# Novel Multiplication and Division by odd squares

1. Dec 1, 2012

### ramsey2879

let $$A_i$$ be an odd integer, $$s_i$$ be the square of $$a_i$$ and $$t_i$$ be the triangular number, $$(s_i -1)/8$$. Same for $$a_j , s_j, t_j, etc$$. Define Multiplication of $$n X A_i , etc$$ to be n * s_i - t_j and division to be the reverse of this process. I found that

n X A_i X A_j X A_k = n X A_k X A_j X A_i = n X A_j X A_k X A_i etc.

for instance ((((4 * 9 - 1)*49 - 6)*25 -3) + 1) / 9 = (4*25-3)*49-6 = (4*49-6) * 25 - 3 = B

8*4-1 = 31 and 8*B - 1 = 31*25*49

Is there a simple way to prove this general result?

2. Dec 1, 2012

### Number Nine

I can barely make this out, since you didn't format it.
If I'm reading it right, this just follows from the commutativity of multiplication.

3. Dec 1, 2012

### Norwegian

Yes, removing all unnecessary notation, the result your are looking for is the following:

Let k be a real number. For real numbers n and x, define n$\circ$x=nx+k(x-1). Then (n$\circ$x)$\circ$y=n$\circ$(xy).

The proof of this is trivial, and your result is the special case of k=-1/8, and where nXa=n$\circ$a2.