8. Find all integer pairs $(a, b)$ such that for any positive integer $n$ we have $n \mid\left(a^{n}+b^{n+1}\right)$.
(Chen Yonggao)
8. The integer pairs $(a, b)$ sought are only $(0,0)$ and $(-1,-1)$.

If one of $a, b$ is 0, obviously, the other must also be 0.
Assume $a b \neq 0$. Take a prime $p$ such that $p>\left|a+b^{2}\right|$.
By Fermat's Little Theorem, we have
$$
a^{p}+b^{p+1}=a+b^{2}(\bmod p) \text {. }
$$

Since $p \mid\left(a^{p}+b^{p+1}\right)$ and $p>\left|a+b^{2}\right|$, then
$$
a+b^{2}=0 \text {. }
$$

At this point, take a prime $q$ such that $q>|b+1|$, and $(q, b)=1$, and set $n=2 q$. Then
$$
\begin{array}{l}
a^{n}+b^{n-1}=\left(-b^{2}\right)^{2 q}+b^{2 q+1} \\
=b^{4 q}+b^{2 q+1}=b^{2 q+1}\left(b^{2 q-1}+1\right) .
\end{array}
$$

Since $n \mid\left(a^{n}+b^{n+1}\right)$ and $(q, b)=1$, we have
$$
q \mid\left(b^{2 q-1}+1\right) \text {. }
$$

Notice that,
$$
b^{2 q-1}+1 \equiv\left(b^{q-1}\right)^{2} b+1 \equiv b+1(\bmod q) .
$$

Since $q>|b+1|$, it must be that $b+1=0$, i.e., $b=-1$, and simultaneously, $a=-b^{2}=-1$.

In summary, the integer pairs sought are only $(0,0)$ and $(-1,-1)$.