How to show that $2+\sqrt{5}i\not | 3$?

Question

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

Answer 1

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 $\mathbb{Z}$.

Answer 2

Hint: If $(a+b\sqrt{5}i)(2+\sqrt{5}i)=3$, then $a^2+5b^2=1$ and so $a=\pm1$ and $b=0$.