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Challenge1 Solutions - MATH 135 Solutions to Challenge Problem Set 1 1 Determine whether there exists a positive integer n such that n 2 n 1 is a

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Unformatted text preview: MATH 135, Solutions to Challenge Problem Set 1 1: Determine whether there exists a positive integer n such that n 2 + n + 1 is a square. Solution: No such integer n exists, as we shall prove. Suppose (for a contradiction) that such an integer does exist, say n 2 + n + 1 = m 2 , where n > 0 and m ≥ 0. Then we have m 2 = n 2 + n + 1 > n 2 , so m > n , and hence m ≥ n + 1. But then m 2 ≥ ( n + 1) 2 = n 2 + 2 n + 1 > n 2 + n + 1 and so m 2 6 = n 2 + n + 1 giving the desired contradiction. 2: Let a = c and for n ≥ 1 let a n = pa n- 1 + q where p 6 = 0. Find a formula for a n in terms of c , p and q . Solution: We have a = c a 1 = pc + q a 2 = p ( pc + q ) + q = p 2 c + pq + q a 3 = p ( p 2 c + pq + q ) + q = p 3 c + p 2 q + pq + q a 4 = p ( p 3 c + p 2 q + pq + q ) + q = p 4 c + p 3 q + p 2 q + pq + q and in general a n = p n c + p n- 1 q + p n- 2 q + ··· + p 2 q + pq + q = p n c + ( p n- 1 + p n- 2 + ··· p 2 + p + 1) q . We can obtain a (non-recursive) formula for the geometric sum p n- 1 + p n- 2 + ··· + p 2 + p + 1 as follows. Let S = p n- 1 + p n- 2 + ··· + p 2 + p + 1 (1). Note that pS = p n + p n- 1 + p n- 2 + ··· + p 2 + p (2). Subtracting (1) from (2) gives S ( p- 1) = p n- 1 and so S = p n- 1 p- 1 . Thus we have a n = p n c + p n- 1 p- 1 q . 3: Let { F n } be the Fibonacci sequence (so F 1 = 1, F 2 = 1, and for n ≥ 3, F n = F n- 1 + F n- 2 ). Prove that F n 2 + F 2 n +1 = F 2 n +1 for all n ≥ 1. Solution: We begin by trying (and failing) to use induction to prove that F n 2 + F 2 n +1 = F 2 n +1 for all n ≥ 1. When n = 1, we have LS = F 1 2 + F 2 2 = 1 2 + 1 2 = 2 and RS = F 3 = F 2 + F 1 = 1 + 1 = 2 = LS , so the equality holds. Let k ≥ 1 and suppose (inductively) that F k 2 + F k +1 2 = F 2 k +1 . Then when n = k + 1 we have LS = F k +1 2 + F k +2 2 = F k +1 2 + ( F k +1 + F k ) 2 = F k +1 2 + F k +1 2 + 2 F k F k +1 + F k 2 = ( F k +1 2 + 2 F k F k +1 ) + ( F k 2 + F k +1 2 ) = ( F k +1 2 + 2 F k F k +1 ) + F 2 k +1 (where the last inequality follows from the induction hypothesis), and we have RS = F 2 k +3 = F 2 k +2 + F 2 k +1 ....
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This note was uploaded on 10/21/2010 for the course MATH 135 taught by Professor Andrewchilds during the Fall '08 term at Waterloo.

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Challenge1 Solutions - MATH 135 Solutions to Challenge Problem Set 1 1 Determine whether there exists a positive integer n such that n 2 n 1 is a

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