= Solution
The multiplicative <field norm> on this ring is
$$
N(a+b\sqrt{-5})=(a+b\sqrt{-5})(a-b\sqrt{-5})=a^2+5b^2.
$$
If an element is a <unit>, its positive integer norm has a multiplicative inverse in the positive integers, hence is one. The only integer solutions of $a^2+5b^2=1$ are $(a,b)=(1,0),(-1,0)$. Therefore \b[the units are exactly $1$ and $-1$].
The standard failure of <unique factorization domain>[unique factorization] is
$$
6=2\cdot3=(1+\sqrt{-5})(1-\sqrt{-5}).
$$
There is no element of norm two or three: $b\ne0$ gives norm at least five, while $b=0$ gives a square. An alleged nontrivial factorization of $2$ (norm four), $3$ (norm nine), or $1\pm\sqrt{-5}$ (norm six) would require a factor of norm two or three. Thus all four displayed factors are <irreducible elements>. No factor in the first factorization is associated to a factor in the second, since associates have the same norm and the norms are respectively four, nine and six. Hence \b[$\mathbb Z[\sqrt{-5}]$ is not a <unique factorization domain>].
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