# Showing two numbers are relatively prime in number fields

Solve $x^3-2=y^2$ in integers.

The standard way to solve this problem is to consider the arithmetic of the ring of algebraic integers $\mathcal{O}_{\mathbb{Q}(\sqrt{-2})}$ and to show that $y+\sqrt{-2}$ and $y-\sqrt{-2}$ are relatively prime. Here are two proofs from Murty's problems in algebraic number theory and titu's introduction to diophantine equations which i dont quite understand, could someone please helps.

It is asy to show that both $x,y$ must be odd.

1st proof: let $d= gcd(y+\sqrt{-2}, y-\sqrt{-2})$, then $d| 2 \sqrt{-2}= -(\sqrt{-2})^3$. So $d$ is a power of $\sqrt{-2}$. On the other hand, if $\sqrt{-2}| y \pm \sqrt{-2}$, then it divides the product of these factors, which is $y^2+2=x^3$ ""But $x$ is odd, hence $\sqrt{-2} \not| d$ and $y+\sqrt{-2}$ and $y-\sqrt{-2}$ are relatively prime.""

2nd proof: With the notation as above, if $d$ divided both, then $d$ would divide $2\sqrt{-2}$ and ""would thus have even norm, which is not possible since $y$ is odd."" So $y+\sqrt{-2}$ and $y-\sqrt{-2}$ are relatively prime.""

I don't understand the arguments inside the "" "". If $2| 2\sqrt{-2}$, then $2\sqrt{-2}=dq$ for some non-unit $q$ (if $y+\sqrt{-2}$ and $y-\sqrt{-2}$ were not relatively prime). Taking norm gives $8=N(d)N(q)$ so that $d$ would have even norm, is that right? But why is this not possilbe?

Thank you so much for helping.

## 1 Answer

Since $d$ divides $y+\sqrt{-2}$, the norm of $d$ divides $y^2+2$, the norm of $y+\sqrt{-2}$. But $y^2+2$ is odd, so $d$ cannot have even norm.