Now, let me show off some of the nice number theory.
From the point of view of Kummer and his "ideal numbers", it is not hard to verify that $2$ behaves like "the square of a prime" in the sense that it satisfies the following two properties in $\mathbb{Z}[\sqrt{-5}]$:
- If $x|a^2b^2$, then $x|a^2$ or $x|b^2$, for all $a,b\in\mathbb{Z}[\sqrt{-5}]$;
- There exists $a\in\mathbb{Z}[\sqrt{-5}]$ such that $x|a^2$ but $x$ does not divide $a$.
In the integers, these two properties together characterize the integers that are squares of primes; there is no such prime in $\mathbb{Z}[\sqrt{-5}]$ whose square is $2$, but let's ignore that annoying fact and "invent" an "ideal" prime number ('ideal' as in "imaginary") $\alpha$ such that $\alpha^2=2$.
Similarly, each of $3$, $1+\sqrt{-5}$, and $1-\sqrt{-5}$ behaves like the "product of two distinct primes", in the sense that they satisfy the following three properties:
- If $x|abc$, then $x$ divides at least one of $ab$, $ac$, and $bc$, for all $a,b,c\in\mathbb{Z}[\sqrt{-5}]$.
- If $x|a^2$, then $x|a$ for all $a\in\mathbb{Z}[\sqrt{-5}]$.
- There exist $a,b\in\mathbb{Z}[\sqrt{-5}]$ such that $x$ divides $ab$, but does not divide $a$ and does not divide $b$.
Again, these three conditions would characterize integers that are products of two distinct primes in $\mathbb{Z}$, but no such elements exist in $\mathbb{Z}[\sqrt{-5}]$. We again "invent" such primes. Since we are hoping that unique factorization might hold in terms of these "ideal" primes, and since $2\times 3 = (1+\sqrt{-5}) \times (1-\sqrt{-5})$ this suggests that there are two of these "ideal primes", $\beta$ and $\gamma$, with $3=\beta\gamma$, $1+\sqrt{-5}=\alpha\beta$, and $1-\sqrt{-5}=\alpha\gamma$. Though it is not easy, one can verify that, at least as far as divisibility is concerned, everything works out nicely, so long as we don't actually go looking for these "ideal primes". Some products of these ideal primes will yield actual numbers, some products will yield "ideal" numbers. (Doing lots of experimentation may even reveal that any product of an even number of "ideal primes" is an actual number, whereas products of an odd number of "ideal primes" is never an actual number).
If this is the case, then $3(1+\sqrt{-5}) = \alpha\beta^2\gamma$, and $3(1-\sqrt{-5})=\alpha\beta\gamma^2$. So the greatest common divisor in terms of these ideal primes will be $\alpha\beta\gamma$. That means that each of $\alpha\beta=1+\sqrt{-5}$, $\alpha\gamma=1-\sqrt{-5}$, and $\beta\gamma=3$ will be common divisors of $3(1+\sqrt{-5})$ and $3(1-\sqrt{-5})$. And indeed, $3$ clearly divides both; $(1+\sqrt{-5})$ divides the first, and $(1+\sqrt{-5})(-2-\sqrt{-5})=3-3\sqrt{-5} = 3(1-\sqrt{-5});$ and similarly, $1-\sqrt{-5}$ divides the second and $(1-\sqrt{-5})(-2+\sqrt{-5}) = 3+\sqrt{-5} = 3(1+\sqrt{-5}).$ So the common divisors in $\mathbb{Z}[\sqrt{-5}]$ are $\pm 1$, $\pm 3$, $\pm(1+\sqrt{-5})$, and $\pm(1-\sqrt{-5})$. (The product of three ideal primes is not an actual number, so we can find no actual counterpart to $\alpha\beta\gamma$ in $\mathbb{Z}[\sqrt{-5}]$). None of them is a multiple of all the others.
In terms of the ideal class group, from the point of view of Dedekind and Kronecker, the ring $\mathbb{Z}[\sqrt{-5}]$ is a Dedekind domain; every ideal can be written as a product of prime ideals in a unique way. It is not hard to show that: $\begin{align*} (2) &= (2,1+\sqrt{-5})^2,& (3)&=(3,1+\sqrt{-5})(3,1-\sqrt{-5}),\\ (1+\sqrt{-5}) &= (2,1+\sqrt{-5})(3,1+\sqrt{-5}) & (1-\sqrt{-5})&=(2,1-\sqrt{-5})(3,1-\sqrt{-5}) \end{align*}$ Thus, $\begin{align*} (3)(1+\sqrt{-5}) &= (2,1+\sqrt{-5})(3,1+\sqrt{-5})^2(3,1-\sqrt{-5})\\ (3)(1-\sqrt{-5}) &= (2,1+\sqrt{-5})(3,1+\sqrt{-5})(3,1-\sqrt{-5})^2. \end{align*}$ So the gcd, in terms of ideals, is the (nonprincipal) ideal $(2,1+\sqrt{-5})(3,1+\sqrt{-5})(3,1-\sqrt{-5}).$ Since the ideal class group of $\mathbb{Z}[\sqrt{-5}]$ is cyclic of order $2$, any product of two nonprincipal ideals is principal; the three possible products of two ideal factors of the gcd above yield the three non-unit common divisors mentioned above.