# I Right Triangle and the Pythagorean Triple Formula

1. Apr 23, 2016

### e2m2a

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?

2. Apr 23, 2016

### Staff: Mentor

What exactly do you call "Pythagorean triple formula"? There are multiple formulas that produce those triples.

3. Apr 23, 2016

### mathman

Usual formulas: $a=2mn,\ b=m^2-n^2,\ c=m^2+n^2$ for integers m>n>0, give all possible integer triplets.

4. Apr 23, 2016

### e2m2a

ok. That's what I wanted to know.

5. Feb 7, 2017

### danieldf

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 wich generates them all with only two variables..?

6. Feb 7, 2017

### Staff: Mentor

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.