Linear Diophantine equation in two variables with additional constraints Given,

$$aX + bY = c$$

where,

$$c > b > a > 0;\quad X, Y > 0;\quad b\nmid c, a\nmid c$$

I want to find out if a solution exists as efficiently as possible (I'm not interested in the solutions). Are there any calculations I can make before (or without the need for) finding $\operatorname{gcd}(a, b)$ that can possibly save some time (even if for only few special cases)? $c, b, a$  can be very large numbers.
"Probably not" still counts as answer for me. You don't have to be 100% certain. I just want to make sure I'm not missing something that's very obvious. 
P.S.,
English is not my first language.
 A: I would say probably not. Finding $\gcd(a,b)$ is computationally fast (with technology). If you get $\gcd(a,b)\nmid c$, then there are no solutions. See this for more general theory and examples.
A: If gcd(a,b) divides c, then there are infinitely many solutions for the equation
aX + bY = c
Divide the equation by gcd(a,b) to get the reduced equation
sX + tY = u
where gcd(s,t)=1
With the extended algorithm for the gcd-calculation, you can calculate a solution (m,n).
All solutions are given by
X = kt + m
Y = -ks + n
for some integer k.
If there is an integer k, such that X and Y satisfy all conditions, a solution is
found. In order to check this, you have to find bounds for the number k by solving
the desired inequalities and check, if the intervals found have a non-empty
intersection.
So, this is the way to find solutions.
Now, I come to your additional restrictions. Since 3x + 4y = 5 is an
equation not having integer solutions with x>0 and y>0, it is not enough
to check, whether gcd(a,b) divides c, even if the additional restrictions hold.
But the way I described is very efficient even for very large numbers, since it only requires modulo-calculations.
I doubt that there are faster ways to decide if there are solutions of the desired form.
