Take a look at multiplication by adding vectors. What I am saying is that F(a,b) = (a+b)*(a+b+1)/2 -a is a function in two variables both integers having the property that a set of vectors (X,Y) can be defined such that F(x1,y1) - F(x2,y2) = x1+y1 -x2-y2 for any pair of two vextors in (X,Y). It can be proven that if (c,d) is an arbitary vector and (xn,yn) is in the set (X,Y) then(adsbygoogle = window.adsbygoogle || []).push({});

F(xn +c,yn +d) = (c+d+1)*F(xn,yn) + K for all vectors (xn,yn) that are in (X.Y). K is a constant that depends upon c and d.

As an example one can define (X,Y) to be all (X,Y) such that F(X,Y) = X + Y. In such a case c = 15, d = -10 is one vector such that c+d +1 = 6 and K = 0.

Corresponding vectors for every integer n are (T(n-1),-T(n-2)) where T(n) = n(n+1)/2 such that K is 0 and the multiplication is by n.

(X,Y) can also be defined by the relation F(x1,y1) - F(x2,y2) = x2+y2 -x1,x2. In this case the multiplyer is (1-c-d) instead of (c+d+1)

Futheremore, other sets can be determined by the relation (X',Y') = (X+j,Y-j) which is simply a diagonal shift of the set of vectors (X,Y) The relations between c,d and K change upon such a shift, but the value of the multiplyer does not.

The proof of all this is based upon the identity T(a+b) = T(a) + T(b) + ab

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# Using vectors to represent integers

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