4. 168 Find the integer solutions of the equation
$$\left(1+\frac{1}{m}\right)^{m+1}=\left(1+\frac{1}{1988}\right)^{1988}$$
［Solution］ The original equation can be transformed into
$$\frac{(m+1)^{m+1}}{m^{m+1}}=\frac{1989^{1988}}{1988^{1988}}$$

If $m>0$, then both sides of (1) are irreducible fractions, so it must be that
$$m^{m+1}=1988^{1988}$$

Notice that when $m \geqslant 1988$, $m^{m+1}>1988^{1988}$;
$$01988$ when, $n^{n}>1988^{1988}$;
$$0<n<1988 \text {, then } n^{n}<1988^{1988} \text {; }$$

Therefore, (4) has only one solution $n=1988$. After calculation, we get
$$m=-1989 .$$

Upon verification, the integer solution to the original equation is $m=-1989$.