117. Find the condition for the divisibility of $(x+1)^{n}+(x-1)^{n}$ by $x$.

To find the condition for the divisibility of $(x+1)^{n}+(x-1)^{n}$ by $x$, we need to determine when the expression is divisible by $x$. This means that when $x=0$, the expression should equal zero.

Let's substitute $x=0$ into the expression:
\[
(0+1)^{n} + (0-1)^{n} = 1^n + (-1)^n
\]

For the expression to be divisible by $x$, the result must be zero:
\[
1^n + (-1)^n = 0
\]

This equation holds true if and only if $n$ is odd. When $n$ is odd, $(-1)^n = -1$, and thus:
\[
1 + (-1) = 0
\]

Therefore, the condition for the divisibility of $(x+1)^{n}+(x-1)^{n}$ by $x$ is that $n$ must be odd.

\[
\boxed{n \text{ is odd}}
\]
Instruction. Transform the divisor $x$ into the form:

$$
x=\frac{1}{2}[(x+1)+(x-1)]
$$