We thought it was so, but actually it is only a parametrisation of some solutions. roninpro brought up that (2,1,3) can't be generated by the proposed system, which I agree with. So we have some sorting out to do in order to know we've gotten all solutions, but in the meantime, you do have an infinite number of solutions at your disposal.
Don't ask how but I believe I've found all solutions. It generalizes roninpro's solution too and it has twice as many parameters.
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 )...
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.
Yes, I should have been more careful about that point.
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
and
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.
I don't mean to be rude, but if you don't see what the answer to the problem is, then you didn't follow what I wrote!
Well I guess that response isn't very helpful to you. To answer your question, the parametrisation is given here:
Whenever we come across a solution with even, we discard it. (I would look at what restrictions to place on and to ensure this, but I don't understand the motivation for restricting to odd integers anyway, so I don't see the worth of it. Maybe it refers to another solution method for which even poses complications.)
To prevent duplicate counting. (The problem statement doesn't require this, but it is a nice thing to have.) A rational point on the ellipse represents a class of solutions for some primitive . Since we defined , restricting ensures that we will count each (positive) rational exactly once.