AA, this is the best I summary I have from my work on the (f-1)/8 method:
How to find the two adjacent Triangle numbers: Take the SQRT of (f-1)/8. This will be the larger of the two Triangle numbers we need, let's call it T2. The smaller staircase number, T1, is simply the T1- 1. Here's an example. For c6107, (f-1)/8 = 16. SQRT 16 = 4. So our first staircase number T2 is 4, and T1 is 3. T2 + T1 = 10 + 6 = 16.
My question is what to do with the remainders. We have remainder from (f-1) mod 8, and sometimes we also have an additional remainder from SQRT ((f-1)/8). So we end up with our two staircase numbers, and at least one remainder to deal with, sometimes two remainders.
The idea is that those two triangle numbers multiply, and that multiples of them will equal our correct (x+n) square. The cool thing about this is that as you work outward from the center of the x+n square, the pattern grows in a very orderly fashion.
I could use some help trying to figure out how to handle the two sets of remainders, but VQC said that we had come VERY close to completing the pattern, which would then lead to the grid shortcut.