HOMEWORK ASSIGNMENT 5 ACCELERATED PROOFS AND PROBLEM SOLVING [MATH08071] Each problem will be marked out of 4 points. Exercise 1 [1, Exercise 10.4]). Show that for all positive integers n, hcf6n + 8, 4n + 5) = 1. Solution. Let n be a positive integer. Since 6n + 8 = 4n + 5) + 2n + 3), we see that hcf6n + 8, 4n + 5) divides 2n + 3. Since 22n + 3) 4n + 5) = 1, we see that hcf6n + 8, 4n + 5) divides 1. Then hcf6n + 8, 4n + 5) = 1. Exercise 2 [1, Exercise 10.6]). Let a, b, c Z such that a, b, c) 0, 0, 0). Define the highest common factor hcfa, b, c) to be the largest positive integer that divides a, b, and c. Prove that there are integers s, t, u, such that hcfa, b, c) = sa + tb + uc. Find such integers s, t, u when a = 91, b = 903, c = 1792. Solution. Let us first prove that for some integers s, t, u. If a = 0, then hcfa, b, c) = sa + tb + uc hcfa, b, c) = hcfb, c) by the definition of hcf. On the other hand, it follows from [1, Proposition 10.3] that there are integers t and u such that hcfb, c) = tb + uc. Then hcfa, b, c) = hcfb, c) = 0 a + tb + uc, which is exactly what we want. If b = 0 or c = 0, then we can proceed in a similar way. Thus, we may assume that none of a, b, and c is zero. Put ) d = hcf hcfa, b), hcfa, c) and d = hcfa, b, c). Then d divides both hcfa, b) and hcfa, c) by [1, Proposition 10.4]. Hence d d by [1, Proposition 10.4]. But d divides hcfa, b) and hcfa, c) by definition of hcf. Then d divides a, b, and c, which implies that d d, because d the largest positive integer that divides a, b, and c. But d d. Then d = d. It follows from [1, Proposition 10.3] that hcfa, b) = xa + yb and hcfa, c) = za + wc for some integers x, y, z, and w. But ) d = d = hcf hcfa, b), hcfa, c) = A hcfa, b) + B hcfa, c) This assignment is due on Thursday 29th October 2015. 1
for some integers A and B. Hence, we see that d = Axa + yb) + Bza + wc), and we can put s = Ax + Bz, t = Ay, and u = Bw. Then d = sa + tb + uc. Now let us apply what we just did for a = 91, b = 903, c = 1792. Since we see that hcf91, 903) = 7. Since 903 = 9 91 + 84, 91 = 84 + 7, 84 = 12 7, 1792 = 19 91 + 63, 91 = 63 + 28, 63 = 2 28 + 7, 28 = 4 7, we see that hcf91, 1792) = 7 as well. Since 7 = 91 84 = 91 903 9 91) = 91 10 903, we can put s = 10, t = 1, and u = 0 to get 7 = sa + tb + uc. Since 7 = 63 2 28 = 63 291 63) = 3 63 2 91 = 21792 19 91) 2 91 = 3 1792 59 91, we also can put s = 59, t = 0, and u = 3 to get 7 = sa + tb + uc. Exercise 3 [1, Exercise 11.2]). a) Which positive integers have exactly three positive divisors? b) Which positive integers have exactly four positive divisors? c) Suppose that n 2 is an integer with the property that whenever a prime p divides n, p 2 also divides n i.e., all primes in the prime factorization of n appear at least to the power 2). Prove that n can be written as the product of a square and a cube. Solution. Let us first find positive integers that have exactly three positive divisors. Let n be a positive integer. If n = p 2 for some prime p 2, then n has exactly three different positive divisors: 1, p, and p 2. This is the only possibility for n to have exactly three positive divisors, because if n = 1, then it has just one positive divisor, if n is prime, then it has exactly two positive divisors: 1 and n, if n is divisible by two different primes p 2 and q 2, then n has at least four different positive divisors: 1, p, q, and pq, if n is divisible by p 3 for some prime p 2, then n has at least four different positive divisors: 1, p, p 2, and p 3. Now let us find positive integers that have exactly four positive divisors. Let n be a positive integer. If n is p 3 for some prime p 2, then n has exactly four different positive divisors: 1, p, p 2, and p 3. If n is a product of two different primes p 2 and q 2, then n has exactly four different positive divisors: 1, p, q, and pq. This are the only two possibilities for n to have exactly four positive divisors, because if n = 1, then it has just one positive divisor, if n is prime, then it has exactly two positive divisors: 1 and n, if n = p 2 for some prime p 2, then n has exactly three different positive divisors: 1, p, and p 2, if n is divisible by p 4 for some prime p 2, then n has at least five different positive divisors: 1, p, p 2, p 3, and p 4, if n is divisible by pq 2 for two different primes p 2 and q 2, then n has at least six different positive divisors: 1, p, q, q 2, pq, and pq 2, if n is divisible by three different primes p 2, q 2, and r 2, then n has at least six different positive divisors: 1, p, q, pq, pr, and qr. 2
Let n be a positive integer such that n 2 and all primes in the prime factorization of n appear at least to the power 2. Let us prove that n can be written as the product of a square and a cube. Put n = p a 1 1 pa 2 2 pat t, where p 1,..., p t are primes such p 1 < p 2 < < p t, and t, a 1, a 2,, a t are positive integers. Then a i 2 for every i {1,..., t}. For every a i {a 1, a 2,, a t }, one of the following possibilities holds: either a i is even, or a i is divisible by 3, or a i = 6k + 1 for some positive k 1, or a i = 6k + 5 for some non-negative k 1. For every a i {a 1, a 2,, a t }, let us construct two positive integers b i and c i as follows: if a i is even, then we put b i = a i /2 and c i = 0, if a i is divisible by 3, then we put b i = 0 and c i = a i /3, if a i = 6k + 1 for some positive k 1, then we put b i = 3k 1 and c i = 1, if a i = 6k + 5 for some non-negative k 1, then we put b i = 3k + 1 and c i = 1. We constructed t non-negative integers b 1, b 2,..., b t, and t non-negative integers c 1, c 2,..., c t. Moreover, it follows from the constructions of these non-negative integers that 2b i + 3c i = a i for every i {1,..., t}. Then n = p a 1 1 pa 2 2 pat t = ) 2 ) 3, p b 1 1 p b 2 2 p bt t p c 1 1 pc 2 2 pct t which shows that n can be written as the product of a square and a cube. Exercise 4 [1, Exercise 11.8]). Find all solutions x, y Z to the following Diophantine equations: a) x 2 = y 3, b) x 2 x = y 3, c) x 2 = y 4 77, d) x 3 = 4y 2 + 4y 3. Solution. Let us first solve x 2 = y 3. If x, y) is a solution, then y 0 and x, y) is a solution as well. Note that 0, 0) and 1, 1) are clearly solutions to x 2 = y 3. Therefore, to find all solutions of x 2 = y 3, we may assume for a while that x and y are positive integers such that x 2 and y 2. Suppose that x, y) is a solution of x 2 = y 3. Put x = p a 1 1 pa 2 2 pat t, where p 1,..., p t are primes such that p 1 < p 2 < < p t, and t, a 1, a 2,, a t are positive integers. Put y = q b 1 1 qb 2 2 qb k k, where q 1,..., q k are primes such q 1 < q 2 < < q k, and k, b 1, b 2,, b k are positive integers. Then x 2 = p 2a 1 1 p 2a 2 2 p 2at t = q 3b 1 1 q 3b 2 2 q 3b k k = y 3, which implies that k = t, p i = q i for every i {1,..., t}, and 2a i = 3b i for every i {1,..., t}. 3
Applying [1, Proposition 10.5] to 2a i = 3b i for every i {1,..., t}, we see that 3 a i and 2 b i for every i {1,..., t}, since 2 and 3 are primes. Put k i = a i 3 Z for every i {1,..., t}. Then k i = b i /2 since 6k i = 3b i for every i {1,..., t}. Put n = p k 1 1 pk 2 2 pkt t, and observe that x = n 3 and y = n 2, since a i = 3k i and b i = 2k i for every i {1,..., t}. Vice versa, for every positive integer n, the pair n 3, n 2 ) is a solution of the equation x 2 = y 3. Thus, all solutions of the equation x 2 = y 3 are given by pairs x, y) = m 3, m 2 ) for any integer m. Now let us find integral solutions of x 2 x = y 3. If x, y) is a solution, then 1 x, y) is also a solution check this). It is clear that x, y) = 0, 0) and x, y) = 1, 0) are solutions. Moreover, these are all solutions with y = 0. Therefore, to find all solutions of x 2 x = y 3, we may assume for a while that x and y are positive integers such that x 2. Suppose that x, y) is a solution of the equation x 2 x = y 3. Since 1 = x 1) x, we have hcfx 1, x) = 1 no matter what x is. Then x = m 3 and x 1 = n 3 for some positive integers n and m by [1, Proposition 11.4]. Then m > n, since x > x 1. Hence, we have 1 = m 3 n 3 = m n)m 2 + nm + n 2 ) m 2 + nm + n 2 3, which is a contradiction. Thus, x, y) = 0, 0) and x, y) = 1, 0) are the only solutions of the equation x 2 x = y 3. Now let us find integral solutions of x 2 = y 4 77. If x, y) is a solution, then x, y), x, y), and x, y) are also solutions. Therefore, to find all solutions of x 2 x = y 3, we may assume for a while that x and y are non-negative integers. Suppose that x, y) is a solution of the equation y 4 x 2 = 77. Since y 2 + x)y 2 x) = y 4 x 2 = 77, both y 2 + x and y 2 x are positive divisors of 77. But we know all positive divisors of 77. They are 1, 7, 11, and 77. Since y 2 x y 2 + x and y 2 + x)y 2 x) = 77, either y 2 + x = 77 and y 2 x = 1, or y 2 + x = 11 and y 2 x = 7. If y 2 + x = 77 and y 2 x = 1, then 2y 2 = 78, which implies that y 2 = 39, which is impossible, because y is an integer. If y 2 + x = 11 and y 2 x = 7, then 2y 2 = 18, which implies that y 2 = 9, which implies that y = 3 and x = 2, since we assumed that y is nonnegative integer. Thus, we see that the only solutions x, y) of the equation x 2 = y 4 77 are the pairs 2, 3), 2, 3), 2, 3), and 2, 3). Now let us find all integral solutions of x 3 = 4y 2 + 4y 3. We can rewrite this equation as x 3 = 2y + 3)2y 1), since 4y 2 + 4y 3 = 2y + 3)2y 1). If x, y) is a solution, then x, 1 y) is also a solution check this). Moreover, if x, y) is a solution, then y 0, since x 3 3 for any integer x. Thus, to find all solutions of x 2 x = y 3, we may assume for a while that y is a positive integer. Suppose that x, y) is a solution of the equation x 3 = 4y 2 + 4y 3. Then x is positive, since we assume that y is positive. Since 4 = 2y + 3 2y 1), we see that hcf2y + 3, 2y 1) 4, which implies that hcf2y + 3, 2y 1) = 1, because 2y + 3 and 2y 1 are odd integers so hcf2y + 3, 2y 1) is not divisible by 2). Then 2y + 3 = m 3 and 2y 1 = n 3 for some 4
positive integers n and m by [1, Proposition 11.4]. Then m > n, since 2y + 3 > 2y 1. Thus, we have 4 = m 3 n 3 = m n)m 2 + nm + n 2 ) m 2 + nm + n 2 > m 2 + n 2, which implies that m = 1 and n = 1, because m and n are positive integers. But m > n. The obtained contradiction implies that the equation x 3 = 4y 2 + 4y 3 does not have solutions with integer x and y. Exercise 5 [1, Exercise 12.4]). Use the idea of the proof of [1, Theorem 12.1] to prove that there are infinitely many primes of the form 4k + 3 where k is an integer). Solution. Suppose the claim is wrong, i.e. there are finitely many primes of the form 4k + 3 where k is a non-negative integer). Let us list them all: 4n 1 + 3, 4n 2 + 3, 4n 3 + 3,..., 4n t + 3 for some non-negative integers n 1, n 2, n 3,..., n t, where t is a positive integer. Put N = 44n 1 + 3)4n 2 + 3)4n 3 + 3) 4n t + 3) 1. The integer N is not divisible by any prime among 4n 1 + 3, 4n 2 + 3, 4n 3 + 3,..., 4n t + 3, because 1 is not divisible by any of them. Then N = p 1 p 2 p 3 p k for some primes p 1 p 2 p 3 p r such that none of them is of the form 4k + 3 where k is an integer), where r is a positive integer. Since N is odd, we see that p 1 2. Then every prime among p 1, p 2,..., p r must be of the form 4k + 1 where k is an integer). Then p 1 = 4k 1 + 1, p 2 = 4k 2 + 1, p 3 = 4k 3 + 1,..., p r = 4k r + 1 for some positive integers k 1, k 2,..., k r. Then N = 4k 1 + 1)4k 2 + 1)4k 3 + 1) 4k r + 1) = 4M + 1 for some positive integer M. Then ) 4 4n 1 +3)4n 2 +3)4n 3 +3) 4n t +3) M = 44n 1 +3)4n 2 +3)4n 3 +3) 4n t +3) 4M = N+1 N 1) = 2, which implies that 2 is divisible by 4. The obtained contradiction shows that there are infinitely many primes of the form 4k + 3 where k is an integer). Note that Exercise 5 is a special case of Theorem 6 Dirichlet). Let a and b be positive coprime integers. Then the sequence contains infinitely many prime numbers. b, b + a, b + 2a, b + 3a, b + 4a, b + 5a,..., This nice result was proved in 1837 by Johann Peter Gustav Lejeune Dirichlet 5
. References [1] M. Liebeck, A concise introduction to pure mathematics Third edition 2010), CRC Press 6