JCSBimp's Hexagonal Orthogonal

Now consider an arbitrary odd positive integer m, its trivial factorization as 1*m, and its corresponding expression as the difference of two squares [(m+1)/2]²-[(m-1)/2]². Since m is odd, either the minuend (the first term) is even and the subtrahend (the second term) is odd, or vice versa, as follows: When m is congruent to 1 modulo 4, (m+1)/2 is odd and (m-1)/2 is even, with the opposite being the case when m is congruent to 3 modulo 4. This will, it turns out, be the case for each minuend and subtrahend corresponding to each distinct factorization of m. The larger number will always be even or odd, and the smaller number will always be odd or even, respectively. The larger number may or may not have the same value modulo 2 as the ceiling root of m; their even/odd natures are essentially independent of each other.. But it does give an indication of whether m can be expressed as the difference between its ceiling square and a lesser perfect square. Since calculating the integer ceiling of the square root of m is fairly straightforward, so in this case is a non-trivial factorization of m, if m is greater than 2 or 3 as stated earlier.

Additionally, the perfect square minuends of all expressions of m as the difference of two squares will each be greater than or equal to the ceiling square of m. As just stated, when the ceiling square of m is even and (m+1)/2 is odd, or vice versa, there will be no expression of m as the difference between the ceiling square of m and some other perfect square. The minuend of any such square-difference expression of m will then be greater than m’s ceiling square.