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Thread: An example of 2 quadratic integers in the same quadratic field

  1. #1
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    An example of 2 quadratic integers in the same quadratic field

    What's an example of two quadratic integers in the same quadratic field for which N(a)|N(b), yet a does not divide b?
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  2. #2
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    Re: An example of 2 quadratic integers in the same quadratic field

    Fix a squarefree integer $\displaystyle D$. Then $\displaystyle \mathbb{Q}(\sqrt{D})$ is a quadratic field. So, $\displaystyle a,b\in \mathbb{Z}(\sqrt{D})\subseteq \mathbb{Q}(\sqrt{D})$. With $\displaystyle N(a)|N(b)$ gives a relationship between them. Let $\displaystyle \alpha_1,\alpha_2,\beta_1,\beta_2 \in \mathbb{Z}$ such that $\displaystyle a = \alpha_1 + \beta_1\sqrt{D}, b = \alpha_2 + \beta_2\sqrt{D}$. Then $\displaystyle N(a) = \alpha_1^2 - D\beta_1^2, N(b) = \alpha_2^2 - D\beta_2^2$. There exists $\displaystyle k \in \mathbb{Z}$ such that $\displaystyle \alpha_2^2-D\beta_2^2 = (\alpha_1^2-D\beta_1^2)k$. If $\displaystyle a$ does not divide $\displaystyle b$, then there exists a unique $\displaystyle q,r\in \mathbb{Z}(\sqrt{D})$ with $\displaystyle b=aq+r$ and $\displaystyle 0<N(r)<N(a)$ (note that $\displaystyle N(r) \neq 0$, as this would imply divisibility). Use these relationships to find an example.
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