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Sun, 17 Jun 2007
Square triangular numbers
I no longer remember how I solved the problem the first time around, but I was tinkering around with it today and came up with an approach that I think is instructive, or at least interesting. We want to find non-negative integers a and b such that ½(a^{2} + a) = b^{2}. Or, equivalently, we want a and b such that √(a^{2} + a) = b√2. Now, √(a^{2} + a) is pretty nearly a + ½. So suppose we could find p and q with a + ½ = b·p/q, and p/q a bit larger than √2. a + ½ is a bit too large to be what we want on the left, but p/q is a bit larger than what we want on the right too. Perhaps the fudging on both sides would match up, and we would get √(a^{2} + a) = b√2 anyway. If this magic were somehow to occur, then a and b would be the numbers we wanted. Finding p/q that is a shade over √2 is a well-studied problem, and one of the things I have in my toolbox, because it seems to come up over and over in the solution of other problems, such as this one. It has interesting connections to several other parts of mathematics, and I have written about it here before. The theoretical part of finding p/q close to √2 is some thing about continued fractions that I don't want to get into today. But the practical part is very simple. The following recurrence generates all the best rational approximations to √2; the farther you carry it, the better the approximation:
Now, we want a + ½ = b·p/q, or equivalently (2a + 1)/2b = p/q. This means we can restrict our attention to the rows of the table that have q even. This is a good thing, because we need p/q a bit larger than √2, and those are precisely the rows with even q. The rows that have q odd have p/q a bit smaller than √2, which is not what we need. So everything is falling into place. Let's throw away the rows with q odd, put a = (p - 1)/2 and b = q/2, and see what we get:
I have two points to make about this. One is that I have complained in the past about mathematical pedagogy, how the convention is to come up with some magic-seeming guess ahead of time, as when pulling a rabbit from a hat, and then at the end it is revealed to be the right choice, but what really happened was that the author worked out the whole thing, then saw at the end what he would need at the beginning to make it all work, and went back and filled in the details. That is not what happened here. My apparent luck was real luck. I really didn't know how it was going to come out. I was really just exploring, trying to see if I could get some insight into the answer without necessarily getting all the way there; I thought I might need to go back and do a more careful analysis of the fudge factors, or something. But sometimes when you go exploring you stumble on the destination by accident, and that is what happened this time. The other point I want to make is that I've written before about how a mixture of equal parts of numerical sloppiness and algebraic tinkering, with a dash of canned theory, can produce useful results, in a sort of alchemical transmutation that turns base metals into gold, or at least silver. Here we see it happen again.
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