I Right Triangle and the Pythagorean Triple Formula

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The discussion centers on whether the Pythagorean triple formula can generate all right triangles with whole number sides. It is established that while the formula produces all primitive triples and some non-primitive ones, it does not account for every possible combination of whole number sides. The common formulas, such as a=2mn, b=m^2-n^2, and c=m^2+n^2, generate specific integer triplets, but some combinations, like (9,12,15), are excluded. The conversation suggests that extending the formula by adding a common factor could cover all triplets, albeit with some duplicates. Ultimately, the Pythagorean triple formula does not uniquely express every right triangle with whole number sides.
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Given: A right triangle and all the sides of the triangle are whole numbers. Does this imply that all the sides of the triangle can only be found by using the Pythagorean triple formula? Another words, is it possible that a right triangle can exist with whole number sides that escape the Pythagorean triple formula? Or put another way, does the Pythagorean triple formula capture every possible right trangle with whole number sides?
 
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What exactly do you call "Pythagorean triple formula"? There are multiple formulas that produce those triples.
 
Usual formulas: a=2mn,\ b=m^2-n^2,\ c=m^2+n^2 for integers m>n>0, give all possible integer triplets.
 
ok. That's what I wanted to know.
 
This formula does not give all possible triples. It gives all primitives and a bunch of the non primitives, right? If you add the multiples of those then you get all. But then they wouldn't be uniquely expressed (I guess..?).

Would it be relevant to find a formula which generates them all with only two variables..?
 
Good point. Some multiples are covered, some are not.

(3,4,5) is generated by n=1, m=2, and (8,6,10) is generated by n=1, m=3, but (9,12,15) is not covered as 15 cannot be written as the sum of two squares.

It is trivial to extend the formula to cover all multiples, just add a common factor k to all three lengths. Then you get some triplets with more than one set of integers, but you cover all triplets at least once.
 
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