<s>[INST] Euclid's Lemma in the proof that $\mathbb{Z}[\sqrt{-5}]$ is not euclidean In proving, I induced that $3\gamma + (2+\sqrt{-5})\delta=1$. Textbook says that multiplying both sides by $2-\sqrt{-5}$ implies $2-\sqrt{-5}$ is multiple of 3. I don't know why. Can you explain the calculation.
 [/INST] If you multiply both sides by $2 - \sqrt{-5}$ you get 
$$ \begin{align} (2 - \sqrt{-5}) \cdot 3 \cdot \gamma + (2 - \sqrt{-5}) \cdot (2 + \sqrt{-5}) \delta &= 2 - \sqrt{-5} \\
(2 - \sqrt{-5}) \cdot 3 \cdot \gamma + 3 \cdot 3 \delta &= 2 - \sqrt{-5} \\
3\cdot ((2 - \sqrt{-5}) \cdot \gamma + 3 \delta) &= 2 - \sqrt{-5} \end{align}$$
From the first to the second line we use $(2 - \sqrt{-5})(2 + \sqrt{-5}) = 4+5 = 9 = 3 \cdot 3$.
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