In a diophantine equation $ax + by = n$ with $(a, b) = 1$, the greatest possible value of $n$ such that both $(x, y)$ are not positive is $ab − b − a$?
This is given in my module (without any proof). I am assuming that "both $(x, y)$ are not positive" means at-least one must be negative. I was wondering how to prove this.
First note that if $x=b-1$ and $y=-1$ then $ax+by=ab-b-a$.
Then what you have to know is that if $(u,v)$ is one solution to $ax+by=n$ then the full set of solutions is given by $x=u+kb$, $y=v-ka$ where $k$ runs through the integers (positive and negative (and zero)).
So to increase $y$, you have to decrease $x$ by $b$ (or more), so you can't get a positive solution when $n=ab-b-a$.
By the way, I'm assuming that $a,b$ are meant to be positive integers.