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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?

darij grinberg
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  • I'm not sure I understand what the question is. Are you asking, given $x,y\in \mathbb Z$ why is it not possible to find integers $a,b$ with $xa+yb=(x-1)(y-1)-1$? But of course it is possible. Take $(x,y)=(2,3)$. then the right hand is $1$ so all we want are integers $a,b$ with $2a+3b=1$, just take $(a,b)=(-1,1)$. Or have I misunderstood? – lulu Aug 06 '17 at 20:14
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    @lulu $a=-1$ is not $\geq0$. – Marja Aug 06 '17 at 20:15
  • @Marja Ah, I missed that condition, thank you. – lulu Aug 06 '17 at 20:15
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    Assume you do have a solution for $ax+by=xy-x-y$. Then $ax+(b+1)y=x(y-1)$. Therefore $x$ divides $b+1$. Let $b=xk-1$ for $k\geq1$ Therefore, $(a+k-1)x=-x$. This is a contradiction since the left hand side is positive and the right had side is not. – Marja Aug 06 '17 at 20:23
  • @Marja, where does $(a+k-1)x=-x$ come from? –  Aug 06 '17 at 20:33
  • I highly recommend that you learn the geometric viewpoint sketched in my answer in the linked thread, since it makes these matters quite obvious. – Bill Dubuque Aug 06 '17 at 20:37
  • There is a missing $y$. Replace $b$ by $xk-1$ in $ax+(b+1)y=x(y-1)$. We get $ax+xky=x(y-1)$. Or $ax+x(k-1)y +xy=xy-x$. From there we get $x(a+(k-1)y)=-x$. – Marja Aug 06 '17 at 20:38

4 Answers4

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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.

Barry Cipra
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1

Suppose $ax+by=xy-x-y.\;$ Then $$ ax+by=xy-x-y\implies x|(xy-x-ax)=(1+b)y\implies x|(1+b)\implies 1+b\geq x.$$ $$\text {and }\quad by\leq ax+by=xy-x-y<(x-1)y\implies b<x-1\implies 1+b<x.$$

DanielWainfleet
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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 $ax + by = xy-x-y$.

Steven Alexis Gregory
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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.

orangeskid
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