Well, I was right about one thing – there was a straightforward solution to this whole Aztec diamond problem. To be fair to myself, my original solution holds up – I neglected to add one somewhere in my equation (we’ll get to that later), an error that was insignificant at the start, and less significant as the Aztec numbers increased. To get to the reveal, it’s worth backing up to our series, OEIS: A046092 again. Somehow, in my haste, I kind of glossed over the fact that these are ‘4 times triangular numbers,’ a fact that became readily apparent to me when I was coming up with the diagrams for the last post. You see…
…our Aztec diamond is made up of four triangles; and it is in fact true that each of our Aztec numbers is 4× the corresponding triangle number. A funny thing about triangle numbers is that if you multiply them by eight and add one, they become perfect squares. This can be demonstrated visually:
This visual only proves that it’s true for the triangle number, 6, but it is universally true and readily proven – this ‘Cool Math Stuff’ post shows it succinctly, and John Conway and Richard Guy discuss it in The Book of Numbers. Rehashing the proof here seems rather pointless. Fascinatingly, I did almost figure this out last time, with the extra unit hypotenuse theory.
So, we know that for our Aztec number, x, x/4 is a triangle number, and for this triangle number, y, 8y+1 is a perfect square. We know that any given side of this square is made up of 2× the length of a side of triangle y plus one, which is ultimately the value that we need to recreate our square from our grid. We can see this rather clearly in the first diagram with the triangle highlighted – the three dots forming the outer side correspond directly to segments of our grid.
Thus, given a triangle number, y, the length of any of its sides is (sqrt(8y+1)-1)/2. Which then leads us to the same thing for our Aztec number, x, (sqrt(2x+1)-1)/2. Now, to solve my problem, I actually need to add one to this. Given that we’re dealing with integers, this can be simplified to ceil(sqrt(2x+1)/2) – precisely what I originally came up with, aside from the +1.
So, my equation was wrong, but it provably works – the off-by-one error is clearly insignificant for a 3×3 square, and, given x, sqrt(x2)-sqrt(x2-1) converges toward zero: