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Homework 2 Section 1.1: 9; Section 1.2: 10, 19, 25a; Section 1.3: 7 (b,d), 8, 9, 17, 26; Also to hand in: Recall that the Fibonacci numbers are given by the sequence f(1) = 1, f(2) = 1, and f(n) = f(n-1) + f(n-2) for all n greater than or equal to 2. (a) Show that every two successive terms of the Fibonacci sequence are relatively prime. (b) Consider all terms of the sequence whose index is a multiple of 3: f(3), f(6), f(9), etc. Compute f(9) and f(12). Use your observations from this computation to prove that for any positive number C, f(3) | f(3C). 1.1 1.2 19. Let p > q be primes. Suppose pq + 1 is a square. Then there exists n in Z such that pq + 1 = n^2. Thus qp=(n+1)(n-1). By the uniqueness of prime factorization we find n+1=1, or n+1=p, or n+1=pq. We can discount the first (which implies n=0) and the last (which implies n-1=1 and hence n=2); so we have that p=n+1 which implies q=n-1. Hence p = q+2 and p and q are twin primes. Conversely, suppose p and q are twin primes, say p = q + 2. Then pq + 1 = (q + 2)q + 1 = q^2 + 2q + 1 = (q+1)^2, and pq + 1 is a square

Homework 2 - University of California, San Diegominnes/archive/fall2010/100a/files/Homework… · Homework 2 Section 1.1: 9; Section 1.2: 10, 19, 25a; Section 1.3: 7 (b,d), 8, 9,

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Page 1: Homework 2 - University of California, San Diegominnes/archive/fall2010/100a/files/Homework… · Homework 2 Section 1.1: 9; Section 1.2: 10, 19, 25a; Section 1.3: 7 (b,d), 8, 9,

Homework 2

Section 1.1: 9; Section 1.2: 10, 19, 25a; Section 1.3: 7 (b,d), 8, 9, 17, 26; Also to hand in: Recall that the Fibonacci numbers are given by the sequence f(1) = 1, f(2) = 1, and f(n) = f(n-1) + f(n-2) for all n greater than or equal to 2.

(a) Show that every two successive terms of the Fibonacci sequence are relatively prime. (b) Consider all terms of the sequence whose index is a multiple of 3: f(3), f(6), f(9), etc. Compute f(9) and f(12). Use your observations from this computation to prove that for any positive number C, f(3) | f(3C).

1.1

1.2

19. Let p > q be primes. Suppose pq + 1 is a square. Then there exists n in Z such that pq + 1 = n^2. Thus qp=(n+1)(n-1). By the uniqueness of prime factorization we find n+1=1, or n+1=p, or n+1=pq. We can discount the first (which implies n=0) and the last (which implies n-1=1 and hence n=2); so we have that p=n+1 which implies q=n-1. Hence p = q+2 and p and q are twin primes.

Conversely, suppose p and q are twin primes, say p = q + 2. Then pq + 1 = (q + 2)q + 1 = q^2 + 2q + 1 = (q+1)^2, and pq + 1 is a square

Page 2: Homework 2 - University of California, San Diegominnes/archive/fall2010/100a/files/Homework… · Homework 2 Section 1.1: 9; Section 1.2: 10, 19, 25a; Section 1.3: 7 (b,d), 8, 9,

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Page 3: Homework 2 - University of California, San Diegominnes/archive/fall2010/100a/files/Homework… · Homework 2 Section 1.1: 9; Section 1.2: 10, 19, 25a; Section 1.3: 7 (b,d), 8, 9,
Page 4: Homework 2 - University of California, San Diegominnes/archive/fall2010/100a/files/Homework… · Homework 2 Section 1.1: 9; Section 1.2: 10, 19, 25a; Section 1.3: 7 (b,d), 8, 9,