4-124 m, n are two different positive integers, find the common complex roots of the equations
$$
\text { and } \quad \begin{aligned}
x^{m+1}-x^{n}+1=0 \\
x^{n+1}-x^{m}+1=0
\end{aligned}
$$
[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
$$
x^{m+1} - x^{n+1} - x^n + x^m = 0,
$$

which simplifies to
$$
x^n (x + 1) (x^{m-n} - 1) = 0.
$$

Since 0 and -1 are not roots of the equations given in the problem, we have
$$
x^{m-n} - 1 = 0,
$$

which implies
$$
x^{m-n} = 1 \text{, and } |x| = 1.
$$

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

which simplifies to
$$
x^n (x - 1) = -1.
$$

From $|x| = 1$ and the above equation, we have
$$
|x - 1| = 1.
$$

Let $x = \cos \theta + i \sin \theta, \theta \in [0, 2\pi)$. Substituting this 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
$$
\begin{array}{l}
\cos \theta = \frac{1}{2}, \\
\theta = \frac{\pi}{3} \text{ or } \theta = \frac{5\pi}{3}.
\end{array}
$$

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

which simplifies to
$$
x = \frac{1}{2} + \frac{\sqrt{3}}{2} i \text{ or } x = \frac{1}{2} - \frac{\sqrt{3}}{2} i.
$$

Both values of $x$ satisfy the relation
$$
x - 1 = x^2,
$$

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

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

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

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

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

which similarly leads to $m = 6k^* + 1$, where $k^*$ is any non-negative integer.
Conversely, when $m$ and $n$ are two different positive integers, and $m = 6k^* + 1$, $n = 6k + 1$, where $k$ and $k^*$ 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 $x^2 - x + 1 = 0$. Therefore, we have
$$
\begin{aligned}
x^{m+1} - x^n + 1 & = x^{6k^* + 2} - x^{6k + 1} + 1 \\
& = x^2 - x + 1 \\
& = 0, \\
x^{n+1} - x^m + 1 & = x^{6k + 2} - x^{6k^* + 1} + 1 \\
& = x^2 - x + 1 \\
& = 0.
\end{aligned}
$$

In summary, when $m = 6k^* + 1$ and $n = 6k + 1$, 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, the two equations given in the problem have no common complex roots.