## Characterize the irreducibles in ZZ[sqrt(2)]

Characterize the irreducible elements in $\mathbb{Z}[\sqrt{2}]$.

Suppose $\zeta \in \mathbb{Z}[\sqrt{2}]$ is irreducible. By our proof of Theorem 8.5, it follows that $(\zeta) \supseteq (p)$ for some unique positive rational prime $p$. That is, $p = \zeta\eta$ for some $\eta \in \mathbb{Z}[\sqrt{2}]$. Now $p^2 = N(\zeta)N(\eta)$, so that either $N(\zeta) = \pm p$ or $N(\zeta) = \pm p^2$. (Recall that $N(\zeta)$ is a rational integer and not 1.) Let $\zeta = a+b\sqrt{2}$.

Suppose $N(\zeta) = \pm p$. Then $\pm p = a^2 - 2b^2$, so that $a^2 - 2b^2$ is a rational prime.

Suppose $N(\zeta) = \pm p^2$. Now $N(\eta) = \pm 1$, so that (by Theorem 7.3) $\eta$ is a unit. (In Theorem 7.9 we saw that the units in $\mathbb{Z}[\sqrt{2}]$ have the form $\pm(1+\sqrt{2})^k$ where $k \in \mathbb{Z}$.) Thus $\zeta$ is an associate of $p$ in $\mathbb{Z}[\sqrt{2}]$. We claim that in this case, there does not exist a solution $(h,k)$ of the equation $h^2 - 2k^2 = \pm p$. If so, then $\pm p = (h+k\sqrt{2})(h-k\sqrt{2})$. But since $\pm p$ is irreducible, either $h+k\sqrt{2}$ or $h-k\sqrt{2}$ is a unit- so $p = \pm 1$, a contradiction.

In summary, if $\zeta = a+b\sqrt{2}$ is irreducible in $\mathbb{Z}[\sqrt{2}]$, then either (1) $N(\zeta) = \pm p$ is a rational prime or (2) $N(\zeta) = \pm p^2$ is the square of a rational prime and the equations $h^2-2k^2 = \pm p$ have no solutions $(h,k)$ in $\mathbb{Z}$.

Conversely, if $N(\zeta) = p$ is prime, then $\zeta$ is irreducible. If $N(\zeta = \pm p^2$ where $p$ is a rational prime and if $h^2-2k^2 = \pm p$ has no solution, then no element of $\mathbb{Z}[\sqrt{2}]$ has norm $\pm p$, so that $\zeta$ is irreducible.