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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.

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  • $\begingroup$ No...if $a=4,b=5$ then you get can $9=4\times 1 + 5\times 1$ but there is no way to get $11$. $\endgroup$ – lulu Mar 8 '17 at 9:54
  • $\begingroup$ There is a point beyond which you get every integer...the idea is to get every possible remainder for the smallest number. Sticking with $1=4,b=5$ you can get $4$ (remainder of $0$), $9$ (remainder of $1$), $14$ (remainder of $2$), $15$ (remainder of $3$)...so you can certainly get every number above $15$ (just subtract a multiple of $4$ to reach one of $4,9,14,15$). $\endgroup$ – lulu Mar 8 '17 at 9:59
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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.

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