## Monday, 23 January 2012

### Composite numbers seen as the difference between two squares Part 4

One quick point I should maybe emphasize. I explained already that the difference between two squares must be a composite number. What I didn't show was that any composite number in the form 6X+/-1 can necessarily be expressed as the difference between two squares (of integers that are separated by more than 1). This is the basis of Fermat factorisation, so well known, but just to run through the logic...

This is because the factors of a composite number of the form 6X+/-1 must themselves be numbers of the form 6X+/-1. And regardless of how many factors a number has it will be possible to express the factors as AB where A and B are numbers in the form 6X+/-1 (and A > B).

Let N = (A - B)/2 + B = A/2 + B/2
Let M = N - B = (A/2 - B/2)

Since A and B are odd numbers, N and M are both positive integers.

N^2 - M^2 = (N+M)(N-M) = (A/2 + B/2 + A/2 - B/2)(A/2 + B/2 - A/2 + B/2) = AB

So unless N - M = 1, and AB is odd, then AB is a composite number.

Thus the chart we started with, with an array of the differences between all square numbers, necessarily contains all composite numbers of  the form 6X+/-1 (if extended infinitely). What we are doing after that is gradually eliminating the parts of that array that can't contain numbers adjacent to 3 x 2^n as the value of n increases.

Part Five