12.94 The equation $x^{n}+(2+x)^{n}+(2-x)^{n}=0$ has a rational solution, the necessary and sufficient condition regarding the positive integer $n$ is what?
(15th Putnam Mathematical Competition, 1955)
[Solution] The necessary and sufficient condition is $n=1$.
When $n=1$, the equation
$$
x^{n}+(2+x)^{n}+(2-x)^{n}=0 .
$$

has a rational solution
$$
x=-4 \text {. }
$$

Next, we prove that $n=1$ is also necessary.
First, because $x, 2+x, 2-x$ cannot all be 0, $n$ cannot be an even number.
When $n$ is an odd number, suppose $x=\frac{p}{q}$ is a solution to equation (1), where $p, q$ are coprime integers. Then from (1) we get
$$
\begin{array}{c}
p^{n}+(2 q+p)^{n}+(2 q-p)^{n}=0, \\
p^{n}+2\left[(2 q)^{n}+C_{n}^{2}(2 q)^{n-2} p^{2}+\cdots+C_{n}^{n-1}(2 q) p^{n-1}\right]=0 .
\end{array}
$$

Since every term except the first can be divided by $2 q$, $p^{n}$ can also be divided by $2 q$. Because $p$ and $q$ are coprime, then $q= \pm 1$.

Assume $q=1$ without loss of generality, otherwise change the signs of $p, q$ simultaneously, the value of $x$ remains unchanged. Since $p^{n}$ can be divided by $2 q=2$, $p$ is even. Let $p=2 r$, substitute into (1), we get
$$
\begin{array}{c}
(2 r)^{n}+(2+2 r)^{n}+(2-2 r)^{n}=0, \\
r^{n}+(1+r)^{n}+(1-r)^{n}=0 .
\end{array}
$$

When $n \geqslant 3$, we have
$$
\begin{array}{l}
r^{n}+2\left(1+C_{n}^{2} r^{2}+C_{n}^{4} r^{4}+\cdots\right)=0, \\
2=-r^{n}-2\left(C_{n}^{2} r^{2}+C_{n}^{4} r^{4}+\cdots\right),
\end{array}
$$

so 2 can be divided by $r^{2}$, thus $r= \pm 1$. But when $r= \pm 1$, (2) is not satisfied, hence $n \geqslant 3$ is impossible.
Therefore, $n$ can only be 1.