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For example, the analogous two dimensional question, x + y = 2n is easy to solve: one can simply consider pairs (i,2n-i) for i=1,2,...,n and thus index every solution, exactly once. We note that we have n such pairs solving x + y = 2n, for every fixed value of positive integer n, and so the cardinality of such a set is equal to n as expected.

However, trying to repeat the same problem for x + y + z = 2n, it is not clear to me how (or if it is possible) to write down a minimal set {(2n-i-j,i,j)} such that varying i and j over particular intervals precisely produces every such triplet, exactly once. It can be shown that the number of elements in such a minimal set would be equal to the nearest integer to n^2/3.

It is not hard to see how one can obtain such an indexing with repetitions, or how one can algorithmically remove repetitions, but what I would like to know is whether there is a clean, general construction, as for the x + y = 2n case. Is this possible, or will one always have to artificially restrict certain values of the parameters on the intervals for which they are defined?

Noam
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  • Iterating through all pairs with i<=j doesn't do it? Aside, for the 2d case, it's (i,2n-i) right? – Argyll Jul 14 '22 at 14:13
  • Thanks- the 2d case had a typo indeed. The iteration you suggested does not work sadly because you will have repetitions due to 2n-i-j eventually going below n (assuming you meant letting 1<=i<=n and 1<=j<=i) – Noam Jul 14 '22 at 14:15
  • I mean 1<=i<=j<=2n-i-j. The 3 numbers always have an inherent ordering anyway – Argyll Jul 14 '22 at 15:47

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