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.
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$(2+\sqrt{-5})(2-\sqrt{-5})=9=3×3$. – Oscar Lanzi Mar 19 '20 at 09:52
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1Thank you for finding my mistake! You have great insight. – Sinaps Mar 19 '20 at 09:55
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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 $(2 - \sqrt{-5})(2 + \sqrt{-5}) = 4+5 = 9 = 3 \cdot 3$.
Oscar Lanzi
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G. Chiusole
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@Sinaps It's just the proof of Euclid's Lemma in this case - see my answer. That's the way you should view it conceptually. – Bill Dubuque Mar 19 '20 at 13:49
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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.
Bill Dubuque
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