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Nipissing University Department of Computer Science & Mathematics MATH 3126 - Number Theory Instructor: Dr. Tzvetalin S. Vassilev Time: 90 minutes Midterm Examination (Solution) October 25, 2011 CLOSED BOOK. NO CALCULATORS. SHOW YOUR WORK 1. [5 pts.] Let tn denote the n-th triangular number. Show that t608 is a product of three consecutive integers. Solution: Since tn = n(n+1) , t608 = 608·609 = 304 · 609 = 2 · 152 · 3 · 203 = 2 · 2 · 76 · 3 · 2 2 7·29 = 2·2·2·38·3·7·29 = 2·2·2·2·19·3·7·29 = (2·2·2·7)·(3·19)·(2·29) = 56·57·58 Please note: no calculator or multiplication of large numbers is required to solve this question. As a matter of fact, any of these is detrimental to solving the question. 2. [5 pts.] Find a formula for an , n ≥ 0, the n-th term of the sequence 3, 10, 21, 36, 55, 78, . . . Solution: Using the finite difference method, we have: 3 10 21 36 55 78 7 11 15 19 23 4 4 4 4 n n n Thus, an = 3 0 + 7 1 + 4 2 = 3 + 7n + 2n(n − 1) = 2n2 + 5n + 3. 3. [5 pts.] Let un denote the n-th Fibonacci number. Prove that u2n+1 − u2n = un−1 un+2 Solution: Direct proof is available here. Those interested in pursuing an inductive approach may do so. u2n+1 − u2n = (un+1 − un )(un+1 + un ) = un−1 un+2 4. [10 pts.] Let tn denote the n-th triangular number. Find the sum of the reciprocals of the triangular numbers, i.e. ∞ X 1 1 1 1 =1+ + + + ... t 3 6 10 n n=1 Solution: Having in mind that k k k X X X 1 2 1 k 1 = =2 =2 =2 1− t n(n + 1) n(n + 1) k+1 k+1 n=1 n n=1 n=1 ∞ k X X 1 1 1 = lim = lim 2 1 − =2 k→∞ k→∞ t t k+1 n=1 n n=1 n 5. [15 pts.] Let tn denote the n-th triangular number. Prove that tm+n = tm +tn +mn. Solution: Recall that tn = n(n+1) . Thus, 2 n(n + 1) n(n + 1) m2 + m n2 + n 2mn + + mn = + + = 2 2 2 2 2 m2 + n2 + 2mn + m + n (m + n)2 + (m + n) (m + n)(m + n + 1) = = = = tm+n 2 2 2 tm + tn + mn = 6. [5 pts.] Prove that 7 divides 32n+1 + 2n+2 , for any positive integer n. Solution: We will give an inductive proof here. Base case: n = 1. 32·1+1 + 21+2 = 33 + 23 = 27 + 8 = 35. 7 divides 35, which establishes the truth of the claim for n = 1. Inductive step: Assume that for some integer k ≥ 1, 7|32k+1 + 2k+2 . We have to prove that 7|32(k+1)+1 + 2(k+1)+2 = 32k+3 + 2k+3 . Further, 32k+3 + 2k+3 = 32 · 32k+1 + 2 · 2k+2 = 9 · 32k+1 + 2 · 2k+2 = 7 · 32k+1 + 2 32k+1 + 2k+2 . But, 7|32k+1 + 2k+2 according to the inductive hypothesis. Also, 7|7 · 32k+1 by definition. Thus, 7|32(k+1)+1 + 2(k+1)+2 = 32k+3 + 2k+3 , which concludes the inductive step. Conclusion: By mathematical induction, ∀n ∈ Z+ , 7|32n+1 + 2n+2 . 7. [5 pts.] Prove that there cannot exist integers a and b such that gcd(a, b) = 3 and a + b = 65. Solution: We know that gcd(a, b) divides any linear combination of a and b. Thus, here 3 has to divide 65, which is absurd. Therefore, no such numbers exist. 8. [5 pts.] If a divides b, determine gcd(a, b) and lcm(a, b). Solution: Obviously, gcd(a, b) = a and lcm(a, b) = b. 9. [5 pts.] Give an example to show that it is not necessarily the case that gcd(a, b, c) · lcm(a, b, c) = abc Solution: Take a = b = c = 2. Then gcd(a, b, c) = lcm(a, b, c) = 2, and therefore gcd(a, b, c) · lcm(a, b, c) = 4, while abc = 8. 10. [5 pts.] Express gcd(481, 299) as a linear combination of 481 and 299. Solution: Here we use the Saunderson’s Algorithm with 481 and 299: 1 · 481 − 0 · 299 = 481 0 · 481 − 1 · 299 = −299 1 · 481 − 1 · 299 = 182 1 · 481 − 2 · 299 = −117 2 · 481 − 3 · 299 = 65 3 · 481 − 5 · 299 = −52 5 · 481 − 8 · 299 = 13 Since gcd(481, 299) = 13 the task is accomplished, 13 = 5 · 481 − 8 · 299. 11. [10 pts.] Show that in any set of (n + 1) different integers selected from the set {1, 2, . . . , 2n} there must exist a pair of coprime integers. Solution: Here we use the Dirichlet Principle. Consider the n pairs of coprime integers (1, 2), (3, 4), (5, 6), . . . , (2n − 1, 2n). Together these pairs comprise the set S = {1, 2, . . . , 2n}. If we select (n + 1) different numbers from S, then by Dirichlet Principle we are guaranteed that we have selected both elements of at least one of the n pairs. Thus, we have selected a pair of coprime numbers. 12. [10 pts.] Prove, for all positive integers n, that n5 n3 7n + + 5 3 15 is an integer. Solution: Another inductive proof here. 5 3 7 3 5 7 Base case: n = 1. 15 + 13 + 7·1 = 51 + 13 + 15 = 15 + 15 + 15 = 1. Thus, the base 15 case is verified. 5 3 Inductive step: Assume that for some integer k ≥ 1, k5 + k3 + 7k is an integer. 15 (k+1)5 (k+1)3 7(k+1) We have to show that 5 + 3 + 15 is an integer. k 5 + 5k 4 + 10k 3 + 10k 2 + 5k + 1 k 3 + 3k 2 + 3k + 1 7k + 7 (k + 1)5 (k + 1)3 7(k + 1) + + = + + = 5 3 15 5 3 15 5 k k 3 7k 7 1 1 4 3 2 2 = + + + + + (k + 2k + 2k + k) + (k + k) + 3 5 15 3 5 15 The expression in the first set of brackets is an integer, according to the inductive hypothesis. The expressions in the second and third set of brackets are integers since k is an integer (i.e. products and sums of integers are integers), the expression in the fourth set of brackets is an integer, too (it equals 1, as shown in the base case). Thus, the entire expression is an integer, being represented as a sum of four integers. 5 3 Conclusion: By mathematical induction, ∀n ∈ Z+ , n5 + n3 + 7n is an integer. 15 13. [15 pts.] Show that if y 2 = x3 + 2, then both x and y are odd. Solution: Assume x is even, that is x = 2k for some integer k. Then, y 2 = x3 +2 = (2k)3 + 2 = 8k 3 + 2 = 4(2k 3 ) + 2. Thus, y 2 has a remainder of 2 when divided by 4. This is absurd. Therefore, x cannot be even, i.e. x is odd. Since x is odd, x3 is odd, and subsequently x3 + 2 is odd. Then y 2 is odd, and y is odd.

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