4- $124 m, n$ are two distinct positive integers, find the common complex roots of the equations
$$x^{m+1}-x^{n}+1=0$$

and
$$x^{n+1}-x^{m}+1=0$$
[Solution] Without loss of generality, assume $m>n$ (if $m<n$, just swap $m$ and $n$ in the following proof). Let $x$ be a common complex root of the two equations given in the problem. Subtracting the two equations, we have

that is, $\square$
$$\begin{array}{c}
x^{m+1}-x^{n+1}-x^{n}+x^{m}=0 \\
x^{n}(x+1)\left(x^{m-n}-1\right)=0
\end{array}$$

Since 0 and -1 are not roots of the equations given in the problem, we have
that is, $\square$
$$x^{m-n}-1=0,$$
$$x^{m-n}=1 \text {, and }|x|=1 \text {. }$$

Therefore, $x^{m}=x^{n}$, and $|x|=1$.
Substituting the above equation into the second equation given in the problem, we get
$$x^{n+1}-x^{n}+1=0,$$

that is, $\square$
$$x^{n}(x-1)=-1$$

From $|x|=1$ and the above equation, we get
$$|x-1|=1 \text {. }$$

Let $x=\cos \theta+i \sin \theta, \theta \in[0,2 \pi)$. Substituting it into the above equation, we get
$$1=|x-1|^{2}=(\cos \theta-1)^{2}+\sin ^{2} \theta=2(1-\cos \theta),$$

Solving the equation, we get
$$\begin{array}{l}
\cos \theta=\frac{1}{2}, \\
\theta=\frac{\pi}{3} \quad \text { or } \theta=\frac{5 \pi}{3} .
\end{array}$$

Thus, we have
$$x=\cos \frac{\pi}{3}+i \sin \frac{\pi}{3} \text { or } x=\cos \frac{5 \pi}{3}+i \sin \frac{5 \pi}{3} ,$$

that is,
$$x=\frac{1}{2}+\frac{\sqrt{3}}{2} i \text { or } x=\frac{1}{2}-\frac{\sqrt{3}}{2} i .$$

These two values of $x$ satisfy the relation
$$x-1=x^{2},$$

Substituting this relation into (2), we get
$$x^{n+2}=-1,$$

that is, $\square$
$$\begin{array}{l}
\left(\cos \frac{\pi}{3} \pm i \sin \frac{\pi}{3}\right)^{n+2}=-1 \\
\cos \frac{(n+2) \pi}{3} \pm i \sin \frac{(n+2) \pi}{3}=-1
\end{array}$$

Thus, we have
$$\frac{(n+2) \pi}{3}=(2 k+1) \pi$$

where $k$ is any non-negative integer. Therefore,
$$n=6 k+1$$

Similarly, substituting (1) into the second equation given in the problem, we get
$$x^{m+1}-x^{m}+1=0$$

Therefore, we can also deduce that $x^{m} \quad m=6 k^{*}+1$, where $k^{*}$ is any non-negative integer.
Conversely, when $m$ and $n$ are two different positive integers, and $m=6 k^{*}+1$, $n=6 k+1$, where $k, k^{*}$ are any non-negative integers, it is easy to verify that
$$x=\cos \frac{\pi}{3} \pm i \sin \frac{\pi}{3}$$

is a common root of the two equations given in the problem.
In fact, we only need to note that when $x=\cos \frac{\pi}{3} \pm i \sin \frac{\pi}{3}$, $x^{3}=1$ and $x^{2}-x+1=0$, so we have
$$\begin{aligned}
x^{m+1}-x^{n}+1 & =x^{6 k^{*}+2}-x^{6 k+1}+1 \\
& =x^{2}-x+1 \\
& =0 \\
x^{n+1}-x^{m}+1 & =x^{6 k+2}-x^{6 k^{*}+1}+1 \\
& =x^{2}-x+1 \\
& =0
\end{aligned}$$

In summary, when $m=6 k^{*}+1, n=6 k+1$, where $k^{*}$ and $k$ are any two different non-negative integers, the two equations given in the problem have two common complex roots.
$$x=\frac{1}{2} \pm \frac{\sqrt{3}}{2} i$$

In other cases of $m$ and $n$, the two equations given in the problem have no common complex roots.