JCSBimp's Hexagonal Orthogonal

These are from studies ancillary to my integer factorization explorations.

Let m = 2n be an even positive integer corresponding to positive integer n, which may or may not be even.

Let s be the Ceiling Root of m, the integer ceiling of m’s square root. s squared is the Ceiling Square of m, the least perfect square greater than m.

Construct a Series of Ceiling Squares of m as follows:

The first element of the Series will be s, the Ceiling Square of m.

s^2 – m will equal some remainder r. If m is a perfect square, r will be 0, and the Series of Ceiling Squares for m will contain only s, with the possibility of 0 added on as an element for completeness’ sake.

If r is not 0, find its Ceiling Square and remainder as above. Proceed like this with successive remainders until the remainder is 0.

Consider the set of the Ceiling Square of m and the successive remainders’ ceiling squares as the Series of Ceiling Squares.

There is an anomaly that occurs if this is done straightforwardly according to only the rules above, and it deserves consideration: If m = 2 (n = 1), or if any subsequent remainder is 2, the Series of Ceiling Squares becomes infinite, since 2 = 2^2 – 2, = 2^2 – (2^2 – 2) and so on, In fact, there is no way to express 2 as the difference of two perfect square integers. This latter observation is true for all integers congruent to 2 modulo 4, since all even perfect squares are congruent to 0 modulo 4 and all odd perfect squares are congruent to 1 modulo 4. Notice that the integers congruent to 2 modulo 4 are exactly those values of 2n where n is odd.

For even m greater than 2 (n greater than 1), it appears so far that there always exists a way to express the Series of Ceiling Squares of m as a finite series. One can consider the first several examples, with possibilities for their Series:

2 = 3^2 – ( 4^2 – 3^2 ). The sequence 3, 4, 3, a not strictly decreasing sequence, terminates.
4 = 2^2 – 0^2.
6 = 3^2 – ( 2^2 – 1^2 ).
8 = 3^2 – 1^2.
10 = 5^2 – ( 8^2 – 7^2 ).
12 = 4^2 – 2^2.
14 = 5^2 – ( 6^2 – 5^2 ).
16 = 5^2 – 3^2.
18 = 5^2 – ( 4^2 – 3^2 ).
20 = 6^2 – 4^2.
22 = 5^2 – ( 2^2 – 1^2 ).

For values of m which are multiples of 4, to find s and t such that s+t=m/2 and s-t=2: let s=m/4+1, and t=m/4-1.

When m=2n for odd n, adjust the Ceiling Square of m to be the smallest odd perfect square greater than m. Then the remainder r will be odd, and will have an expression as the difference of integer squares [(r+1)/2]^2 – [(r-1)/2]^2, giving m an expression in terms of three squares as in the examples above.