= Solution
Since $10\equiv2\pmod4$, the <ring of integers> is $\mathcal O_K=\mathbb Z[\sqrt{10}]$, with <field norm> $N(x+y\sqrt{10})=x^2-10y^2$. Norms $\pm2$ and $\pm3$ are impossible: modulo five they would require a square congruent to 2 or 3. An <algebraic integer> is a unit exactly when its norm is $\pm1$.
If 2 factored into two nonunits, their absolute norms would both be 2, since $N(2)=4$; this is impossible. The same argument for $N(3)=9$ excludes a factorization of 3. Finally $\epsilon+1=4+\sqrt{10}$ has norm 6, so a nonunit factorization would require absolute norms 2 and 3, also impossible. Hence \b[2, 3 and $\epsilon+1$ are irreducible]. The conjugate $\overline\epsilon+1$ is irreducible by the same argument.
The equality $6=2\cdot3=(4+\sqrt{10})(4-\sqrt{10})$ gives inequivalent factorizations into irreducibles: no norm-six factor is associated to either 2 or 3, whose norms are 4 and 9. A <principal ideal domain> is a <unique factorization domain>. Consequently \b[$\mathcal O_K$ is not a <principal ideal domain>].
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