2. Determine all positive integers $n$, such that the equation $x^{n}+(2+x)^{n}+(2-x)^{n}=0$ has integer solutions.
2. Solution: Clearly, if $n$ is even, then the equation has no real roots. If $n=1$, then $x=-4$.

When $n \geqslant 3$ and is odd, since the left side of the equation is a polynomial with the leading coefficient 1 and the constant term $2^{n+1}$, therefore, its integer solutions can only be of the form $-2^{t}$ (where $t$ is a non-negative integer):

If $t=0,1,2$, then $x=-1,-2,-4$ are not roots of the equation;
If $t \geqslant 3$, then $-2^{n t}+\left(2-2^{t}\right)^{n}+\left(2+2^{t}\right)^{n}=0$
$$
\begin{array}{l}
\Leftrightarrow-2^{n(t-1)}+\left(1-2^{t-1}\right)^{n}+\left(1+2^{t-1}\right)^{n}=0, \\
\quad \because-2^{n(t-1)}+\left(1-2^{t-1}\right)^{n}+\left(1+2^{t-1}\right)^{n} \equiv
\end{array}
$$
$2(\bmod 4), \therefore$ the left side $\neq 0$.
In summary, the original equation has a unique integer solution $x=-4$ only when $n=1$.