How to prove that $\langle 3+8\sqrt{2},7\rangle = \langle 3+\sqrt{2}\rangle$ in the ring $\mathbb{Z}[\sqrt{2}]$? How to prove that the ideals $\langle 3+8\sqrt{2},7\rangle$ and $\langle 3+\sqrt{2}\rangle$ are equal in the ring $\mathbb{Z}[\sqrt{2}]$?
I tried using the factors and reducing to the form of the other and vice versa, but it did not work.
If there was not two factors involved in the first I would've tried the associativity of generators.
A hint would be appreciated to find out.

Hint: show that each contains the generators of the other.
Solution:

 Note that $7=(3+\sqrt{2})(3-\sqrt{2})$, so $7 \in (3+\sqrt{2})$, hence $7\sqrt{2}$, and $3+\sqrt{2}+7\sqrt{2}=3+8\sqrt{2}\in(3+\sqrt{2})$. Thus $(7,3+8\sqrt{2})\subset (3+\sqrt{2})$. Conversely, since $7\in(7,3+8\sqrt{2})$, $3+8\sqrt{2}-7\sqrt{2} = 3+\sqrt{2}\in (7,3+8\sqrt{2})$. Hence $(3+\sqrt{2}) \subset (7,3+8\sqrt{2})$ as well. Thus $(7,3+8\sqrt{2})=(3+\sqrt{2})$. 

