266. When division is exact. For which positive integer $n$ will the quantity $n^{4}+n^{2}$ be divisible by $2 n+1$?
266. 

$$
\begin{aligned}
f(n) & =\frac{n^{4}+n^{2}}{2 n+1}=\frac{n^{2}\left(n^{2}+1\right)}{2 n+1}=\frac{n^{2}}{4}\left[\frac{4 n^{2}+4}{2 n+1}\right]= \\
& =\left(\frac{n}{2}\right)^{2} \cdot\left[2 n-1+\frac{5}{2 n+1}\right]
\end{aligned}
$$

It is obvious that the greatest common divisor of the numbers $n$ and $2n+1$ is 1. Therefore, $f(n)$ can be an integer only when $\frac{5}{2n+1}$ is an integer, which occurs when $n=2, 0, -1$, or $-3$. Consequently, the only positive integer for which $f(n)$ will be an integer is $n=2$. [In this case, $f(n)=4$.]