ATRACTOR

If \(a+bi\sqrt{5}\) is a divisor of \(6\) in \(\mathbb{Z}\left[i\sqrt{5}\right]\), there exists \(c+di\sqrt{5}\) such that \(\left(a+bi\sqrt{5}\right)\left(c+di\sqrt{5}\right)=6\).

hence \[\overline{\left(a+bi\sqrt{5}\right)}\overline{\left(c+di\sqrt{5}\right)}=\overline{6},\] so \[\left(a-bi\sqrt{5}\right)\left(c-di\sqrt{5}\right)=6.\]

From \[\left(a+bi\sqrt{5}\right)\left(c+di\sqrt{5}\right)=6\] and \[\left(a-bi\sqrt{5}\right)\left(c-di\sqrt{5}\right)=6\] we get \[\begin{array}{rcc} \left(a+bi\sqrt{5}\right)\left(a-bi\sqrt{5}\right)\left(c+di\sqrt{5}\right)\left(c-di\sqrt{5}\right) & = & 36\\ \left(a^{2}+5b^{2}\right)\left(c^{2}+5d^{2}\right) & = & 36 \end{array}\]

Since \(a^{2}+5b^{2}\) and \(c^{2}+5d^{2}\) are integers, \(a^{2}+5b^{2}\) is \(1,2,3,4,6,9,12,18\) or \(36\).

As \(a\) and \(b\) are integers, it is easy to see that \(2,3,12,18\) are excluded. It remains to examine \(1,4,6,9,36\).

• \(a^{2}+5b^{2}=1\) ; we can only have \(a=\pm1\), \(b=0\)

• \(a^{2}+5b^{2}=4\); we can only have \(a=\pm2\), \(b=0\)

• \(a^{2}+5b^{2}=6\); we can only have \(a=\pm1\), \(b=\pm1\)

• \(a^{2}+5b^{2}=9\); we can only have \(a=\pm2\), \(b=\pm1\) or \(a=\pm3\), \(b=0\)

• \(a^{2}+5b^{2}=36\); we can only have \(a=\pm6\), \(b=0\) or \(a=\pm4\), \(b=\pm2\)

So the only possible divisors of \(6\) are \[\begin{array}{c} 1,-1,2,-2,1+i\sqrt{5},1-i\sqrt{5},-1+i\sqrt{5},-1-i\sqrt{5},2+i\sqrt{5},2-i\sqrt{5},\\ -2+i\sqrt{5},-2-i\sqrt{5},3,-3,6,-6,4+2i\sqrt{5},4-2i\sqrt{5},-4+2i\sqrt{5},-4-2i\sqrt{5} \end{array}\]

By checking every case we see that the divisors of \(6\) are \(\pm1,\pm2,\pm\left(1+i\sqrt{5}\right),\pm\left(1-i\sqrt{5}\right)\)

(For example \(1+i\sqrt{5}\) is a divisor of \(6\) because \(\frac{6}{1+i\sqrt{5}}=1-i\sqrt{5}\in\mathbb{Z}\left[i\sqrt{5}\right]\);

\(2+i\sqrt{5}\) is not a divisor of \(6\) because \(\frac{6}{2+i\sqrt{5}}=\frac{6\left(2-i\sqrt{5}\right)}{9}=\frac{12-6i\sqrt{5}}{9}\notin\mathbb{Z}\left[i\sqrt{5}\right]\);)

What are the divisors of \(2+2i\sqrt{5}\) in \(\mathbb{Z}\left[i\sqrt{5}\right]\)?

If \(a+bi\sqrt{5}\) is a divisor of \(2+2i\sqrt{5}\) in \(\mathbb{Z}\left[i\sqrt{5}\right]\), there exists \(c+di\sqrt{5}\) such that \(\left(a+bi\sqrt{5}\right)\left(c+di\sqrt{5}\right)=2+2i\sqrt{5}\)

Then \[\overline{\left(a+bi\sqrt{5}\right)}\overline{\left(c+di\sqrt{5}\right)}=\overline{2+2i\sqrt{5}},\] hence \[\left(a-bi\sqrt{5}\right)\left(c-di\sqrt{5}\right)=2-2i\sqrt{5}\]

From \[\left(a+bi\sqrt{5}\right)\left(c+di\sqrt{5}\right)=2+2i\sqrt{5}\] and \[\left(a-bi\sqrt{5}\right)\left(c-di\sqrt{5}\right)=2-2i\sqrt{5}\] we get \[\begin{array}{rcc} \left(a+bi\sqrt{5}\right)\left(a-bi\sqrt{5}\right)\left(c+di\sqrt{5}\right)\left(c-di\sqrt{5}\right) & = & 24\\ \left(a^{2}+5b^{2}\right)\left(c^{2}+5d^{2}\right) & = & 24 \end{array}\]

Since \(a^{2}+5b^{2}\) and \(c^{2}+5d^{2}\) are integers, \(a^{2}+5b^{2}\) is \(1,2,3,4,6,8,12,24\) or \(36\).

As \(a\) and \(b\) are integers, it is easy to see that \(2,3,12\) are excluded. It remains to examine \(1,4,6,24\).

• \(a^{2}+5b^{2}=1\) ; we can only have \(a=\pm1\), \(b=0\)

• \(a^{2}+5b^{2}=4\) ; we can only have \(a=\pm2\), \(b=0\)

• \(a^{2}+5b^{2}=6\) ; we can only have \(a=\pm1\), \(b=\pm1\)

• \(a^{2}+5b^{2}=9\) ; we can only have \(a=\pm2\), \(b=\pm1\) or \(a=\pm3\), \(b=0\)

• \(a^{2}+5b^{2}=24\) ; we can only have \(a=\pm2\), \(b=\pm2\)

• \(a^{2}+5b^{2}=36\) ; we can only have \(a=\pm6\), \(b=0\) or \(a=\pm4\), \(b=\pm2\)

So the only possible divisors of \(2+2i\sqrt{5}\) are: \[\begin{array}{c} 1,-1,2,-2,1+i\sqrt{5},1-i\sqrt{5},-1+i\sqrt{5},-1-i\sqrt{5},\\ 2+i\sqrt{5},2-i\sqrt{5},-2+i\sqrt{5},-2-i\sqrt{5} \end{array}\]

By checking every case we see that the divisors of \(2+2i\sqrt{5}\) are \(\pm1,\pm2,\pm\left(1+i\sqrt{5}\right),\pm\left(2+2i\sqrt{5}\right)\)

(For example \(1+i\sqrt{5}\) is a divisor of \(2+2i\sqrt{5}\) because \(\frac{2+2i\sqrt{5}}{1+i\sqrt{5}}=2\in\mathbb{Z}\left[i\sqrt{5}\right]\);

\(1-i\sqrt{5}\) is not a divisor of \(2+2i\sqrt{5}\) because \(\frac{2+2i\sqrt{5}}{1-i\sqrt{5}}=\frac{\left(2+2i\sqrt{5}\right)\left(1+i\sqrt{5}\right)}{6}=\frac{-8+4i\sqrt{5}}{6}\notin\mathbb{Z}\left[i\sqrt{5}\right]\); )