Problem
of the
Week
PROBLEM 122

Voilà!

1 + 4 = 2 + 3

1 + 4 + 6 + 7 = 2 + 3 + 5 + 8
12 + 42 + 62 + 72 = 22 + 32 + 52 + 82

1 + 4 + 6 + 7 + 10 + 11 + 13 + 16 = 2 + 3 + 5 + 8 + 9 + 12 + 14 + 15
12 + 42 + 62 + 72 + 102 + 112 + 132 + 162 = 22 + 32 + 52 + 82 + 92 + 122 + 142 + 152
13 + 43 + 63 + 73 + 103 + 113 + 133 + 163 = 23 + 33 + 53 + 83 + 93 + 123 + 143 + 153

Find a way to divide the integers from 1 to 32 into two disjoint sets of sixteen integers each so that the sum of the k'th powers of the elements in each set are equal, for k = 1, 2, 3, 4, or explain why no such partitioning of these numbers is possible.

Can you state (and solve?) a general conjecture about this situation?


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ã2001 Alberto L. Delgado