Uni maths: divisibility

Let a + b = xd , a - b = yd , were d = (a + b , a - b).

We have

a = (x + y) d / 2
b = (x - y) d / 2

Note that x + y and x - y are both odd or even , so (x ± y)/2 are both integers or not.

When (x ± y)/2 are integers both , d must be integer 1 since (a , b) = 1.

When (x ± y)/2 are not integers both , d/2 must be integer 1 since (a , b) = 1 , i.e. d = 2

Thus , d = (a + b , a - b) = 1 or 2 .

This is trivial.

If d|(a+b) and d|(a-b), then d|2a and 2b.

So, (a+b, a-b) | 2．(a, b) = 2 (given that (a, b) = 1).

Hence, (a+b, a-b) = 1 or 2.