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Sums of Squares Differences

IntermediatesJust as the triangular numbers are the partial sums of the sequence { 1 2 3 4 … }, so also the squares are the partial sums of the sequence { 1 3 5 7 … }. In other words, if you know n^{2} for some n, you can obtain (n+1)^{2} simply by adding 2n+1. That's never really worked for me, though … for some reason it's always seemed like too much of a hassle to stop, figure out what n is, and compute 2n+1. If I think of 2n+1 as n + (n+1), though, then it's easy. If I know that 900 is 30^{2}, I can add 30 to get 930, add 31 to get 961, and that's 31^{2}. The intermediate numbers that show up in the process I'm cleverly going to call “intermediates”; here's a little picture of how they work.
In that role alone, intermediates wouldn't have been worth writing about, but then recently I realized that the same numbers play at least four other roles. The first other role is entirely trivial, but I like it. If we imagine that each square n^{2} has associates n^{2} ± n at its left and right hands, forming a group of three numbers, then it's kind of neat how the groups fit together.
The expression n^{2} + n suggests the next role, especially if you factor it as n(n+1): the intermediate that comes after n^{2} is 2T(n), i.e., twice the n^{th} triangular number. You can see the same thing in the picture above, where the middle figure can be broken down into two triangles of size three, and hopefully you can also see it in the diagram below, where the increments are grouped in two different ways.
If instead of factoring the expression n^{2} + n, you complete the square, you get (n+½)^{2}  ¼ … which shows that intermediates are essentially the squares of halfintegers. Those turn up fairly often in math and physics, notably when you're doing angular momentum calculations in quantum mechanics. Another place you can see halfinteger behavior is in the table of squares, in the squares of numbers that end in “5”. The fourth other role of intermediates follows directly from the third. I've talked about the Babylonian multiplication algorithm in two other essays, In Mathematics and Multiplication, and each time I used a different equation to represent it. Now, here's another variant.
ab = [ (a+b) / 2 ]^{2}  [ (ab) / 2 ]^{2} Consider the two numbers a+b and ab. The difference between them is an even number (2b), so the two are either both even or both odd. If they're both even, the product ab is a difference of squares; and if they're both odd, it's a difference of halfinteger squares. But, in the second case, we can write each halfinteger square as an intermediate plus the constant ¼, then the constants cancel, and the product is a difference of intermediates. In practice, it's probably easier to do as I did in Multiplication, and think in terms of the center c and difference d.
(cd)(c+d) = c^{2}  d^{2} Now, though, we know that when the center and difference are halfintegers, we can replace the squares with the corresponding intermediates. If, for example, we want to compute 36 times 43, we just find the center (39½) and difference (3½), work out the intermediates (1560 and 12), and subtract to get 1548.

See AlsoDifferences Multiplication Table, The Repunits @ July (2005) 