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.
 A: 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 $(2 - \sqrt{-5})(2 + \sqrt{-5}) = 4+5 = 9 = 3 \cdot 3$.
A: It's an instance of  Euclid's Lemma. Let $\,w = 2\!+\!\sqrt{-5}.\,$ so $\,\bar w = 2\!-\!\sqrt{-5}.\,$ Then we have
$$ \overbrace{\gcd(\color{#c00}{3,w})=1}^{\text{by Bezout equation}},\ \ \color{#c00}3\mid \overbrace{\color{#c00}w\:\!\bar w}^{\large 9}\,\overset{\rm\color{#0a0}{EL}}\Longrightarrow\, 3\mid \bar w,\, \ \ \text{by $\,\rm\color{#0a0}{EL}=$ Euclid's Lemma}\qquad$$
The hinted proof is precisely the common (Bezout-based) proof of Euclid's Lemma in this instance. Of course it is better conceptually to invoke this fundamental property (Euclid's Lemma) by name rather than repeat its proof inline when applying it.
