I don't know how to post in Math Fun, but maybe this will reach someone who does. The identity appears trivial to me. It is certainly trivial if the one set consists of the integers 1 to n and the other consists of the integers n+1 to 2n. The n differences will be (n + i) - i , i = 1 to n and the sum is just n x n. Call this trivial case C. Given any partition of the 2n integers arranged as Proizvolov specifies -- call this case P. Start with 1 in whichever set it occurs in P. It is paired with an integer, say x. In C x was matched with x + n mod 2n. Continue the sequence (1, x, (x + n mod 2n) --). In P n mod 2n is matched with an integer y. Continue the sequence (1,x, y --). y is matched with z = y + n mod 2n in C etc. Continue in this way until 1 appears again. Either the sequence spans all 2n integers, or it closes. If it spans all of the integers, the entries can be grouped by pairs with the first and last entries paired to give a sum of n^2 -- since that is just the C sum, The other grouping of paired differences is the P sum which must be the same. If the extension does not span all of the integers, start a new sequence with the first unused integer etc. Consider the case n = 8 C is 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Let P be (for example) 1 2 4 7 10 13 14 16 15 12 11 9 8 6 5 3 The sequences will be (1-15-7-9) (2-12-4-11-3-16-8-10) (5-14-6-13) and the sum will be 2x8 + 4x8 + 2x8 = 64. The actual P differences in the three cycles are (14, 2), (10, 7, 13, 2) and (9, 7) Gus Simmons __________ Information from ESET Smart Security, version of virus signature database 10304 (20140823) __________ The message was checked by ESET Smart Security. http://www.eset.com