$4 \cdot 247$ Find all positive integers $n$ such that the following equation has an integer solution.
$$
x^{n}+(2+x)^{n}+(2-x)^{n}=0 \text {. }
$$
[Solution] When $n=1$, we get $x+(2+x)+(2-x)=0, x=-4$ as its integer solution.
When $n$ is even, the original equation has integer solutions if and only if:
$$
\left\{\begin{array}{l}
x=0, \\
2+x=0, \\
2-x=0 .
\end{array}\right.
$$

But (1) is a contradictory equation, hence when $n$ is even, the original equation has no integer solutions.
When $n$ is an odd number greater than 1, using the binomial theorem to expand the left side and then combining and organizing, we get a polynomial $p(x)$, whose highest degree coefficient is 1, all coefficients are non-negative integers, and the constant term is $2^{n+1}$. By Vieta's theorem, the possible integer solutions of $p(x)=0$ can be set as $-2^{t}, t$ being a non-negative integer:
If $t=0$, then $x=-1$, the left side of the original equation is the sum of three odd numbers, so it cannot be zero.
If $t=1$, then $x=-2$, the left side of the original equation is $(-2)^{n}+0+4^{n}$, which also cannot be 0.
If $t \geqslant 2$, let $t=p+1, p \geqslant 1$, then the left side of the original equation is
$$
\begin{aligned}
& 2^{n}\left[-2^{p n}+\left(1-2^{p}\right)^{n}+\left(1+2^{p}\right)^{n}\right] \\
= & 2^{n}\left\{-2^{p n}+2\left[1+\binom{n}{2} 2^{2 p}+\binom{n}{4} 2^{4 p}+\cdots\right]\right\} .
\end{aligned}
$$

But $n \geqslant 3$, the value inside the large braces is congruent to 2 modulo 4, so it cannot be 0, thus the left side cannot be 0.
Therefore, when $n$ is an odd number greater than or equal to 3, the original equation has no integer solutions.
In summary, the equation $x^{n}+(2+x)^{n}+(2-x)^{n}=0$ has integer solutions if and only if $n=1$.