Prove that in the ring $Z[\sqrt2]$: $\langle3+8\sqrt2, 7\rangle$ = $\langle3+\sqrt2\rangle$ So, 
I know that $\langle a\rangle$ = $aR$ and $\langle a,b\rangle$ = $aR + bR$ for any ring R.
I then multiplied each side by $\sqrt2$ and computed them.
These are the steps that I took:
$\langle3+8\sqrt2, 7\rangle$ = $\langle3+\sqrt2\rangle$
$(3+8\sqrt2)\sqrt2 + 7\sqrt2 = (3+\sqrt2)\sqrt2$
$3\sqrt2 + 16 + 7\sqrt2 = 3\sqrt2 + 2$
$16 + 10\sqrt2 = 2+3\sqrt2$
I'm not really sure where to go from here.
Any help would be much appreciated. 

Hint: To show $\ \langle 3+8\sqrt{2}, 7\rangle=\langle 3+\sqrt{2}\rangle\ $, you need to show:


*

*$\ 3+8\sqrt{2} = r_1\left(3+\sqrt{2}\right)\ $ and $\ 7 = r_2\left(3+\sqrt{2}\right)\ $ for some $\ r_1\ $ and $\ r_2\ $ in $\ \mathbb{Z}\left[\sqrt{2}\right]\ $; and

*$\ 3+\sqrt{2}=s_1\left(3+8\sqrt{2}\right) + s_2 7\ $ for some $\ s_1\ $ and $\ s_2\ $ in $\ \mathbb{Z}\left[\sqrt{2}\right]\ $.


The first part is very straightforward: simply calculate $\ r_1=\frac{3+8\sqrt{2}}{3+\sqrt{2}}\ $ and $\ r_1=\frac{7}{3+\sqrt{2}}\ $ in $\ \mathbb{Q}\left[\sqrt{2}\right]\ $, and check to see whether the results both lie in $\ \mathbb{Z}\left[\sqrt{2}\right]\ $.  The second is not much more difficult.  A solution for $\ s_1\ $ a small integer, and $\ s_2\ $ a small multiple of $\ \sqrt{2}\ $, is fairly easy to find by trial and error.
