Why is the gcd of $61+35\sqrt{3} $ and $170+32\sqrt{3}$ is $19 + 11\sqrt{3}$? As a completion or correction to the solution of my question here:
Find gcd of $a = 170 + 32\sqrt{3}$ and $b = 61 + 35\sqrt{3}.$ Then find $f,g \in \mathbb{Z}[\sqrt{3}]$ such that $af + bg = d$ using norm function.
I was told by my professor that:  the gcd of $61+35\sqrt{3} $ and $170+32\sqrt{3}$ is $19 + 11\sqrt{3}$ and that we should not use the norm function and that we should use the ordinary procedure for calculating the gcd to get it. And my professor added that you will even get that  $$19 + 11\sqrt{3} = (61 + 35\sqrt3) (718 - 413\sqrt{413}) +(170+32\sqrt{3})(-3 + \sqrt{3}),$$ As I mentioned in the above post I do not know how to complete after this step:
$$\frac{170 + 32 \sqrt{3}}{61 + 35 \sqrt{3}} = \frac{3505}{23} - \frac{1999 \sqrt{3}}{23},$$ specifically could anyone show me what should be the remainder in my previous line please?
 A: Division with remainder for number of these forms should be done as follows:
$$\frac{170 + 32 \sqrt{3}}{61 + 35 \sqrt{3}} = \frac{3505}{23} - \frac{1999 \sqrt{3}}{23}$$
This would be the fraction. Now to get the integral quotient, take the closest integer for the coefficients of the rational and irrational par
$$\frac{3505}{23} =152 + \frac{9}{23} \\
\frac{1999}{23} = 86 + \frac{21}{23} = 87 -\frac{2}{23}$$
Therefore, the integral quotient is
$$152 - 87 \sqrt{3}$$
and so
$$170 + 32 \sqrt{3} = (61 + 35 \sqrt{3})(152- 87 \sqrt{3})+ 33 + 19 \sqrt{3}$$
So $33 + 19 \sqrt{3}$ is the remainder.
Now divide $61 + 35 \sqrt{3}$ by $33 + 19 \sqrt{3}$. We have
$$\frac{61 + 35 \sqrt{3}}{33 + 19 \sqrt{3}} = 3 - \frac{2}{3}\sqrt{3}$$.  Therefore the integral quotient is
$3 - \sqrt{3}$.  The remainder is
$$(33 + 19 \sqrt{3}) \cdot\frac{\sqrt{3}}{3}=19 + 11 \sqrt{3}$$
Notice that the last remainder $19 + 11 \sqrt{3}$ divides the last divider $33 + 19 \sqrt{3}$. Therefore, we stop here, the $\gcd$ is $19 + 11 \sqrt{3}$.
Note:  the division process is very similar to the usual division for integers.
