128 Solutions to Exercises 3) Let k ::::: 2 and assume P(k) is true. We must prove that P(k -1) is true. So we suppose that aI, ... , ak-l are k -1 nonnegative numbers. The idea is now to apply P( k) to the k numbers aI, a2, ... , ak-l, k-.yal··· ak-l since the kth root of the product of these numbers is Thus, and so k/ _1 k/ H_1 y a l···ak_l(al···ak_d k-1 = y(al ... ak-l) k-1 ~(al ... ak_I)6 = (al·· .ak_d l / k-l = k-.yal·· ·ak-l· k-.yal··· ak-l = ~(al·.· ak-l) k .yal ... ak-l al + a2 + . .. + ak-l + k-.yal··· ak-l ::; k ' usingP(k) ( 1) al+···+ak-l k-.yal ... ak-l 1 -k::; k ( k-1) al+···+ak-l k-.yal ... ak-l -k-::; k al + . .. + ak-l k-1 which is P(k -1), as desired. 41. (With thanks to M. Oleson) (a) Assume to the contrary that mn - m -n = am + bn where a and b are nonnegative integers. Then m(n -1 -a) = n(l + b), so n I m(n -1 -a). Since m and n are relatively prime, n I (n - 1 -a) by Corollary 4.2.10, which is impossible since 0 < n - 1 -a < n. (b) There exist integers
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