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**ThePerfectHacker** You proved that 2 is irreducible. You can show that the other two are irreducible by a similar argument. If $\displaystyle \sqrt{-n} = \alpha \beta$ for non-units, then by taking norms $\displaystyle n = N(\alpha)N(\beta)$. Notice that $\displaystyle 1 < N(\alpha) < n$. Let $\displaystyle \alpha = a + b\sqrt{-n}$ then $\displaystyle N(\alpha) = a^2 + nb^2$. For this to be less than $\displaystyle n$ it is necessary for $\displaystyle b=0$. Therefore, $\displaystyle \alpha$ must be a real number. By a similar argument $\displaystyle \beta$ must be a real number. However, it is impossible to have $\displaystyle \alpha \beta$ to be an imaginary number. Thus, it is not possible to write $\displaystyle \sqrt{-n} = \alpha \beta$ for non-units.

Say that $\displaystyle 1+\sqrt{-n} = \alpha \beta$ for non-units. By taking norms, $\displaystyle 1 + n = N(\alpha)N(\beta)$. Thus, we require that $\displaystyle 1 < N(\alpha) < n+1$. If $\displaystyle \alpha = a + b\sqrt{-n}$ then $\displaystyle 1< a^2 + nb^2 < n+1$. We cannot have $\displaystyle |b|\geq 2$ so $\displaystyle b^2=0,1$. Likewise for $\displaystyle \beta = c + d\sqrt{-n}$. If $\displaystyle b^2=1$ then $\displaystyle 1 < a^2 + n < 1 \implies 1 < a^2 < 1$ which is a contradiction. Therefore, $\displaystyle b=0$, and now repeat the same argument as above.