I haven't read it yet but
https://www.physicsforums.com/showthread.php?t=367945", is most certainly relevant to your question.
Here is my attempt, which to be honest is a complete overkill since it resorts to a more advanced theorem!:), but it was pretty much the first thing that came into my head.
First we note that if we have one solution to the equation, say (x,y), then (x+nb, y+na) is also a solution. Therefore if we can find one solution, we can always, from this solution, construct another solution for which both coordinates are positive.
With the aim of finding a single solution, we note that it is clear that the problem is equivalent to (where I have redefined the meaning of the variables a,b,c):
<br />
ax - by = c<br />
For a>0, b>0, and where a and b are coprime [I have divided both sides of the equation by the gcd(a,b), and then redefined the variables (e.g. c now represents the result of this division on the rhs etc..likewise for a,b)].
It is also clear that if we can find a solution to the equation :
<br />
ax - by = 1<br />
(where a>0, b>0, and where a and b are coprime), then we can immediately find a solution to the equation above (simply by multiplying both sides of the equation by c).
We proceed by noting that from a "[URL of fermat's little theorem[/URL], and given the coprimality of a and b :
<br />
\exists\ p\ s.t.\ a^p = 1 (mod\ b)<br />
In particular p is given by http://en.wikipedia.org/wiki/Euler%27s_totient_function" : p=\phi(b). Therefore :
<br />
a^p = Nb + 1\ for\ some\ integer\ N<br />
So that
<br />
a^p - N b = 1<br />
Therefore the solution to :
<br />
ax - by = 1<br />
(where a>0, b>0, and where a and b are coprime) is given by :
<br />
(x,y) = (a^{p-1}, N)<br />
where x, y are both integers, and therefore we have achieved the desired result