# Proof for primitive pythagorean triples

I was wondering if there was a proof that any integer pythagorean triple can be represented as a positive integer multiple of a primitive pythagorean triple. This seems quite related to the fundamental theorem of arithmetic, so I was wondering if the proof would be similar.

• How are you defining a primitive pythagorean triple? – lulu Feb 2 '16 at 15:48
• This looks like a circular argument: either the numbers in the triple are not coprimes and you can simplify, or they are coprimes and you call them primitive. – Yves Daoust Feb 2 '16 at 15:51

Hint: If you have a pythgorean triple $(a,b,c)$ divide each number by $gcd(a,b,c)$ and remember if $p$ divides any two of the new numbers it must divide the third.

This is trivial from the definition.

We say that $(a,b,c)$ is a primitive Pythagorean triple if it is a Pythagorean Triple and there do not exist $(a',b',c')$, also a Pythagorean Triple, such that $\exists k$ such that $a=ka'$ etc.

Let $(a,b,c)$ be a Pythagorean triple. If it's primative, $k=1$ and we are done. If it's not, the. $\exists k$ such that...

It is known all the pythagorean triples are given by $$(\lambda(s^3-t^2), \lambda(2st),\lambda(s^2+t^2))$$ where $(s,t)=1$ and $\lambda\in \mathbb N$. The answer to your question it clearly follows.

A multiple of a Pythagorean triple is also a Pythagorean triple, $$a^2+b^2=c^2\iff(na)^2+(nb)^2=(nc)^2.$$

There is nothing more to say.

• The downvoter is wrong. – Yves Daoust Feb 2 '16 at 20:32