Novel Multiplication and Division by odd squares

[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?

 

[Number Nine]

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

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.

 

[Norwegian]

Is there a simple way to prove this general result?

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\circx=nx+k(x-1). Then (n\circx)\circy=n\circ(xy).

The proof of this is trivial, and your result is the special case of k=-1/8, and where nXa=n\circa².