## Exhibit a pair of algebraic integers in QQ(sqrt(-5)) which are relatively prime but for which Bezout’s identity does not hold

Show that $3$ and $1 + \sqrt{-5}$ are relatively prime as algebraic integers in $\mathbb{Q}(\sqrt{-5})$, but that there do not exist algebraic integers $\lambda,\mu \in \mathbb{Q}(\sqrt{-5})$ such that $3\lambda + (1+\sqrt{-5})\mu = 1$.

Let $a+b\sqrt{-5}$ be an arbitrary integer in $\mathbb{Q}(\sqrt{-5})$. Since $-5 \equiv 3 \not\equiv 1$ mod 4, $a$ and $b$ are integers. Note that $N(a+b\sqrt{-5}) = a^2 + 5b^2$. If this element has norm 3, then taking this equation mod 5 we have $a^2 \equiv 3$. Note, however, that the squares mod 5 are 0, 1, and 4. In particular, no algebraic integer in $\mathbb{Q}(\sqrt{-5})$ has norm 3.

Now $N(3) = 9$ and $N(1+\sqrt{-5}) = 6$; if these elements have a factorization, then some factor must be a unit. Thus both $3$ and $1 + \sqrt{-5}$ are irreducible as integers in $\mathbb{Q}(\sqrt{-5})$. By Theorem 7.7 in TAN, the units in this ring are precisely $\pm 1$, so that 3 and $1 + \sqrt{-5}$ are not associates.

If $\delta|3$ and $\delta|(1+\sqrt{-5})$, then since $N(\delta) \neq 3$, $N(\delta) = \pm 1$, and thus by Lemma 7.3 in TAN $\delta$ is a unit. That is, every common divisor of $3$ and $1+\sqrt{-5}$ is a unit, so that these elements are relatively prime.

Suppose now that there are integers $a,b,c,d$ such that, with $\lambda = a+b\sqrt{-5}$ and $\mu = c+d\sqrt{-5}$, we have $3\lambda + (1+\sqrt{5})\mu = 1$.Comparing coefficients, we have $3a+c-5d = 1$ and $3b+c+d = 0$. Mod 3, these equations reduce to $c+d \equiv 1$ and $c+d \equiv 0$, a contradiction. So no such $\lambda$ and $\mu$ exist.