Why is $xy-x-y$ not purchasable with coins of denominations $x$ and $y$? As asked here:

For integers $n, x,y > 0$, where $x,y$ are relatively prime, 
      every $n \ge (x-1) (y-1)$ can be expressed as $xa + yb$, with nonnegative integers $a,b \ge0$.

And a proof by Hagen von Eitzen conclude that for $n \ge (x-1) (y-1)$ there are $a,b \ge0$ such that $xa + yb = n$ hold. 
Trying for several numerical example I found out that the lower bound $n_0=(x-1) (y-1)$ is the least upper bound, i.e. there is no $a,b \ge0$ such that $xa + yb = (x-1) (y-1)-1$ hold. But I can't prove it. using the same method by Hagen von Eitzen's answer I came to the conclusion that $b\ge-1$ which is not a contradiction since if $b\ge-1$ then $b$ can be any positive number as well. If I had concluded that $b=-1$ or $b\le-1$ or $b \in B$ for any $B \subset \mathbb{Z^-}$ that would be something. How there is no $a,b \ge0$ such that $xa + yb = (x-1) (y-1)-1$ hold?     
 A: If $xy-x-y=ax+by$, then $x(y-a-1)=(b+1)y$, hence $x\mid(b+1)y$.  Since $x$ and $y$ are relatively prime, we must have $x\mid b+1$, i.e., $b+1=kx$ for some integer $k$.  Assuming $b\ge0$ and using the fact that $x\gt0$, we see that $k\ge1$, and thus $b=kx-1\ge x-1$.  Using the fact that $y\gt0$ as well, we find that
$$xy-x-y=ax+by\ge ax+(x-1)y$$
from which we conclude $-1\ge a$.  Thus $xy-x-y=ax+by$ has no solutions with $a$ and $b$ both non-negative.
A: Suppose $ax+by=xy-x-y.\;$ Then $$ ax+by=xy-x-y\implies x|(xy-x-ax)=(1+b)y\implies x|(1+b)\implies 1+b\geq x.$$  $$\text {and }\quad by\leq ax+by=xy-x-y<(x-1)y\implies b<x-1\implies 1+b<x.$$  
A: If you let $a=y-1$ and $b=-1$, then $xa+yb=xy-x-y$.
Since $x$ and $y$ are relatively prime, then any integer solution must be of the form $a=y-1-yt$ and $b=-1+xt$.
If there exists a $t$ such that both $a$ and $b$ are positive, then
$a \ge 0 \implies  t \le 1 -\dfrac 1y \implies t \le 0$
$b \ge 0 \implies t \ge \dfrac 1x \implies t \ge 1$
Since this can't happen, there is no $a \ge 0$ and $b \ge 0$ such that $ax + by = xy-x-y$.
A: HINT:
$$a x + b y = x y - x - y \Leftrightarrow (a+1) x + (b+1) y = x y $$
Now, if $\alpha x + \beta y = x y$, then $\alpha x$ is divisible by $y$, so $\alpha$ is ( since $(x,y) = 1$). It should be easy now.
