In what follows let all variables be in some Euclidean Domain $R$.
- Suppose $n = kq + r$ s.t. $\deg(r) < \deg(k)$ or $r = 0$.
- Now suppose that $k \mid n$ so that $n = kq_1$ for some $q_1 \in R$.
Question 1: Since we have that $n = kq + r = kq_1 + 0$, can we somehow conclude that $r = 0$ and $q_1 = q$?
Question 2: If yes, do we even need $R$ to be a Euclidean Domain, or can it just be a UFD?
You have $$kq_1=n=kq+r$$ Rewrite as $$k(q_1-q)=r$$ But you also have some condition relating $r$ and $k$ involving degree. I'm a bit unsure what you mean by "deg" since in a general Euclidean domain it's a Euclidean function not degree; but whatever you've got implies that $q_1-q$ must be zero to avoid contradiction.
In answer to question 2, absent a Euclidean function you have no contradiction, since $k$ and $r$ might be associates, and $q-q_1$ a unit.