8. Given that $a>b$ are positive integers. Let the quotient when $a^{2}+b^{2}$ is divided by $a+b$ be $q$, and the remainder be $r$. If $q^{2}+2 r=2020$, then the ordered pair of real numbers $(a, b)=$ $\qquad$ .
8. $(52,29)$ or $(53,15)$.

Notice that, for positive integers $a>b$, we have $(a+b)^{2}>a^{2}+b^{2} \geqslant \frac{(a+b)^{2}}{2}$.
Thus, $\frac{a+b}{2} \leqslant q2020$
$\Rightarrow q=44$ or $q=43$.
Upon inspection, when $q=43$, $2 r=2020-43^{2}=$
171 is odd, which is a contradiction; when $q=44$, $r=42$.
Therefore, $a^{2}+b^{2}=44 a+44 b+42$
$\Rightarrow(a-22)^{2}+(b-22)^{2}=1010$.
By enumeration, we get $31^{2}+7^{2}=1010$.
Also, $a+b>42$, hence, there are two solutions that satisfy the requirements: $(a, b)=(53,29)$ or $(53,15)$.