I have $$7+i,\quad\quad 5+3i\in\mathbb{Z}[i]$$
The answer is given as $$7+i=(5+3i)\cdot 1+(2-2i)$$
$$5+3i = (2-2i)\cdot 2i+(1-i)$$
$$2-2i=(1-i)\cdot 2$$
So $hcf(7+i,5+3i) = 1-i$
What I don't understand is how they're finding what to multiply by, like in the first line, they multiply $5+3i$ by $1$, in the second line they multiply $2-2i$ by $2i$. I tried using the fraction $\frac{7+i}{5+3i}$, and then rationalising the denominator but it got complicated and didn't look anything like what they've gotten. Could someone explain their steps in more detail?
related
column on the RHS of this page. But apparently the MSE search function doesn't know how to find $\mathbb{Z}[i],$, either. – dxiv May 10 '17 at 00:58