I’m gathering data for a hobby project and notating where the triangles in the data correspond to Pythagorean triples, but sometimes it doesn’t seem clear to me with certain data.

Can there exist Pythagorean triples in which the leg lengths are not coprime with each other but both are coprime with the hypotenuse? i.e., a right triangle in which (leg1, leg2, hypotenuse) = (a * n, b * n, c), in which a, b, c, and n are whole numbers and n is not a factor of c?

How can I determine if a right triangle with given lengths can scale to be a Pythagorean triple? If any of the values in (leg1: leg2: hypotenuse) are irrational, that does indeed mean the values cannot scale to be whole numbers?

Once it is determined that the triangle can scale to a Pythagorean triple, what is the best method of scaling the values to three whole numbers?

Thanks for any help

Edit: I’ve found an effective way to determine primitive Pythagorean triples from given leg lengths. Using a calculator that can output in fractional form, such as wolfram alpha, input leg1 / leg2 and the output will be a fraction with the numerator and denominator denoting the leg lengths of a primitive Pythagorean triple. Determining the hypotenuse is then simply using the Pythagorean Theorem.

  • PlanterTreeEnglish
    1·
    3 months ago

    I try to answer your question with an example, maybe I misunderstood.

    16 + 9 = 25
4*4 + 3*3 = 5*5

    4, 3 and 5 are prime numbers, and form the Pythagorean triple (3, 4, 5).

    Both legs 4 and 3 are coprime: gcd(3, 4) = 1. and both are coprime with the hypotenuse gcd(3, 5)=1 and gcd(4, 5) = 1. Now lets multiply this triple by 7. Obviously a new Pythagorean triple is formed (7*3, 7*4, 7*5).

    But coprime-ness of the legs is lost because of the multiplication, this is not a counter example to your theory:

    1. Leg lenghts are not co-prime with each other ✅ gcd(7*3, 7*4) = 7 > 1
    2. Leg lengths are coprime with the hypotenuse ☠ No.

    I try to translate this example to a proof.

    Given a Pythagorean triple (a, b, c) where a, b and c are integers that fullfill the equation (I) a*a+b*b = c*c. Further say a and b share a greatest common factor x>1. This means gcd(a, b) = x. This means a and b are not coprime, which is your first constraint.

    This means we can write a as x*u and b as x*v, where u and v are integers. Now place this in equation (I)

    (x*u)*(x*u) + (x*v)*(x*v) = c*c
↔ x*x*u*u + x*x*v*v = c*c | commutative and associative law
↔ x*x (u*u+v*v) = c*c

    Since x, u, v and c are integers, this means c has to share the common factor x with a and b. Which means gcd(a, c) >= x, which is a contradiction with gcd(a, c) = gcd(b, c) = 1 (legs are co-prime with hypotenuse).

    Can someone write this proof better please lol, and pls correct if wrong.