )2 is a square. I claim that p r 1 1 p r is square-free. Suppose it were not and had a square factor x 2 p r 1 1 p r, with x 1. Then some prime q divides x and so q 2 divides p r 1 1 p r. This is a contradiction, as the only possible primes dividing p r 1 1 p r the p i, and as r i 1, we have p 2 i does not divide p r 1 1 p r Thus n can be written as the product of a square and a square-free integer. 2. Section Problem 10 Write a = p e 1 1 p e and b = p f 1 1 p f where the p i are distinct primes and e i, f i 0. As a 3 b 2 we have 3e i 2f i and hence e i 2 3 f i f i for each i. Thus a b. 3. Section Problem 32 (5th edition). Problem 30 (6th edition) For distinct primes p i we have (p e 1 1 p e, p f 1 1 p f ) = pmin{e 1,f 1 } 1 p min{e,f } [p e 1 1 p e, p f 1 1 p f ] = pmax{e 1,f 1 } 1 p max{e,f } We use this to compute the greatest common divisors and least common multiples. a) ( , ) = ( , ) = b) ( , ) = 7 ( , ) = c) ( , ) = ( , ) = d) ( , ) = ( , ) = Section Problem 36 (5th edition). Problem 34 (the 6th edition) Let us assume that a and b are positive (we will multiply each by ±1 to get all integer solutions at the end). Note that (a, b) = 18 = and [a, b] = 540 = Thus the only prime divisors of a and b are 2, 3 and 5. Write a = 2 d 2 3 d 3 5 d 5 and b = 2 e 2 3 e 3 5 e 5, where d 2, d 3, d 5, e 2, e 3, e 5 are non-negative integers. The smaller of the two exponents, min(d 2, e 2 ) equals 1 (the exponent of 2 in 18) and max(d 2, e 2 ) equals 2 (the exponent of 2 in 540). After possibly interchanging a and b, we may assume d 2 = 1 and e 2 = 2. Similarly, min(d 3, e 3 ) = 2 (the exponent of 3 in 18), max(d 3, e 3 ) = 3 (the exponent of 3 in 540), and min(d 5, e 5 ) = 0 (the exponent of 5 in 18), min(d 5, e 5 ) = 1 (the exponent of 5 in 540). Thus (d 3, e 3 ) = (2, 3) or (3, 2) and (d 5, e 5 ) = (0, 1) or (1, 0). This gives rise to four possibilities:. are

2 (i) a = = 18 b = = 540, (ii) a = = 90 b = = 108, (iii) a = = 54 b = = 180, and (iv) a = = 270 b = = 36. All other possibilities are obtained from these four by interchanging a and b and multiplying a or b (or both) by Section Problem 60 (50th edition). Problem 56 (6th edition) We will argue by contradiction. Suppose there are finitely many primes of the form 6 + 5, where is a positive integer. Let them be p 1,..., p. Let N = 6p 1 p 1. We have N > 1, as p 1 = 5 > 0. Let q be an arbitrary prime dividing N. By the division algorithm we have q = 6a + r for some integer a and r {0,..., 5}. If r {0, 2, 4} then q is even and hence is 2. But 2 does not divide N, as 2 divides N + 1. Thus r 2, If r = 3, then 3 q = 3(2a + 1) and thus q = 3. Again, 3 does not divide N, as 3 divides N + 1. Thus q 3. If r = 5, then q is of the form By our assumption p 1,..., p are the only primes of this form. Thus q = p i for some i. But p i divides N + 1, so does not divide N. Thus every prime q dividing N is of the form 6a + 1 for some integer a. For any two integers of this form we have (6a+1)(6b+1) = 6(6ab+a+b)+1, which is also in this form. As we showed above, N is the product of primes, say, q 1,..., q r (not necessarily distinct), and each of these primes is of the form Thus N = q 1... q r must also be of this form. This is a contradiction, as N is of the form This contradiction shows that there are infinitely many primes of the form Section Problem 72 (5th edition). Problem 68 (6th edition) Let p i be one of the primes in the list. Suppose i m, then we have p i Q. Supose p i (Q + R), then p i (Q + R Q) = R. This is a contradiction, as p i does not appear in the prime factorization of R. Thus p i does not divide Q + R. If m < i n, then p i R, and by the same reasoning we have p i does not divide Q + R. Thus none of the primes in the list divide Q + R. If there were finitely many primes p 1,..., p n, then as every integer > 1 has a prime factor, one of the p i would have to divide Q + R. This is a contradiction and so there are infinitely many primes. 7. Section Problem 2 a) As (3, 4) = 1 there are infinitely many solutions. As x = y = 1 gives a particular solution, all solutions are given by x = 1 + 4n, y = 1 3n, where n is an integer. b) As (12, 18) = 6 which does not divide 50, there are no solutions c) First we find a particular solution to 30x + 47y = 1 by using the Euclidean algorithm. 74 = =

3 17 = = Bac substitution: 1 = = 13 (17 13) 3 = = (30 17) = = 30 4 (47 30) 7 = Thus x 0 = 11, y 0 = 7 is a particular solution to 30x+47y = 1. To get a particular solution to 30x + 47y = 11, multiply both numbers by 11: x 1 = 121, y 1 = 77. The general solution is thus x = t, y = 77 30t, where t ranges over the integers. d) As (25, 95) = 5 which divides 970, there are infinitely many solutions. Divide both sides by 5: our equation reduces to 5x + 19y = 194. x 0 = 4, y 0 = 1 is a particular solution to 5x + 19y = 1. x 1 = = 776, y 0 = 194 is a particular solution to As x = 16, y = 6 gives a particular solution, all solutions are given by x = t, y = 194 5t, where t ranges over the integers. e) Euclidean algorithm: 1001 = = = = = = Thus (102, 1001) = 1, and the Diophantine equation 102x y = 1 has infinitely many solutions. To find a particular solution, use bac substitution: 1 = = 5 (7 5) 2 = 7 ( 2) = 7 ( 2) + (19 7 2) 3 = = 19 3 ( ) 8 = 83 ( 8) = 83 ( 8)+(102 83) 35 = = ( ) 43 = 1001 ( 43) Thus x = 422, y = 43 gives a particular solution, and the general solution is x = t, y = t, where t ranges over the integers. 8. Section Problem 6 This problem can be posed as finding a nonnegative integer solution to the linear diophantine equation 63x + 7 = 23y, where x is the number of plantains in a pile, and y is the number of plantains each traveller receives. We find a particular solution x = 5, y = 14 and then, noting that (63, 23) = 1, all integer solutions are given by x = 5+23n, y = 14+63n, where n is an integer. These are both positive if and only if n 0. Thus the number of plantains in a pile could be any positive integer of the form n. 9. Section Problem 8 This problem can be posed as finding a nonnegative integer solution to the linear diophantine equation 6o + 33g = 549, where o is the number of oranges and g is the number of grapefruits. Since (18, 33) = 3, and 549 is divisible by 3, we see that there will be infinitely many solutions. To simplify our equation, let us divide both sides by 3: 6o + 11g = 183. A particular solution to 6o + 11g = 1 is o = 2, g = 1. A particular solution to 2o+11g = 183 is thus o = = 366, and g = ( 1) 183 = 183. General solution: o = t, g = 183 6t, where t is an integer.

4 To get positive solutions, we need to choose t so that t 0 and 183 6t 0. From the first inequality, t = , and from the second, t = The only integer values of t satisfying these inequalities are 31, 32, and 33. t = 31 correspond to o = 25, g = 3, t = 32 correspond to o = 14, g = 9, t = 33 correspond to o = 3, g = 15. The total number of pieces of fruit purchased in these three cases are = 28 if t = 31, = 23 if t = 32, and = 28 if t = 33. The smallest number of pieces of fruit purchased is 28: 3 oranges and 15 grapefruits. 10. Section Problem 10 We consider the linear diophantine equation 11l + 8c = T, where l represents the number of lobster dinners bought, c represents the number of chicen dinners bought, and T represents the total bill. If T = 1, then l = 3, c = 4 gives a particular solution. As (8, 11) = 1, for an arbitrary integer T the general solution is given by l = 3T + 8n, c = 4T 11n, where n is an integer. Both solutions are non-negative if 3T + 8n 0 and c = 4T 11n 0, i.e., n [ 3 8 T, 4 11 T ]. a) If T = 777, then 3 8 T = , 4 11 T = , and the integers n in the interval [ 3 8 T, 4 11 T ] are 291, 290,..., 283. Thus there are 9 possibilities for l = n and c = n: (1) l = 3 and c = 93 if n = 291, (2) l = 11 and c = 82 if n = 290, (3) l = 19 and c = 71 if n = 289, (4) l = 27 and c = 60 if n = 288, (5) l = 35 and c = 49 if n = 287, (6) l = 43 and c = 38 if n = 286, (7) l = 51 and c = 27 if n = 285, (8) l = 59 and c = 16 if n = 284, and (9) l = 67 and c = 5 if n = 283. b) If T = 96, then 3 8 T = 36, 4 11 T = , and the integers n in the interval [ 3 8 T, 4 11 T ] are 36 and 35. Thus there are 2 possibilities for l = n and c = n: (1) l = 0 and c = 12 if n = 36, (2) l = 8 and c = 1 if n = 35.

5 c) If T = 69, then 3 8 T = 257 8, 4 11 T = 25 1, and there are no integers in the interval 11 [ 3 8 T, 4 T ]. It is impossible for the total bill to be $69. 11

SOLUTIONS TO PROBLEM SET 1. Section = 2 3, 1. n n + 1. k(k + 1) k=1 k(k + 1) + 1 (n + 1)(n + 2) n + 2,

SOLUTIONS TO PROBLEM SET 1. Section = 2 3, 1. n n + 1. k(k + 1) k=1 k(k + 1) + 1 (n + 1)(n + 2) n + 2, SOLUTIONS TO PROBLEM SET 1 Section 1.3 Exercise 4. We see that 1 1 2 = 1 2, 1 1 2 + 1 2 3 = 2 3, 1 1 2 + 1 2 3 + 1 3 4 = 3 4, and is reasonable to conjecture n k=1 We will prove this formula by induction.