I am trying to get the primitive solutions under a certain limit.
The origin of this idea was micromass' suggetion:
In the interest of generating the primitives efficiently, and
I am trying to eliminate the computation of gcd's if I can,
because there are too many degenerate (multiple) cases...
I did a bit more experiments and found that I can get the range check down to
order of forth root: 2*limit^(0.25) to be precise.
Also using gcd = 1, skips many candidates.
Focusing on signs and parity also cuts it down.
I kept a hash of the results to avoid repeats,
but if I a nice...
It might be that the only problem is efficiency:
How can we efficiently generate all solutions?
i.e. how can we avoid spending time on degenerate solutions,
because it seems like there are lots of them...
I guess I am not very good at this either
(otherwise I would have found the solution by now :smile:)
I think they are both the same. I want to know where to stop
finding the primitives and jump to finding the all other non-primitive solutions.
If the answer is that m,n should span...
Yes, I checked it. Somehow only they (this paper and others)
only confidently talk about primitives in case of the basic Pythagorean triples.
I want to know :
1) if I am supposed to extent to negative m,n values
2) what ranges of m & n will give me all the primitives
For example if...
I also read a lot of papers on that subject.
Most just state that these generating functions generate infinitely many solutions,
but they are not all the solutions, meaning there might be more.
They sometimes use modular arithmetic to show the non-existence of a solution.
I would like to...
By trial and error, I found that the cube root of limit is enough for the range.
However this might be by chance or might depend on that particular limit.
And that is still a lot of (primitive) starting points...
How can I judge? Can you be more specific?
I simplified all results using the gcd's.
then I tried randomly picking primitives,
and all seemed legitimate...
Yes, I tried something similar to that, but it produces a lot of overlaps.
Let's say we want to find all values below 10K. Since we got to this point
by simplifying the very original equation, getting back to that set requires
to multiply by every integer, after we find the primitives...
After we solve a problem, all hard things look simple and trivial. I can't blame haruspex for that.
I think this is punishing all, for the behavior of the few...
And it looks like SeventhSigma doesn't want to share his solution in any case:
I believe the purpose of any forum is to have a 2...
Hi SeventhSigma,
I am always fascinated by solutions to Diophantine equations.
I am glad that you figured it out, but we haven't learned anything
from this discussion. Can you please share your solution with us?
Thanks,
LittleNewton