Irreducibility of $x^2+y^2+z^2-xyz-2$ Are there some general criteria for deciding the irreducibility of polynomials? For example the one in the title?
 A: If this polynomial were to factor, it would have to factor into a polynomial of degree 2 and a polynomial of degree 1.
\begin{align*}
x^2 + y^2 + z^2 - xyz - 2
&= \big([\text{terms of degree } 2] + [\text{terms of degree } \le 1] \big) \\
&\quad \big([\text{terms of degree } 1] + [\text{terms of degree } \le 0] \big) \\
\end{align*}
The terms of degree $2$ in the first factor and the terms of degree $1$ in the second factor must multiply to $-xyz$.
By symmetry in $x,y,$ and $z$, we may assume they are therefore $-xy$ and $z$, respectively.
$$
x^2 + y^2 + z^2 - xyz - 2= (-xy + ax + by + cz + d)(z + e)
$$
But there is no $x^2$ or $y^2$ term on the RHS, so this is impossible.
General Method
Some discussion of this in the case that the polynomial is irreducible over ANY field, can be found here.
There are also some tricks in the case of two variables here.
One possible approach is to consider any polynomial $p(x,y,z)$ (in general, any number of variables) as a polynomial in $x$ over the ring $\mathbb{Z}[y,z]$ (in general, $\mathbb{Z}$ could be any unique factorization domain).  Note that $\mathbb{Z}[y,z]$ is a unique factorization domain itself, so things like Gauss's lemma and Eisenstein apply.  In this case, we would have to have
$$
x^2 + (-yz) x + (y^2 + z^2 - 2) = (x + q_1(y,z))(x + q_2(y,z))
$$
for polynomials $q_1$ and $q_2$,
which reduces to the problem of showing $y^2 + z^2 - 2$ is irreducible over $\mathbb{Z}[y,z]$.  Then in turn we could consider $y^2 + z^2 - 2$ as a polynomial in $y$ over the ring $\mathbb{Z}[z]$, to get
$$
y^2 + z^2 - 2 = (y + r_1(z))(y + r_2(z))
$$
for polynomials $r_1$ and $r_2$ in one variable.
Then we have that $r_1(z) + r_2(z) = 0$, so $r_1(z) = -r_2(z)$,
so $z^2 - 2 = -r_1(z)^2$, so $2 - z^2$ is a perfect square.
So we have $2 - z^2 = (az + b)^2$, and we conclude $a^2 = -1$, $ab = 0$, and $b^2 = 2$.  This is of course impossible, not just over $\mathbb{Q}$ but over any field of characteristic $ \ne 2$.
