"Michael J Hardy" <firstname.lastname@example.org> wrote in message news:email@example.com... > Darrell (firstname.lastname@example.org) wrote: > >> For argument sake I will assume a!=b. Since the smallest positive >> integer >> is 1, you can always make a-bk=1 by letting |a-b|=1 and |k|=1. IOW, >> a=b+1 >> or b=a+1 with k=+/- 1 >> accordingly. > > > That is nonsense. You can't just "let |a - b| = 1. > a and b are given. The answer depends on them. > > E.g., if a = 20 and b = 3, then a - bk = 20 - 3k. In that case, > the integers of the form a - bk are those of the form 20 - 3k, > namely 20, 17, 14, 11, 8, 5, 2, -1, -4, ... etc. (and also 23, 26, > 29,...). The smallest positive integer among those is 2, and the > value of k that achieves it is 6. > > The answer depends on a and b, so you've got to say what function > of a and b it is (I've given the answer in only one case). -- Mike Hardy > >
Perhaps you should read the rest of the thread where I already retracted my response.