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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.

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First you need to show that a solution exists for $c$ in the given range; that is done e.g. here.

Then because $c<ab$, $0\le x<b$ and $0\le y<a$. 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.

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