## Factor some given algebraic integers in a quadratic field

Factor $5 + \sqrt{3}$ in $\mathbb{Q}(\sqrt{3})$ and $7 + \sqrt{-5}$ in $\mathbb{Q}(\sqrt{-5})$.

Note that $N(5+\sqrt{3}) = 5^2 - 3 = 22 = 2 \cdot 11$. If $\alpha = \beta\gamma$, then $N(\beta)N(\gamma) = 2 \cdot 11$, so (without loss of generality) $N(\beta) = 2$. Suppose $\beta = h+k\sqrt{3}$; then $h^2-3k^2 = 2$. Mod 3, we have $h^2 \equiv 2$. This is a contradiction since 2 is not a square mod 3. That is, no algebraic integer in $\mathbb{Q}(\sqrt{3})$ has norm 2. (Similarly, no integer has norm 11.) Thus $5+\sqrt{3}$ is irreducible as an integer in $\mathbb{Q}(\sqrt{3})$.

Evidently, $(1+\sqrt{-5})(2-\sqrt{-5}) = 7+\sqrt{-5}$. Now $N(1+\sqrt{-5}) = 6$ and $N(2-\sqrt{-5}) = 9$. We claim that no integer in $\mathbb{Q}(\sqrt{-5})$ has norm 3; to that end, suppose $h^2+5k^2 = 3$. We must have $k = 0$, so that $h^2 = 3$, a contradiction. In particular, both $1+\sqrt{-5}$ and $2-\sqrt{-5}$ are irreducible in $\mathbb{Q}(\sqrt{-5})$.