06-26-2013, 06:42 AM
(06-26-2013, 06:18 AM)AFtExploision Wrote: So, I have found that:
n^2 = (n-1)^2 + 2(n-1) + 1
(n+1)^2 = n^2 + 2n +1
is this essentially what you are saying?
Sort of, my theory relies on you knowing a square and using the known square to calculate an unknown square of relative position on the number line. What you show is kind of just a breakdown of how to solve a square. I figured out my theory really quite simply, so if here's the thought process. 1^2 = 1, 2^2 = 4, 4-1 = 3. 3^2 = 9, 9-4 = 5. 4^2 = 16, 16-9 = 7. 7-5 = 2. 5-3 = 2. So, starting after the difference between 1 squared and 2 squared, the difference increases by 2 each time, so if you take 2 times the lesser number subtract 1, then you get the difference, but 3 off, so add 3 to the end of that and you get the difference between the two squares. Pretty much it was just a random coincidence that I found the pattern. Yours works too, mine is just easier assuming you know the square previous to it, or one relatively close.