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challenge1 Solutions

# challenge1 Solutions - MATH 135 Solutions to Challenge...

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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 = n 2 + n + 1 giving the desired contradiction. 2: Let a 0 = c and for n 1 let a n = pa n - 1 + q where p = 0. Find a formula for a n in terms of c , p and q . Solution: We have a 0 = 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 .

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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 . If we could show that ( F k +1 2 + 2 F k F k +1 ) = F 2 k +2 then we would have LS = RS and our induction proof would work. We shall modify this abortive proof by proving two equalities at once.
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