While watching basketball tonight, I noticed that for 3, 4, and 6, $(6 \times 3) - (4 \times 3) = 18 - 12 = 6$. I thought this was a cool relationship and it led me to the following question:
For some positive integer (greater than 1), x, how many pairs of positive integers (also greater than 1), a and b, satisfy $|(ax - bx)| = a$ or $b$. So for any given x, how many pairs satisfy the equation?
My intuition says it is infinite no matter what x you choose, but I can't think of a way to prove this.
Thanks!
PS: if dropping the absolute value would make a solution easier, add in the assumption that a is greater than or equal to b.
Suppose $x$ is non-negative, else consider $-x$.
Suppose $|ax-bx|=a$, else we can swap over the roles of $a$ and $b$.
Let us vary $b$. It is clear that $a$ is non-negative and is a multiple of $x$. In fact, any positive $a=kx$ can be obtained by setting $b=a-k$.
Verifying, we have $|ax-bx|=|ax-(a-k)x|=|kx|=a$.
So the only case where it is finite is when $x=0$, the only solution is $a=0$ or $b=0$.