I think you will end up with something similar to generating Pythagorean triples. I'll have to think about it/review some sources. Or maybe someone will post a nice solution in the meantime.
Hello All, I am trying to parameterize solutions in integers to the following Diophantine Equation:
(x^2) + 5*(y^2) = (z^2)
This comes from an additional review portion of my text. It states the following tip:
Consider only solutions that satisfy GCD[x,y,z]=1, and that x,y,z > 0 and that x is odd. If two cases appear, combine them into one by using absolute value.
Does anybody have any ideas? Any help is appreciated!
I think you will end up with something similar to generating Pythagorean triples. I'll have to think about it/review some sources. Or maybe someone will post a nice solution in the meantime.
Pythagorean triples are typically given as a^2 + b^2 = c^2, as on the Wikipedia page.
So your (x,y,z) became (a,b,c). m and n are positive integers that we iterate over. k is a multiplier. It is explained in more detail in the Wikipedia article. However, since gcd(x,y,z) = 1, you will not need the multiplier.
Oh, sorry, I meant to put , , .
You allow to range freely throughout the integers. Doing this gives you infinitely many solutions. Now, we have a little bit of a problem: we might not catch all of the solutions. To fix this, we throw in a parameter , which also ranges through the integers. This gives
, ,
This should give you every possible solution.
I'm not entirely sure that will work; one solution that I found was , which gives - we have some trouble with the 5.
I think that some insight for the proof comes from Dr. Math: Math Forum: Ask Dr. Math FAQ: Pythagorean Triples
By "only solutions", Chip meant that the solutions to the Diophantine equation have the form given in our equation. And actually, we may have a problem. I don't think that we can represent for some . So, I suppose that some modification to the formula is required.
Maybe you can have a look at that Dr. Math article and try a similar argument.