Show the ideal $I=\langle4,2+2\sqrt{-29}\rangle$ in $\mathbb{Z}[\sqrt{-29}]$ satisfies the equality $\langle8\rangle=I^{2}$ of ideals in $\mathbb{Z}[\sqrt{-29}]$.
I tried to factorise $x^{2}+29$ over $\mathbb{F}_{2}$ but cannot manage to reach $I$ this way.
Thank you for any help in advance.
Note first that $I = (2)J$, where $J = (2, 1 + \sqrt{-29})$. Then it suffices to show $(2, 1 + \sqrt{-29})^2 = (2)$. If we square this ideal, we have $$J^2 = (4, 2 + 2\sqrt{-29}, -28 + 2\sqrt{-29}) = (2)(2, 1 + \sqrt{-29}, -14 + \sqrt{-29}).$$ We need to show that the ideal $(2, 1 + \sqrt{-29}, -14 + \sqrt{-29})$ is really all of $\mathbb{Z}[\sqrt{-29}]$. To see this, we just play with the generators: \begin{align*} (2, 1 + \sqrt{-29}, -14 + \sqrt{-29}) &= (2, 1 + \sqrt{-29}, \sqrt{-29})\\ &= (2,1,\sqrt{-29})\\ &= \Bbb{Z}[\sqrt{-29}]. \end{align*} Then the desired equality follows: $J^2 = (2)\mathbb{Z}[\sqrt{-29}] = (2)$, so $I^2 = (4)(2) = (8)$.