solving a linear Diophantine equation

Question

How to show that there exist non-negative integers $x,y$ such that

$ax+by=ab+k$ where $a,b$ are co-prime whole numbers is true for all non zero integers $k$.

PS: Sorry for missing the key information that $a$ and $b$ are coprimes...

Answer 1

This is not true in general, e.g $2x+2y=4+1$. It is true only if $k$ is a multiple of the $\gcd(a,b)$.

Answer 2

This is false as stated, since $\rm\,a,b,x,y\in \Bbb N\Rightarrow ax+by \ge 0,\,$ but $\rm\,ab+k < 0\, $ if $\rm\,k < -ab.\:$ Probably you meant: if $\rm\,a,b\in\Bbb N$ are coprime then $\rm\:ax+by = n > ab\!-\!a\!-\!b\,$ has a solution $\rm\,x,y\in \mathbb N.\:$

Indeed, by Bezout there are $\rm\,j,k \in \Bbb Z\:$ with $\rm\,aj+bk = 1,\:$ so scaling by $\rm\,n\,$ there are $\rm\,x,y \in \Bbb Z\,$ with $\rm\,ax+by = n.\:$ Since $\rm\,ax+by = a(x\pm b) + b(y\mp a)\,$ we can repeatedly add/subtract $\rm\,b\,$ from $\rm\,x\,$ till we get a solution with $\rm\,0\le x \le b\!-\!1.\:$ Then, necessarily, $\rm\,y\ge 0,\:$ since otherwise $\rm\:y\le -1\:$ so then $\rm\:n = ax+by \le a(b\!-\!1)+b(-1) \le ab-a-b,\:$ contra hypothesis. Thus $\rm\,x,y\ge 0,\,$ as claimed.

For a more geometrical viewpoint see here, whihc also includes links to the literature (Frobenius and postage stamp or coin chnaging problems).

Answer 3

Hints:

• Two integers $a,b$ are coprime iff $ac+bd=1$ for some $c,d\in\mathbb Z$ (see this).
• For coprime $a,b$ and any $n\in\mathbb Z$ the equation $ax+by=n$ has a solution ($x=cn$...).
• If $ax+by=n$ then $\forall z\in\mathbb N, \ \ \ a(x-bz)+b(y+az)=a(x+bz)+b(y-az)=n.$
• If $ax+by=n$ and $x>0,y<0$ (the same when $x<0,y>0$) then $\exists z\in\mathbb N:$ $x-bz>0, \ y+az>0 \iff \exists z\in\mathbb N: -\dfrac{y}{a}<z<\dfrac xb \iff \dfrac xb + \dfrac{y}{a}>1 \iff ...n> ab.$