MHB Show GCD of x,y,z is 1: Wave Hello!

[evinda]

Hello! (Wave)

We suppose that the integers $x,y,z$ satisfy $x^2+2y^2=z^2$ and $(x,y)=1$ . I want to show that $(x,z)=(y,z)=1$, and that $x$ is odd and $y$ even.

I have tried the following:

Let $(x,z)=d>1$. Then there exists a prime number $p$ such that $p \mid d$.
Since $d \mid x$ and $d \mid z$, we get that $p \mid x$ and $p \mid z$. So $p \mid x^2$, $p \mid z^2$.
Thus $p \mid z^2-x^2=2y^2$. But then how can we deduce that $p \mid y^2$, so that we could get a contradiction? (Thinking)

 

[Olinguito]

First show that $x$ is odd. Suppose it was even. Then $x^2+2y^2=z^2\implies z$ would be even; hence $z^2=x^2+2y^2$ would be divisible by $4$; hence $y$ would be even. (If $y$ were odd, $x^2+2y^2\equiv2\pmod4$.) Hence $x$ and $y$ would be both even, contradicting $\gcd(x,y)=1$. Thus $x$ must be odd.

Therefore the $p$ in your working must be odd (since it divides $x$, which is odd). Then $p\mid2y^2$ should imply $p\mid y^2$, giving your contradiction. Showing that $\gcd(y,z)=1$ is similar (and more straightforward).

Finally, note that $x$ odd $\implies\ x^2\equiv1\pmod8$. If $y$ were odd, then $z^2=x^2+2y^2\equiv3\pmod8$, which is impossible.

 

[evinda]

First show that $x$ is odd. Suppose it was even. Then $x^2+2y^2=z^2\implies z$ would be even;

$z^2$ would be even and this would imply that $z$ is even, right? (Thinking)

(If $y$ were odd, $x^2+2y^2\equiv2\pmod4$.)

Wouldn't we have that $x^2+2y^2 \equiv 3 \pmod{4}$ ?

 

[Olinguito]

$z^2$ would be even and this would imply that $z$ is even, right? (Thinking)

That is right. (Smile)

Wouldn't we have that $x^2+2y^2 \equiv 3 \pmod{4}$ ?

No, at this juncture we are assuming $x$ is even to get a contradiction.

 

[I like Serena]

Hello! (Wave)

We suppose that the integers $x,y,z$ satisfy $x^2+2y^2=z^2$ and $(x,y)=1$ . I want to show that $(x,z)=(y,z)=1$, and that $x$ is odd and $y$ even.

I have tried the following:

Let $(x,z)=d>1$. Then there exists a prime number $p$ such that $p \mid d$.
Since $d \mid x$ and $d \mid z$, we get that $p \mid x$ and $p \mid z$. So $p \mid x^2$, $p \mid z^2$.

Hey evinda!

As an alternative to Olinguito's approach, let's follow your reasoning a bit further.
We also have more specifically that $p^2 \mid x^2$ and $p^2 \mid z^2$, don't we? (Wondering)

Thus $p \mid z^2-x^2=2y^2$. But then how can we deduce that $p \mid y^2$, so that we could get a contradiction? (Thinking)

Thus $p^2 \mid 2y^2$.
If $p=2$ we must have $p\mid y$, and otherwise we must also have that $p\mid y$, don't we? (Wondering)