Choose such that and we get
I've convinced myself that these cover all solutions but can't/don't feel like saying why (to much typing (Surprised))...
Assuming this is right, this can be generalized to the equation .
I can't speak for chiph588@'s parametrisation although it looks interesting.
I have a proof that roninpro's parametrisation can be modified to obtain all solutions. I will post it for review when I have time, but here is the resolution to the sticking point brought up earlier. Consider that when m = 5 and n = 1 we have (x,y,z) = (20,10,30); dividing by gcd(x,y,z) = 10, we obtain (x,y,z) = (2,1,3). Following this procedure, I claim all solutions are obtainable.
I wrote out a lengthy proof on paper involving rational points on an ellipse and got exactly roninpro's parametrisation as the end result; it will take a while to type out, won't get to it for some hours.
So we have
and suppose we are interested in all integer solutions. First note that there are trivial solutions for all integers . Without loss of generality we can consider just solutions in positive integers for the non-trivial solutions. It can be seen that if satisfies, then satisfies for any positive integer , thus we can obtain all non-trivial solutions from the primitive solutions having . (This is standard use of the term primitive.) As for the problem statement restricting to odd; I don't know what the motivation is for that, but we can just discard any solutions for which is even at the very end.
Now for the actual proof/derivation I mentioned earlier. Disclaimer is that my solution method is not original.
Let and , from which we obtain
This is an equation of an ellipse; solutions to the problem correspond with points on the ellipse. Consider the point on the ellipse. A parametrisation for all non-vertical lines through is , where is the slope of the line. The intersection of the line with the ellipse is given by
It can be seen that if is rational then and are rational. Replace with and we have
which simplifies to
Now we can take the parametrisation . Let . Then we have a parametrisation for primitive solutions . We must choose positive integers and such that
Restricting , we obtain all primitive triples uniquely.
Well I guess that response isn't very helpful to you. To answer your question, the parametrisation is given here: