1
$\begingroup$

I'm almost done proving that for any odd numbers $s, t : s > t \geq 1, \ \ \gcd(s, t) = 1$, one can generate PPTs (Primitive Pythagorean Triples) with the formula $$\displaystyle (a, b, c) = \left(st, \frac{s^2-t^2}{2}, \frac{s^2+t^2}{2}\right)$$

However, the only piece I'm missing, is to show that these three expressions, given the initially stated conditions on $s$ and $t$, will be coprime.

I.e. I need to show that $\gcd\left(st, \frac{s^2-t^2}{2}, \frac{s^2+t^2}{2}\right) = 0$.

The book hints at using the fact that if a prime divides a product, it divides at least one factor in the product. In other words $p \mid xy \Rightarrow p\mid x \vee p \mid y$. But I haven't been able to piece that together into a proof.

I've tried a proof by contradiction, where using the definition of divisibility, I tried assuming that there exists a prime $p$ that divides both $st$ and $\frac{s^2- t^2}{2}$, and seeing if I could get a blatant contradiction, but I haven't seen any yet. Or at least not noticed.

I'd especially favor any tip/solution that involves my idea of a proof by contradiction, as it will at least validate some of the efforts I've made so far.

But any help would be appreciated, of course!

$\endgroup$
3
$\begingroup$

We know $p=2$ cannot divide all the factors because $s$ and $t$ are stipulated to be odd, so $2 \nmid st$.

Let $p$ be an odd prime. Suppose $p|st$. By the fact you mentioned, $p|s$ or $p|t$, and not both because $s$ and $t$ are coprime; let's say $p \mid s$ and $p \nmid t$. Then $p \nmid (s+t)$ and $p \nmid (s-t)$, so $p \nmid (s+t)(s-t) = s^2 - t^2$, and thus $p \nmid \frac{s^2-t^2}{2}$.

$\endgroup$
  • $\begingroup$ Ah nice, got it. But does this also cover our favorite even prime? I suspect it's trivial, but I'm not seeing it. $\endgroup$ – Alec Jan 10 '18 at 16:13
  • $\begingroup$ @Alec Nope, because of the division by two. But since you stipulated $s$ and $t$ to be odd, we know $2 \nmid st$ $\endgroup$ – BallBoy Jan 10 '18 at 16:15
  • $\begingroup$ @Alec I've added this to the answer. $\endgroup$ – BallBoy Jan 10 '18 at 16:22
0
$\begingroup$

You may or may not be able to generate primitive triples with your function because it reads the same as Euclid's formula except with $A$ and $B$ reversed and all the terms divide by $2$. For Euclids $(m,n)=(2,1)$ you would get $(2,1.5,2.5)$ as a triple. Let's try your equation with terms multiplied by $2$ and with $A,B$ reversed.

$\text{We are given }\quad A=m^2n^2\quad B=2mn\quad C=m^2+n^2$

Let $x$ be the GCD of $m,n$ and let $p$ and $q$ be the cofactors of $m$ and $n$ respectively. Then we have

$$A=(xp)^2-(xq)^2\quad B=2xpxq\quad C=(xp)^2+(xq)^2$$ $$A=x^2(p^2-q^2)\quad B=2x^2(pq)\quad C=x^2(p^2+q^2)$$

If $GCD(m,n)=1$, then $GCD(A,B,C)=1$ and $(A,B,C)$ is a primitive triple. This means that $m$ and $n$ must be co-prime to generate a primitive.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.