ATRACTOR

Something similar in \(\mathbb{Z}\left[i\sqrt{5}\right]\) to what has been done in \(\mathbb{Z}\left[i\right]\) and \(\mathbb{Z}\left[\sqrt{2}\right]\) would be to consider a function \(v\) defined by \(v(a+bi)=a^{2}+5b^{2}\). We would like, for any \(a+bi\sqrt{5}\), \(c+di\sqrt{5}\), to find \(q_{1}+q_{2}\sqrt{5}\) and \(r_{1}+r_{2}\sqrt{5}\) such that: \[a+bi\sqrt{5}=\left(c+di\sqrt{5}\right)\left(q_{1}+q_{2}i\sqrt{5}\right)+r_{1}+r_{2}i\sqrt{5},\] with \(v\left(r_{1}+r_{2}i\sqrt{5}\right)<v\left(c+di\sqrt{5}\right)\) or \(r_{1}+r_{2}i\sqrt{5}=0\).

Similar calculations to the ones for \(\mathbb{Z}\left[i\right]\) and \(\mathbb{Z}\left[\sqrt{2}\right]\) make us look for \(q_{1}+q_{2}\sqrt{5}\) and \(r_{1}+r_{2}\sqrt{5}\) such that: \[\frac{r_{1}+r_{2}i\sqrt{5}}{c+di\sqrt{5}}=\left(\frac{ac+5bd}{c^{2}+5d^{2}}-q_{1}\right)+\left(\frac{bc-ad}{c^{2}+5d^{2}}-q_{2}\right)i\sqrt{5}\] and \[v\left(\frac{r_{1}+r_{2}i\sqrt{5}}{c+di\sqrt{5}}\right)<1.\]

Now, the best we can guarantee (once that \(q_{1}\) and \(q_{1}\) are integers) is that \[\left|\frac{ac+5bd}{c^{2}+5d^{2}}-q_{1}\right|\leq\frac{1}{2}\] and \[\left|\frac{bc-ad}{c^{2}+5d^{2}}-q_{2}\right|\leq\frac{1}{2}.\]

But then we can only guarantee that \[\begin{array}{cl} v\left(\frac{r_{1}+r_{2}i\sqrt{5}}{c+di\sqrt{5}}\right) & \leq\left(\frac{ac+5bd}{c^{2}+5d^{2}}-q_{1}\right)^{2}+5\left(\frac{bc-ad}{c^{2}+5d^{2}}-q_{2}\right)^{2}\\ & \leq\left(\frac{1}{2}\right)^{2}+5\left(\frac{1}{2}\right)^{2}=\frac{3}{2}, \end{array}\] which is not enough.