5. From this unit (10), we know that the given condition is
$$\frac{(m+k) m}{(m+k, m)}=\frac{(n+k) n}{(n+k, n)}$$

Since $(m+k, m)=(m, k),(n+k, n)=(n, k)$, we get from the above equation
$$\frac{(m+k) m}{(m, k)}=\frac{(n+k) n}{(n, k)}$$

Let $(m, k)=d_{1}$, then $m=m_{1} d_{1}, k=k_{1} d_{1}$, where $\left(m_{1}, k_{1}\right)=1$.
Let $(n, k)=d_{2}$, then $n=n_{1} d_{2}, k=k_{2} d_{2}$, where $\left(n_{1}, k_{2}\right)=1$. Thus, equation (1) becomes
$$\left(m_{1}+k_{1}\right) m_{1} d_{1}=\left(n_{1}+k_{2}\right) n_{1} d_{2}$$

Multiplying both sides of the above equation by $k_{1}$ and using $k_{1} d_{1}=k_{2} d_{2}(=k)$, we obtain
$$\left(m_{1}+k_{1}\right) m_{1} k_{2}=\left(n_{1}+k_{2}\right) n_{1} k_{1}$$

The left side of the above equation is a multiple of $k_{2}$, so $k_{2}$ also divides the right side, i.e., $k_{2} \mid k_{1} n_{1}^{2}$. But $\left(k_{2}, n_{1}\right)=1$, hence $\left(k_{2}, n_{1}^{2}\right)=1$, which implies $k_{2} \mid k_{1}$. Similarly, we can prove $k_{1} \mid k_{2}$. Combining these, we get $k_{1}=k_{2}$, i.e., $(m, k)=(n, k)$. Therefore, from (1) we know $(m+k) m=(n+k) n$, from which it is easy to conclude $m=n$.