Prove that if $x = 2uv$ and $y = u^2 - v^2$, show that $(x, y, z)$ is a primitive Pythagorean triple if and only if $\gcd(u, v) = 1$.

The direction $\gcd(u, v) \neq 1$ implies $(x, y, z)$ is a non-primitive Pythagorean triple – this is straightforward. But I need the converse too, for which I'm completely stumped.

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    $\begingroup$ If $u=5, v= 3$ then $(x,y,z) = (30,16,34)$ $\endgroup$ – Neil W May 4 '14 at 1:02
  • $\begingroup$ What is $z$?${}$ $\endgroup$ – Jack M May 4 '14 at 1:10
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    $\begingroup$ $z=u^2+v^2$, as in the general form of a Pythagorean triple $\endgroup$ – Zubin Mukerjee May 4 '14 at 1:11
  • $\begingroup$ @ZubinMukerjee Sure, but the OP doesn't say that anywhere in their question. Typo? $\endgroup$ – Jack M May 4 '14 at 1:13
  • $\begingroup$ I suppose the wording "... a Pythagorean triple $(x,y,z)$ is primitive if and only if ..." would work better, since the values of $x$ and $y$ determine the value of $z$. $\endgroup$ – Zubin Mukerjee May 4 '14 at 1:15

As has been pointed out in the answer by Zubin Mukerjee, the result as stated is not correct. We state and prove a correct version.

Theorem: Let $u$ and $v$ be positive integers, with $v\lt u$. Let $x=2uv$, $y=u^2-v^2$, and $z=u^2+v^2$. Then $(x,y,z)$ is a primitive Pythagorean triple if and only if $\gcd(u,v)=1$ and $u$ and $v$ are of opposite parity.

It is clear that $x$, $y$, and $z$ are positive integers. It is also easy to verify that $(2uv)^2+(u^2-v^2)^2=(u^2+v^2)^2$.

(i) We show that if $\gcd(u,v)\ne 1$ or $u$ and $v$ are of the same parity, than the triple $(x,y,z)$ is not primitive. Suppose that $\gcd(u,v)=d\gt 1$. Then $d^2$ divides all of $2uv$, $u^2-v^2$, and $u^2+v^2$, so the triple $(x,y,z)$ is not primitive.

Next we show that if $u$ and $v$ are of the same parity, then $(x,y,z)$ is not primitive. This is because in that case all of $2uv$, $u^2-v^2$, and $u^2+v^2$ are even.

(ii) Next we show that if $\gcd(u,v)=1$ and $u$ and $v$ are of opposite parity, then $(x,y,z)$ is primitive.

Suppose to the contrary that some $d\gt 1$ divides both $y$ and $z$. Then some prime $p$ divides both $y$ and $z$. So $p$ divides $u^2-v^2$ and $p$ divides $u^2+v^2$. Note that that since $u$ and $v$ are of opposite parity, it follows that $u^2+v^2$ is odd. So $p$ is odd.

Since $p$ divides $u^2-v^2$ and $u^2+v^2$, it follows that $p$ divides their sum and difference $2u^2$ and $2v^2$. since $p$ is odd, $p$ divides $u^2$ and $v^2$, and since $p$ is prime, $p$ divides $u$ and $v$. This contradicts the fact that $\gcd(u,v)=1$, and completes the proof.

  • $\begingroup$ This implicitly makes use of prime factorisation and the (generalised) Euclid's lemma... but I don't think we can help that, at least the latter! Oh well. Thanks regardless $\endgroup$ – Noldorin May 4 '14 at 16:58
  • $\begingroup$ It's worth noting that part (ii) can at least be formulated easily without proof by contradiction, by the way. :) $\endgroup$ – Noldorin May 4 '14 at 22:23

The reason you're completely stumped might be that the converse is false :)

It is not true that

If $\gcd{(u,v)}=1$, then $(2uv, u^2-v^2, u^2+v^2)$ is a primitive Pythagorean triple.

Counterexample(s): $u=7$, $v=3$ corresponds to the triple $(42, 40, 58)$, which is clearly not primitive. Indeed, for any $u$ and $v$ both odd, every value in the triple will be even, making it non-primitive.

  • $\begingroup$ Hmmm, you're right. But how odd. This question came out of a printed book... I would normally be loath to think it contains such an inaccuracy! $\endgroup$ – Noldorin May 4 '14 at 1:16
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    $\begingroup$ As noted by @Neil in the 1st comment. $\endgroup$ – Gerry Myerson May 4 '14 at 1:19
  • $\begingroup$ Anyway, this suggests I might have to modify the condition on the RHS to be $(\gcd(u, v) = 1) \wedge (u\ \text{even} \oplus v\ \text{even})$, perhaps? $\endgroup$ – Noldorin May 4 '14 at 1:20
  • $\begingroup$ If $u$ and $v$ are relatively prime and of opposite parity, you do get a primitive triple, and you get all primitive triples in this way. $\endgroup$ – André Nicolas May 4 '14 at 2:38
  • $\begingroup$ @AndréNicolas: Indeed. How can I show this though, using only super elementary methods? (i.e. no Euclid's lemma yet even.) $\endgroup$ – Noldorin May 4 '14 at 14:27

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