Suppose $\displaystyle \mathbb{Q}(\sqrt{d})$ is a unique factorization domain, and $\displaystyle \alpha$ is an integer in $\displaystyle \mathbb{Q}(\sqrt{d})$ so that $\displaystyle \alpha$ and $\displaystyle \bar{\alpha}$ have no common factor, but $\displaystyle N(\alpha)$ (norm of alpha) is a perfect square in $\displaystyle \mathbb{Z}$. Show that $\displaystyle \alpha$ is a perfect square in the quadratic integers in $\displaystyle \mathbb{Q}(\sqrt{d})$.

I'm not sure if I'm approaching this correctly but if $\displaystyle \alpha$ and $\displaystyle \bar{\alpha}$ have no common factor, then $\displaystyle \alpha=\pi_{1}\pi_{2}\cdots\pi_{k}$ and $\displaystyle \bar{\alpha}=\pi'_{1}\pi'_{2}\cdots\pi'_{j}$ where $\displaystyle \pi_{i}$ and $\displaystyle \pi'_{i}$ are prime in $\displaystyle \mathbb{Q}(\sqrt{d})$. But $\displaystyle N(\alpha) = \alpha\bar{\alpha}=\pi_{1}\pi_{2}\cdots\pi_{k}\pi' _{1}\pi'_{2}\cdots\pi'_{j}=n^{2}$ where $\displaystyle n \in \mathbb{Z}$. Somehow I need to show that $\displaystyle \alpha=\beta^{2}$ where $\displaystyle \beta$ is a quadratic integer in $\displaystyle \mathbb{Q}(\sqrt{d})$ (i.e., $\displaystyle \alpha$ is a perfect square in the quadratic integers in $\displaystyle \mathbb{Q}(\sqrt{d})$).