I am trying to understand this answer from a previous post. Particularly the last line. Why $2\mid d$ and $d\mid 4$ follows that $d\in\{\pm 1, \pm 2\}$? I think it should be $d\in\{\pm 1, \pm 2 \pm 4\}$ and therefore if $d=4$, $d$ could satisfy $(1+\sqrt{-3})\mid d$
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1Why not post a comment directly to that answer so the author can reply? – lulu May 16 '18 at 23:26
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2Well, you’re talking about natural integers ($\Bbb Z$) only. So no radicals. But $2$ does not divide $d=\pm1$. – Lubin May 17 '18 at 01:02
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@Lubin then $d$ should only be $\pm 2$? Why not $\pm 4$, since $2|4$ and $4|4$? – John Keeper May 17 '18 at 12:02
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1I didn’t say that at all. Of course $\pm2$ and $\pm4$ are the possibilities. – Lubin May 17 '18 at 13:00
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Then that answer is wrong, right? Since if $d\in{\pm 2 \pm 4}$ ,$d$ could satisfy $(1+\sqrt{-3})\mid d$ for $d=4$. – John Keeper May 17 '18 at 17:01