Problem 5.29 The following Lemma is true, but the proof given for it below is invalid. Lemma. For any prime p and positive integers n, x1,x2,...,xn, if p | x1x2xn, then p|x; for some 1 i n. = ak.) (Recall that a | b means that there exists an integer k such that b = False Proof. Proof by strong induction on n. The induction hypothesis P(n) is that the Lemma for n. Base case n = 1: When n = 1, we have p | x, therefore we can let i = 1 and conclude p | x. Induction step: Now assuming the claim holds for all kn, we must prove it for n + 1. Suppose p|x1x2*Xn+1. Let Yn = XnXn+1, SO X1X2Xn+1 = x1x2xn-1Yn. Since the right-hand side of this equation is a product of n terms, we have by induction that p divides one of them. If p xi for some i