# $ax+by=c$ has unique solution for positive integers

Let $$a$$, $$b$$ be positive, relative prime integers. Let $$c$$ be another positive integer such that $$a \nmid c$$, $$b \nmid c$$ and $$ab − a − b < c < ab$$. We want to show that $$ax+by=c$$ has unique solution for positive integers. My idea was that, given that $$a$$ and $$b$$ are coprimes, the solutions must be of the form $$x=cx_0+bt$$, $$y=cy_0-at$$, where $$(x_0, y_0)$$ is a solution for $$ax+by=1$$ and $$t$$ is any integer. If we set the condition that $$x$$ and $$y$$ must be nonnegative, we bound the possible values of $$t$$ in $$[-cx_0/b, cy_0/a]$$. From there on I've tried in different ways to see that this interval is narrower than $$2$$ using both $$ax_0+by_0=1$$ and $$ab − a − b < c < ab$$, but I don't really succeed.

First you need to show that a solution exists for $$c$$ in the given range; that is done e.g. here.
Then because $$c, $$0\le x and $$0\le y. But the smallest vector (by norm) we can add to $$(x,y)$$ to generate another solution is, since $$(a,b)=1$$, $$(-b,a)$$. If we add this vector $$x$$ will become negative; if we subtract it $$y$$ will become negative, and both are disallowed. Thus the solution that we proved exists for $$ax+by=c$$ is unique.