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Jan 7, 2016 at 9:14 comment added Pietro Majer But those two sums are not equal in general (The answer was just meant to show that the problem is somehow trivial, even if you ask that all $x_i$'s be distinct positive integers)
Jan 7, 2016 at 3:32 comment added JMP let $\sum_{i=1}^{p-1}x_i^2=\sum_{i=p+1}^{p+q-1}x_i^2=X$, then the brackets are trivial, as we have $(X-X\pm1)$
Jan 7, 2016 at 1:06 history answered Pietro Majer CC BY-SA 3.0