Thanks for the info on the "imaginary" numbers, I really appreciate it.

We may have misinterpreted our view on the formula I had posted, but I'm sure that was my fault due to disorganized and underinforming you guys of the patterns I had noticed when dealing with consecutive/chronological/incremental (by one) of the values of both "N" and "X" (or "Y"). Let me start by introducting the patterns I had discovered:

"N1" = 0, (0 x 0) or (0 squared) = 0
"N2" = 1, (1 x 1) or (1 squared) = 1
The above two numbers, after being squared, have a difference of "1," indicating that the SUM of both "N1" and "N2" before being squared are equal to "1", the figure that is equal to the sum (or difference in this example) of both of the numbers after being squared. (This is the only time this pattern occurs, the exact SUM and DIFFERENCE of the original numbers before squaring being equal to the SUM and DIFFERENCE of the numbers after squaring.)

"N1" = 1, (1 x 1) or (1 squared) = 1
"N2" = 2, (2 x 2) or (2 squared) = 4
The above two numbers, after being squared, have a difference of "3," (4 - 1 = 3, or "N2" - "N1" = 3) indicating that the SUM of both "N1" and "N2" BEFORE being squared are equal to "3," the figure that is equal to the DIFFERENCE of "N2" and "N1" ("N2" - "N1" = 3) after both numbers have been squared. Notice a pattern yet? Here it is: THE DIFFERENCE OF THE TWO CONSECUTIVE NUMBERS SQUARED IS ONLY TWO NUMBERS DIFFERENT (+ 2) COMPARED TO THE PREVIOUS EXAMPLE (which occured in a chronological order between the first and second examples, i.e., example one had "N1" = 0 and "N2" = 1 before (and after) being squared-the only two numbers that will give the exact original number both before and after squaring, THEN example two-before being squared-used "N2" from example one as "N1" in this new example, with "N1" now being equal to 1 and a new constant for "N2," being ("N1" + 1) the number 2. Thier DIFFERENCE after both numbers were squared? 3. This had shown me that the DIFFERENCE between the two consecutive squares was equal to their original sums before squaring. A few more to help you guys understand the underlying pattern here...

"N1" = 2, (2 x 2) or (2 squared) = 4
"N2" = 3, (3 x 3) or (3 squared) = 9
Once again, the two numbers' DIFFERENCE ("N2" - "N1" = 5) after squaring is equal to the SUM of the original numbers before being squared. Can you spot the UNDERLYING pattern yet? Here's a clue: Example 1 = (0 + 1 = 1), Example 2 = (1 + 2 = 3), and Example 3 = (2 + 3 = 5). Here's ONE more example before I reveal the pattern I had mentioned...

"N1" = 3, (3 x 3) or (3 squared) = 9
"N2" = 4, (4 x 4) or (4 squared) = 16
Yet again, the DIFFERENCE of the two numbers AFTER being squared ("N2" - "N1" = 7) is equal to the SUM of the two original numbers BEFORE being squarred (["N1," or 3] + ["N2," or 4] = 7). If you haven't already noticed it, each example continues to prove that the SUM of the original numbers-BEFORE being squared is equal to the DIFFERENCE of the two numbers after being squared. Each and EVERY TIME this occurs with consecutive or chronological rational (whole) numbers, the difference between examples is equal to two (2).

I cannot claim credit for this until I can confirm that it has NOT been mentioned or discovered before; that's why I'm asking you guys-probably the most highly educated group of people on any forum I've found (don't blush-you guys deserve it for all your hard work). That's when I came up with the formula-after discovering this pattern.

0 x 0 = 0 (Reference point)
1 x 1 = 1 (0 + 1 = 1) (1 - 0 = 1)
2 x 2 = 4 (1 + 2 = 3) (4 - 1 = 3)
3 x 3 = 9 (2 + 3 = 5) (9 - 4 = 5)
4 x 4 = 16 (3 + 4 = 7) (16 - 9 = 7)
5 x 5 = 25 (4 + 5 = 9) (25 - 16 = 9)
6 x 6 = 36 (5 + 6 = 11) (36 - 25 = 11)
7 x 7 = 49 (6 + 7 = 13) (49 - 36 = 13)
8 x 8 = 64 (7 + 8 = 15) (64 - 49 = 15)
9 x 9 = 81 (8 + 9 = 17) (81 - 64 = 17)
10 x 10 = 100 (9 + 10 = 19) (100 - 81 = 19)
11 x 11 = 121 (10 + 11 = 21) (121 - 100 = 21)
12 x 12 = 144 (11 + 12 = 23) (144 - 121 = 23)

Look at the numbers on the far right hand side of each parenthetical equation beside the squared numbers above-going down the list, each of the numbers immediately beside the ")" on both equations representing consecutive or chronological numbers continue to increase by two (2) as you read down. Reading up, the numbers in the same location continue to decrease by two (2). This realization told me that a simple equation could be used with the number preceeding or proceeding ANY selected number to determine it's square, based on that information alone...

I'd be thrilled if nobody else had figured this out and had come up with an equation similar to mine or one that provided and prove the same relationship, but I'm almost sure it's been done before-even though I'd never heard of it until I had attempted to learn a method to help me remember larger squares and identify the square root of larger numbers without the use of a calculator...

Does this get the basic point across, RSOCOR01? I'm sure I (or anybody else) could revise and further simplify the formula/equation, but I had just posted what I had (believed to have) discovered at the time. I've got a formula to help found squares for ANY number, no matter how far away, based upon only the knowledge of a single rational number and it's square, and of course the number you need to know or find the square of. I'll try to revise it and simplify it as much as possible before posting it here, but it may be a few days away. I'll get behind in classes if I don't bust my ass within the next 30 hours to finish a pile of ill-defined problems that I have to have completed and turned in Tuesday morning...

BTW-Thanks for everybody's time, help, and cooperation/consideration on this matter as I attempt to research it further in the small window of free time I do have at the present moment...I greatly appreciate your help...

Sincerely,