Uniqueness of the remainder and quotient in an Euclidean domain 
Let $R$ be a Euclidean ring with Euclidean norm $N$. Let $a,b\in R\setminus\{0\}$ and let $q,r\in R$  such that $a=bq+r$ with $r=0$ or $N(r)<N(b)$.
  Prove that $r$ and $q$ are unique if and only if $N(a+b)\le\max\{N(a),N(b)\}$.

I do not see how to relate both.
If I suppose the second:
let $q'$ and $r'$ be  such that 
$a=bq'+r'$ with $r'=0$ or $N(r')<N(b)$ 
It is sufficient to prove that $r'=r$, because in this case $a=bq'+r'=bq+r$ implies $bq'=bq$ and then $q'=q$.
How $bq'+r'=bq+r$ then  $r'-r=b(q-q')$.
By the properties of Euclidian norm:
$$
N(r')-N(r)\le N(r)<N(b)
\quad\text{and}\quad
N(r-r')=N(b(q-q'))\ge N(b)
$$
I got other inequalities but I do not think they are any good.
For the converse I have no idea.
 A: We can assume that $N$ satisfies the property $N(a)\le N(ab)$ for all $a,b\in R\setminus \{0\}$ (for details check this).
Using the above property it's easy to prove that $ N(x)=N (x') $ if $ x $ and $ x'$ are associates. 
$\Longrightarrow$) We're going to prove it by contrapositive. Let's suppose that $N(a+b)>\max\{N(a),N(b)\}$, then $a+b\neq 0$ and so we can divide $a$ by $a+b$. We have $$a=(a+b)0+a$$ with $N(a)<N(a+b)$ and $$a=(a+b)1+(-b)$$ with $N(-b)=N(b)<N(a+b)$.
Therefore the quotient and the remainder are not unique.
$\Longleftarrow$) As you noted, it's enough to prove that $q=q'$ or $r=r'$. If $r=0$ or $r'=0$ the result follows immediately. So, from now on we assume that $r,r'\neq 0$ and we're going to prove that $q=q'$. If $r=r'$ we're done. Otherwise, let's suppose that $q\neq q'$. Since $b(q-q')=r'-r$ we have $$N(b)\le N(b(q-q'))=N(r'-r)=N(r-r')\le \max\{N(r),N(-r')\}<N(b).$$ 
So we got $N(b)<N(b)$, an absurd. Hence, it must be $q=q'$ and therefore $r=r'$.
$-------------------------------------$
The above result is used to characterize the euclidean domains having unique quotient and remainder. Check for example these papers:
i) A Characterization of Polynomial Domains Over a Field by Tong-Shieng Rhai
ii) Uniqueness in the Division Algorithm by M. A. Jodeit, Jr.
