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. MATH 307, SPRING 2005, SOLUTION of Midterm 1 Note that for solutions of numbers 3 (a), (b), and 4, 5, I gave necessary hints for you only, not the complete solutions as students should show in their work. 1. The Mathematical Induction (a) (4 points) Use the mathematical induction to prove the formula: 12 + 22 + 32 + · · · + n2 = n(n + 1)(2n + 1) . 6 Solution. For n = 1 : 12 = 1, and 1(n+1)((2)(1)+1) = 1, correct. 6 Assume that the formula is correct for n = k ≥ 1, that means for this k, 12 + 22 + 32 + · · · + k 2 = k(k + 1)(2k + 1) . 6 Now we prove the formula for n = k + 1. Consider 12 + 22 + 32 + · · · + k 2 + (k + 2 1)2 = k(k+1)(2k+1) + (k + 1)2 = k(k+1)(2k+1) + 6(k+1) = (k+1)[k(2k+1)+6(k+1)] = 6 6 6 6 (k+1)(2k2 +7k+6) 6 = (k+1)(k+2)(2k+3) = (k+1)[2(k+1)+1][(k+1)+1] . Compare with the given 6 6 formula we see that it is correct for n = k + 1. Hence the formula is proved. (b) (5 points) Use the mathematical induction to prove that for n > 0, 15 is a divisor of (24n − 1). Solution. For n = 1, 2(4)(1) − 1 = 15, and 15|15, correct. Assume that the statement is correct for n = k ≥ 1, that means 15|24k − 1. For n = k + 1 we see 24(k+1) − 1 = 24k .24 − 1 = 24k .24 − 24 + 24 − 1 = 24 (24k − 1) + 24 − 1 = 24 (24k − 1) + 15. But by our induction assumption, (24k − 1) is divisible by 15. Hence 24(k+1) − 1 is divisible by 15. The proof is complete. Typeset by AMS-TEX 1 2 2. The Binomial Theorem (a) (3 points) Write the (a + b)n binomial formula: n n n n Solution. (a + b)n = an + an−1 b + · · · + b 0 1 n (b) (4 points) Prove that n n n + +···+ = 2n . 0 1 n Solution. Using (a) for a = b = 1. 3. Prime numbers (a) (5 points) Let n > 1 be an integer. Prove that 23n + 1 is not a prime number. Hint: 23n = (2n )3 . Solution: 23n + 1 = (2n )3 + 1 = (2n + 1)[(2n )2 − 2n + 1] for n > 1 we see each factor is greater than 1. Hence 23n + 1 is NOT prime. (b) (4 points) Show that for any integer n > 1, n5 + 1 is not a prime number. Solution. Use the formula n5 + 1 = (n + 1)(n4 − n3 + n2 − n + 1). For n > 1 each factor is greater than 1. (c) (6 points) Let p be a prime number. Find all values of p such that 5p + 1 is a perfect square. Solution. Since 5p + 1 is a pefect square, 5p + 1 = a2 for some positive integer a. Hence 5p = a2 − 1 = (a − 1)(a + 1). Therefore p = (a−1)(a+1) . Now, since 5 is a 5 prime, we must have either 5|(a − 1) or 5|(a + 1). This together with the fact that p is a prime gives either a − 1 = 5 or a + 1 = 5. The first case yields a = 5 + 1 = 6 and so p = 7. The second case gives a = 5 − 1 = 4, and so p = 3. Answer: p = 7 or p = 3. Remark. From the previous part (c), we can state a general problem: Let q, p be prime numbers such that qp + 1 is a perfect square. Find q and p. Using the solution of (c) above, you can try to answer this problem. A list of such q, p less than 106 was found today (4/27) by my younger son (an undergraduate computer science student at HTC, OU). There are 78498 prime numbers in the range [2, 1000000] in which 8169 prime-pairs that satisfy the condition pq + 1 = n2 . View attachment for details. 3 4. Greatest common divisor (gcd) and least common multiple (lcm) (a) (4 points) Find gcd and lcm of 2030, and 4246. Solution. Use Euclidian algorithm, or write these numbers in the product of ab primes to find the answers for gcd. For lcm use the formula lcm(a, b) = gcd(a,b) (b) (5 points) Find x and y such that gcd(48, 138) = 48x + 138y. Solution. Use Euclidian Algorithm. Answer: x = 3, y = −1 5. Linear Diophantine Equations Find the general solution of the following equations: (a) (5 points) 24x + 36y = 156. Solution. Use Euclidian Algorithm to find gcd(36, 24) = 12 and a specific solution x0 = −13, y0 = 13. Then the general solution is: x = −13 + 3t, y = 13 − 2t for any t ∈ R. (b) (5 points) 102x + 1003y = 221. Solution. Do the same way as part (a). Answer: x0 = 130, y0 = −13, gcd(102, 1003) = 17. Hence the general solution is: x = 130 + 59t, y = −13 − 6t for all t ∈ R.