Show that $(3, 1+\sqrt{-5}) \neq \mathbb{Z}[\sqrt{-5}]$ Show that it is not possible to find "integers" $x, y$ of this kind which satisfy the equation $3x-(1+\sqrt{-5})y=1$, even though $3$ and $1+\sqrt{-5}$ are "primes", and therefore their greatest common divisor is 1.
My effort: I let  $x$ be $a+b\sqrt{-5}$ and $y$ be $c+d\sqrt{-5}$ and then substituted this into $3x-(1+\sqrt{-5})y$.
I ended up with this being $(3a-c+5d)+\sqrt{-5}(3b-d-c)$, and for this to be $1$, I postulated that $(3a-c+5d)$ must equal $1$ and $\sqrt{-5}(3b-d-c)$ must equal $0$.
I end up with two equations with four unknowns, giving infinite solutions. What have I done wrong? Also, what is the significance of this question in terms of learning the difference between the natural numbers and "integers" of this kind?
Thanks.
 A: Steven's comment is a good way to finish your solution. Here's an approach that will help with understanding the significance of your question: Multiply both sides by $1 - \sqrt{-5}$ to get
$$3x - (1 + \sqrt{-5})y = 1 \quad\implies\quad 3(1 - \sqrt{-5})x - 6y = 1 - \sqrt{-5}$$
The left-hand side can be factored as $3[(1 - \sqrt{-5})x - 2y]$. For any choice of "integers" $x$ and $y$, the quantity in brackets will be an "integer" $a + b\sqrt{-5}$, so the left-hand side is $3$ times an "integer", i.e. $3a + 3b\sqrt{-5}$. The right-hand side isn't of this form, so the equation has no solution.
These "integers" are the elements of the ring $\mathbb{Z}[\sqrt{-5}] = \{a + b\sqrt{-5} : a, b \in \mathbb{Z}\}$. You can think of rings like $\mathbb{Z}[\sqrt{-5}]$ as analogous to the integers $\mathbb{Z}$ in that you can add/subtract/multiply two elements of the ring and get another element in the ring. But this question shows that Bézout's identity holds in $\mathbb{Z}$ but not in $\mathbb{Z}[\sqrt{-5}]$. What this means is that you can't think about primes and factorization the same way. The essential difference between $\mathbb{Z}$ and $\mathbb{Z}[\sqrt{-5}]$ is that prime factorization is unique in $\mathbb{Z}$ but not in $\mathbb{Z}[\sqrt{-5}]$; note that our solution above relied on the fact that $6$ can be factored two different ways: $$6 = 3 \cdot 2 = (1 + \sqrt{-5}) (1 - \sqrt{-5})$$ If you study algebraic number theory, you'll learn how to salvage this situation by defining primes in $\mathbb{Z}[\sqrt{-5}]$ in a slightly different way so that prime factorization is unique. This allows us to do number theory in rings other than $\mathbb{Z}$, which is interesting in its own right and also helps us solve many classical problems over the integers, for example determining which prime numbers are expressible in the form $x^2 + ny^2$.
A: Hint: $\,\ w := 1+\sqrt{-5}\,\Rightarrow\, w\bar w = \color{#7cf}6\,\Rightarrow\, \color{#c00}{(w,3)(\bar w,3)} = \color{#7cf}3(\color{#7cf}2,w,\bar w,3) = \color{#c00}{(3)},\,$ but if $\,(3,w)=(1)\,$ then conjugating yields $\,(3,\bar w)=(1)\,$ so $\,(w,3)(\bar w,3) = \color{#c00}{(1)},\,$ contra above (more detail here).
Similarly $\, \color{#0a0}{(w,2)(w,3)=(w)},\ \ \color{#90f}{(\bar w,2)(\bar w,3) = (\bar w)},\ \ (w,2)(\bar w,2) = (2)$
Using the above, the nonunique factorization $\, \color{#0a0}w\color{#90f}{\bar w} = 2\cdot\color{#c00}3\,$ of numbers can refined to a unique factorization of ideals using the below ideal analog of $\rm 4NT$ = Four Number theorem.
$$  \overbrace{\color{#0a0}{(w,2)(w,3)}}^{\textstyle (\color{#0a0}w)}\overbrace{(\color{#90f}{\bar w,2)(\bar w,3)}}^{\textstyle (\color{#90f}{\bar w})} \,=\, \overbrace{(w,2)(\bar w,2)}^{\textstyle (\color{c00}2)}\overbrace{\color{#c00}{(w,3)(\bar w,3)}}^{\textstyle (\color{#c00}3)}$$
If those ideals were all principal $(x,y) = (z)$ then we could read those ideals as number gcds $\gcd(x,y)\approx z\,$ and then the above refinement is just the usual Four Number Theorem (which is true in any UFD or GCD domain, and yields uniqueness of factorizations into irreducibles).
So the significance of the nonprincipality of $(3,w)$  is that it implies that we have to pass from number gcds to ideal gcds in order to regain unique factorization.
Note: the gcd of ideals $I,J\,$ is defined to be $\,I+J,\,$ so $\,(w,2) = (w)+(2)\,$ is the ideal gcd of $(w)$ and $(2),\,$ etc. The reason for this is that the ideal sum satisfies the universal property of the gcd in rings where contains = divides for ideals (e.g. PIDs and Dedekind domains), viz.
$$\begin{align} A\supseteq B,C &\iff A\supseteq B+C,\qquad\ \text{universal property of ideal sum}\\[.2em]
\rightsquigarrow\ \ A\ \mid\ B,C&\iff A\ \mid\  B+C,\qquad\ \text{when contains = divides}\\[.2em] 
A\ \mid\ B,C&\iff A\ \mid \gcd(B,C),\ \ \ \ \text{universal property of gcd}\end{align}$$
as explained in the prior linked answer.
