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}).

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