JCSBimp's Hexagonal Orthogonal

When expressing a number in terms of its Ceiling Square and Remainder, some time ago I decided it would be good to come up with a systematic technique that would work with even integers as well as odd. I have what might be a refinement of the part of that technique that deals with the multiples of 4. This is setting up for some other ideas I’m working on tonight so that, when and if I post about them, I will have a systematic approach to describe, across the board.

When describing an odd integer m in terms of the square of its Ceiling Root and the Remainder (see previous posts for definition/discussion of these concepts), I have chosen the Ceiling Root c to be ceil(sqrt(m)) or ceil(sqrt(m))+1 according to which of those will make c congruent modulo 2 to (m+1)/2 modulo 2.

The idea I am playing with that will serve as the subject of a possible future post – I apologize for teasing – now has me looking at determining the ceiling square for an even integer value of m. The context in which I am pondering this has the additional implication that the m in question is divisible by 4, or, equivalently, is congruent to 0 modulo 4. I want to make a choice between ceil(sqrt(m)) and ceil(sqrt(m))+1 for a Ceiling Root/Remainder expression of even-valued m that meshes nicely, in my mind at least, with the Ceiling Root calculation for odd-valued m.

The reason that I am looking at even values of m is that I wish to take a Ceiling Root/Remainder pair associated with an odd integer, and find a sensible Ceiling Root value, not of the integer itself, but of its own Remainder, which could well be even. However, due to the choices I have made above in choosing the original ceiling root, the even number Remainder will always be congruent to 0 modulo 4. I assure the reader, this will all be more understandable when/if it bears fruit in the Bigger Revelation Post I keep teasing you about.

So we have it that even m will always, in my application of this method anyway, be a multiple of 4. I have found that the multiples of 4 can all be expressed uniquely as differences of perfect squares whose square roots differ by 2. The perfect squares whose difference is 4n for some positive integer n are (n+1)² and (n-1)². They are either both even or both odd, and so it is neatly consistent to choose the Ceiling Root as being even or odd to match.

That’s the brief insight. Hang on for the bigger one! It is – I hope – coming soon. (The only reason it might not come is if the big idea I have comes to nought. But I think it will still be worth a blog post anyway, as a lesson-learned at the very least.)