8. Let $\varphi(n)$ denote the Euler's function, then all positive integer pairs $(m, n)=$ $\qquad$ such that $\varphi(m n)=\varphi(m)+\varphi(n)$ holds.
8. $(2,2),(3,4),(4,3)$.

Notice,
$$
\begin{array}{l}
\frac{\varphi(m n)}{m n}=\prod_{p \mid m n}\left(1-\frac{1}{p}\right) \\
=\frac{\prod_{p \mid m}\left(1-\frac{1}{p}\right) \prod_{p \mid n}\left(1-\frac{1}{p}\right)}{\prod_{p \mid(m, n)}\left(1-\frac{1}{p}\right)} \\
=\frac{\frac{\varphi(m)}{m} \cdot \frac{\varphi(n)}{n}}{\varphi(d)} .
\end{array}
$$

Let $d=(m, n)$. Then
$$
\begin{array}{l}
\varphi(m n)=\frac{d \varphi(m) \varphi(n)}{\varphi(d)}=\varphi(m)+\varphi(n) \\
\Rightarrow \frac{d}{\varphi(d)}=\frac{1}{\varphi(m)}+\frac{1}{\varphi(n)} \\
\Rightarrow d=\frac{\varphi(d)}{\varphi(m)}+\frac{\varphi(d)}{\varphi(n)} \\
\Rightarrow d=\frac{1}{a}+\frac{1}{b}\left(a=\frac{\varphi(m)}{\varphi(d)}, b=\frac{\varphi(n)}{\varphi(d)} \in \mathbf{Z}_{+}\right) .
\end{array}
$$

Therefore, only $(a, b, d)=(1,1,2)$ or $(2,2,1)$.
When $(a, b, d)=(1,1,2)$,
$$
\varphi(m)=\varphi(n)=1 \Rightarrow m=n=2 \text {; }
$$

When $(a, b, d)=(2,2,1)$,
$$
\begin{array}{l}
\varphi(m)=\varphi(n)=2 \\
\Rightarrow(m, n)=(3,4) \text { or }(4,3) .
\end{array}
$$

Hence, $(m, n)=(2,2),(3,4),(4,3)$.