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How can I prove that there is no $x\in \mathbb{Z}[\sqrt{-5}]$ such that $x\cdot (2+\sqrt{5}i)= 3$ ?

New2Math
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2 Answers2

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Write $x=a+ib\sqrt{5}$. You have
$$x.(2+\sqrt{5}i)=(a+ib\sqrt{5})(2+\sqrt{5}i)=(2a-5b)+i\sqrt{5}(a+2b)$$

This is equal to $3$ iff $a+2b=0$ and $2a-5b=3$. This is easy to see that there are no solution in $\mathbb{Z}$.

TheSilverDoe
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  • Easier - simply rationalize the denominator - see my comment on the other answer. Please strive to close dupes of FAQs, not add more dupe answers. – Bill Dubuque Nov 05 '20 at 21:26
  • I am sorry, @BillDubuque, king of duplicates. – TheSilverDoe Nov 06 '20 at 07:30
  • I cannot delete it because people already answered it, why you are so mad you all know that searching for duplicates on this site is impaired – New2Math Nov 06 '20 at 08:38
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Hint: If $(a+b\sqrt{5}i)(2+\sqrt{5}i)=3$, then $a^2+5b^2=1$ and so $a=\pm1$ and $b=0$.

lhf
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  • Why the downvote? – lhf Nov 05 '20 at 21:10
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    Not my downvote, but maybe they don't see how to finish (or maybe they don't like the question quality, or maybe they think experienced users shouldn't be answering dupes of trivial FAQs). To finish note that the hint $\Rightarrow, 3\mid 2+\sqrt{-5},,$ contradiction. This is clearer by rationalizing the denominator $, v/w = (v\bar w)(w\bar w),$ yielding $$ 2!+!\sqrt{-5}\mid 3\ \ {\rm in}\ \ \Bbb Z[\sqrt{-5}]\ \Rightarrow\ \dfrac{3}{2!+!\sqrt{-5}} = \dfrac{2!-!\sqrt{5}}3\in \Bbb Z[\sqrt{-5}]\ \Rightarrow!\Leftarrow$$ – Bill Dubuque Nov 05 '20 at 21:20
  • @BillDubuque Your argument is indeed quicker and maybe easier, but to apply it, you go outside from $\mathbb{Z}[\sqrt{-5}]$ (in $\mathbb{C}$, or at least in $\mathbb{Q}[\sqrt{-5}]$)... In my opinion, this is less elegant than proving directly the result. But it works well, though ! (and not my downvote either on lhf's answer) – TheSilverDoe Nov 06 '20 at 07:41
  • @TheSilverDoe It's trivial to eliminate the use of fractions: rewrite is as $,w\mid n\iff w\bar w\mid n\bar w.,$ Then rat'ing the denom becomes rat'ing the divisor. More generally see the method of simpler multiples. $\ \ $ – Bill Dubuque Nov 06 '20 at 09:05