JCSBimp's Hexagonal Orthogonal

I’ve updated the testing “scratchpad” for the first example I’ve been working, and have added a second one. I’ll add a third one if I need more help in teasing a general algorithm out of all of this. That’s entirely possible.

First, for m=1501, with more of the process written in:

M = 1501 = 39^2 - 20, remainder between 4^2 and 5^2,
  giving Ceiling Root C0=39 and putting M between
  A0 = 1496 = 39^2 - 5^2 and
  B0 = 1505 = 39^2 - 4^2.
  M = A0 + 5 = B0 - 4.
  B0 - A0 = 9.
  Setting A0 as origin, perfect square marks at 0 9 16 ...
  Row polynomial is x(10-x).
  Now calculate 2 or 3 "higher remainders" as follows: 
  (C0+1)^2-M = 1600-1501 = 99, between 9^2 and 10^2.
  A1 = 1500 = 40^2 - 10^2.
  B1 = 1519 = 40^2 - 9^2.
  M = A1 + 1 = B1 - 18.
  B1 - A1 = 19.
  Setting A1 as origin, perfect square marks at 0 19 36 ...
  Row polynomial is x(20-x).
  (C0+2)^2-M = 1681-1501 = 181, between 13^2 and 14^2.
  If we choose these two perfect squares, A2 = 1485 = 41^2 - 14^2.
  However, we can choose instead A2 = 1456 for uniformity with A0 and A1.
  A2 = 1456 = 41^2 - 15^2.
  B2 = 1485 = 41^2 - 14^2.
  (Currently doubting how much the B values matter aside from B-A span.)
  M = A2 + 45 = B2 + 16.
  B2 - A2 = 29.
  Setting A2 as origin, perfect square marks at 0 29 56 ...
  Row polynomial is x(30-x).
  (C0+3)^2-M = 1764-1501 = 263, between 16^2 and 17^2.
  But if we follow the linearly-increasing pattern already established,
  A3 = 1364 = 42^2 - 20^2.
  B3 = 1403 = 42^2 - 19^2.
  M = A3 + 137 = B3 + 98.
  B3 - A3 = 39.
  Setting A3 as origin, perfect square marks at 0 39 76 ...
  Verified that the differences between A's and M constitute an integer
  sequence corresponding to the values of quadratic polynomial:
  F(X) = 24X^2-28X+5, where X is the integer added to C0.
  Discriminant of F(x) is 304=2^4*19.

  Searching for the positive integer value of X so that F(X) is of the form
  y(10+10X-y), for some positive integer y.

And then, for m=172451, with a tad more hand-waving:

M = 172451 = 416^2 - 605, putting it between
  A0 = 172431 = 416^2 - 25^2 and
  B0 = 172480 = 416^2 - 24^2.
  C0 = 416.
  M = A0 + 20 = B0 - 29.
  Perfect square marks at 0 49 96 ...
  Row polynomial is x(50-x).
  (C0+1)^2-M = 1438, between 37^2 and 38^2.
  A1 = 172445 = 417^2 - 38^2.
  B1 = 172520 = 417^2 - 37^2.
  M = A1 + 6 = B1 - 69.
  Perfect square marks at 0 75 148 ...
  Row polynomial is x(76-x).
  Now we progress linearly rather than using C0+2.
  A2 = 172123 = 418^2 - 51^2.
  B2 = 172224 = 418^2 - 50^2.
  M = A2 + 328 = B2 + 227.
  Perfect square marks at 0 101 200 ...
  Row polynomial is x(102-x).
  
  F(X) = 168X^2-182X+6.
  Discriminant is 29092 = 2^2*7*1039.

  Searching for the positive integer value of X so that F(X) is of the form 
  y(50+26X-y), for some positive integer y.