466. For what number and for which natural $n$ is the fraction $\frac{n^{2}+1}{n+1}$ reducible? Find all solutions.
$\triangle$ Let

$$
\text { GCD }\left(n^{2}+1, n+1\right)=d \text {. }
$$

Since $n+1$ is divisible by $d$, then the product

$$
n(n+1)=n^{2}+n
$$

is also divisible by $d$. Now we get:

$$
\begin{aligned}
& \left(n^{2}+n\right): d \\
& -\left(n^{2}+1\right): d \\
& (n-1): d
\end{aligned}
$$

Therefore, the numbers $n+1$ and $n-1$ are divisible by $d$. But the greatest common divisor of two numbers differing by 2 is 1 or 2 (see the statements of problems 455 and 456). We are interested in the case $d=2$. This is possible for all odd $n$.

Answer: by 2 for all odd $n$.