$x^2+y^2-1$ irreducible in $\mathbb{Q}[x,y]$?

Question

I want to show that $p(x,y)=x^2+y^2-1$ is irreducible in $\mathbb{Q}[x,y]$. There is already a similar problem posted:

$\mathbb{Q}[x,y]/\langle x^2+y^2-1 \rangle$ is an integral domain, and its field of fractions is isomorphic to $\mathbb Q(t)$

but this question uses the irreducibility of $p(x)$ to show that it is prime and therefore an integral domain. The part of the problem I am stuck on is showing that $x^2+y^2-1$ is irreducible over $\mathbb{Q}$ in the first place.

I have considered taking the approach used in the above link except for backwards, but I am unsure of whether or not the converses of the theorems used still apply.

Edit: I see that this question has been marked as a duplicate, but I think I have explained how I am asking a different question than the one that was posted previously.

Answer 1

Suppose that polynomial is a product $f(x,y)g(x,y)$. Then $f,g$ both must be linear because $p = x^2+y^2-1$ is of largest degree $2$. Now write down what $p = fg $ means and see what $f,g$ can actually be.