Show that $α$ is a perfect square in the quadratic integers in $\mathbb Q[\sqrt{d}]$ Question:
Suppose that $\mathbb Q[\sqrt{d}]$ is a UFD, and $α$ is an integer in $\mathbb Q[\sqrt{d}]$ so that $α$ and $\barα$ have no common factor, but $N(α)$ is a perfect square in $\mathbb Z$. How can I show that $α$ is a perfect square in the quadratic integers in $\mathbb Q[\sqrt{d}]$?
What I have Done:
I'm not sure if I'm approaching this correctly but if  $\alpha$ and  $\bar{\alpha}$ have no common factor, then $\alpha=\pi_{1}\pi_{2}\cdots\pi_{k}$ and  $\bar{\alpha}=\pi'_{1}\pi'_{2}\cdots\pi'_{j}$ where $ \pi_{i}$ and  $\pi'_{i} $ are prime in  $\mathbb{Q}(\sqrt{d})$. But $N(\alpha) = \alpha\bar{\alpha}=\pi_{1}\pi_{2}\cdots\pi_{k}\pi'  _{1}\pi'_{2}\cdots\pi'_{j}=n^{2}$ where  $n \in \mathbb{Z}$. Somehow I need to show that $\alpha=\beta^{2}$ where  $\beta$ is a quadratic integer in $ \mathbb{Q}(\sqrt{d})$ (i.e.,  $\alpha$ is a perfect square in the quadratic integers in  $\mathbb{Q}(\sqrt{d})$)
 A: [I guess you mean $\mathbb Z[\sqrt d]$ instead of $\mathbb Q[\sqrt d]$. The latter is a field, and it's not really meaningful to talk about unique factorisation in a field.]
You can complete your strategy by proving that (with your notation) the $\pi_i$ and $\pi_j'$ are different.
Lemma. Let $A$ be a UFD and $x$, $y$ be two coprime elements such that $xy$ is associated to a square. Then $x$ and $y$ are associated to squares.
Proof. If $x = \epsilon\, p_1^{e_1} \cdots p_r^{e_r}$ and $y = \eta \,q_1^{f_1}\cdots q_s^{f_s}$ are their decompositions in irreducible elements, (with $\epsilon, \eta \in A^\times$) then the $p_i$ and the $q_j$ are disjoint (because $x$ and $y$ are coprime) and the decomposition of $xy$ is then $xy = \epsilon\,\eta\, p_1^{e_1} \cdots p_r^{e_r}q_1^{f_1}\cdots q_s^{f_s}$.
Because $xy$ is a associated to a square, all the $e_i$ and all the $f_j$ are even integers, which proves that
$$x = \epsilon\, \left( p_1^{e_1/2} \cdots p_r^{e_r/2}\right)^2 \qquad \text{and}\qquad y = \eta \,\left( q_1^{f_1/2} \cdots q_s^{f_s/2}\right)^2.$$
This answers your question.
And to answer Alonso del Arte's concern: $\alpha = (1+2i)^2 \in \mathbb Q(i)$ works. Then $\alpha$ and $\overline{\alpha} = (1-2i)^2$ are coprime ($1 \pm 2i$ are irreducible elements and they aren't associated because $\pm 1$ and $\pm i$ are the only units) and $\alpha \overline{\alpha} = \left( (1+2i)(1-2i) \right)^2 = 25.$ You can do such examples every time a prime number $p$ (5 in this example) splits in $\mathbb Z[\sqrt d]$.
