Let $a$, $b$ and $c$ be positive integers such that $\gcd(a,b)=1$ and $a$ and $b$ do not divide $c$. Prove that if $ab − a − b < c < ab$ then the equation $ax+by=c$ has only one positive solution $x,y\in\mathbb{ Z}$.
I know that it has infinitely many solutions, but I don't know how to prove that it only has one positive solution.
Existence: let $(x,y)\in\mathbb N\times\mathbb Z$ be such that $ax+by=c$ and $x$ minimal. Then, $x-b<0$ hence $by=c-ax\ge c-a(b-1)>-b$ hence $y\ge0.$
Uniqueness: if there were two positive solutions, there would exist two solutions of the form $(x,y)\in\mathbb N^2$ and $(x-b,y+a)\in\mathbb N^2.$ But then $c=ax+by\ge ab+b0$ would contradict the assumption $c<ab.$
