A Proizvolov arrangment of the 2n integers, 1 through 2n, consists of an arbitrary subset A of n of the integers arranged in increasing order, matched in pairs with the remaining n integers, subset B, arranged in decreasing order. The Proizvolov identity says that Sum/ai - bi/ = n2. Without regard for whether the smaller integer in a pair was in set A or set B, index pairs by the integers 1 through n in increasing order. The other integer in each pair is the one the index was paired with in the Proizvolov arrangment. Call this ordered set C. The only way construction C could fail to be possible is if some Proizvolov pair consisted of two integers, both less than or equal to n. Let x be the largest integer less than or equal to n in A. Assume it is in position i, indexed from the left. Then there are n - i integers less than or equal to n to be placed in ascending order – from the right – in B. But this says the smallest integer that could be paired with an integer less than or equal to n is n+ 1. Therefore construction C is always possible.. Clearly Sum/ci - i/ = Sum/ai - bi/ since at most the order of the summands has been altered. If the Proizvolov arrangement had been the trivial one in which no integer was derranged the sum of the integers in any pair would have been 2n + 1 and the difference would have been 2n - i, where i is the index. Sum(2n - i) = n2 in this case. If ci is not 2n - i + 1, interchange ci with 2n - i + 1 in the pair, say (j, 2n - i + 1), in which it occurs. This cannot affect Sum(ci - i ) since the ci by construction are all larger than n, so the change in the value of ci - i is exactly offset by the change in cj - j. Continue in this way interchanging the ci where necessary to cause the ci to all be 2n - i + 1. Since no interchange affects the sum, the final result is the trivial Proizvolov arrangement and Sum/ai - bi/= n2 as was to be shown. 　 　 　 Here is an example to illustrate the procedure. Let the Proizvolov arrangment be: 1 3 7 8 9 11 12 15 16 14 13 10 6 5 4 2 The C arrangement will be 1 2 3 4 5 6 7 8 16 15 14 12 11 9 13 10 The first sum preserving interchange needed is for index 4 when 12 and 13 must be interchanged. At index 5, 11 and 12 must be interchanged etc. __________ Information from ESET Smart Security, version of virus signature database 10315 (20140826) __________ The message was checked by ESET Smart Security. http://www.eset.com