If $gcd(a, b) = 1$ and $ax + by = c$ has a positive integer solution, then does $ax + by = d$ when $d > c $?
It's pretty obvious that both of these have a solution as gcd(a,b) = 1 and 1 divides everything so it will divide both c and d. I'm unsure what it would mean for there to be a positive solution.
An INTEGRAL solution of $$ax+by=c$$ exists if and only if $c$ is divisible by $gcd(a,b)$. But to find out whether a solution with positive $x$ and $y$ exists, you need a further restriction.
If $(u/v)$ is a solution, the general solution is $(u-bt/v+at)$ with $t\in \mathbb Z$. You get two inequalities and you have to verify whether they can be solved simultaneously.