>>4594

For those who like algebra, we can show that this formula evaluates to 0 for correct inputs:

(x+n)^2 - (f-1) - 2d(n-1) - n^2 = 0

(x+n)^2 - f + 1 - 2dn + 2d - n^2 = 0

-f + 1 + 2d + d^2 = d^2 + 2dn + n^2 - (x+n)^2

(d+1)^2 - f = (d+n)^2 - (x+n)^2

c = (d+n)^2 - (x+n)^2

We end up with the original formula that defines what n and x are.

>>4707

I believe that's intentional… the matching colors show how different parts of the large square add up to form the small square, with c being left over as the difference between them.

I don't think the red area is equal to x^2, though.

>>4709

Nah, no games, just a regular anon.

I see the label but suspect it is a mistake. The sides of the overall square are x+n, so x^2 should only be the upper left quadrant, diagonally opposite from the n^2 section.

>>4772

May want to triple-check that, since it should be exactly the same as the similar equation using 'e'. Can you post an example where it fails?

>>4774

Well, for example take 39 = 3 * 13 = 8^2 - 5^2. So d=6, e=3, f=10, n=2, x=3.

The formula says:

(x+n)^2 = n^2 + 2d(n-1) + f - 1

25 = 4 + 12 * 1 + 10 - 1

25 = 25

Could you try that one in your code, or any other specific example we can verify by hand?

>>4776

Great, thanks for double-checking!