Do I understand right? That this is based on this formula for quadratics:

(n+1)^2=n^2+2n+1; let m=n+1

m^2-n^2-1=2n

Thus consecutive squares are always 2n+1 appart.

That formule you mentioned is equivilent to this code:

Code:

Let x = 0
For i = 1 to n
x = x + i + i - 1
Next i
REM Returns in x, the square of n using recursive addition instead of multiplication.

The optimizes trivially without multiplication to:

Code:

Let x=0
For i = 1 to n
x = x + i + i
Next i
x = x - n

You can further remove the For...Next loop using an identity but you get this:

Let x = 2 * n * (n + 1) / 2 - n

Or:

Code:

Let x = n * (n + 1) - n

Or:

Code:

Let x = n^2 + n - n

As you can see, this leads you back to your original equation of x=n^2.

EDIT:

You should note that for m=n+1:

m^3 - n^3 - 1 =

(n+1)^3 - n^3 - 1 =

n^3 + 3n^2 + 3n + 1 - n^3 - 1 =

3(n^2 + n) =

3n(n+1)