Solution. a) We have

$$
\begin{aligned}
A & =\frac{a^{2}-1}{n^{2}+a n}\left(\frac{1}{1-\frac{1}{n}}-1\right) \frac{a-a n^{3}-n^{4}+n}{1-a^{2}} \\
& =\frac{1}{n(n+a)}\left(\frac{n}{n-1}-1\right) \frac{-\left(n^{3}-1\right)(a+n)}{-1} \\
& =\frac{1}{n(n+a)} \frac{n-n-1}{n-1} \frac{-\left(n^{3}-1\right)(a+n)}{-1} \\
& =\frac{1}{n} \frac{1}{n-1}(n-1)\left(n^{2}+n+1\right) \\
& =\frac{n^{2}+n+1}{n}=n+1+\frac{1}{n}
\end{aligned}
$$

b) Using the result from a), we have that $A$ is a natural number if and only if $\frac{1}{n}$ is a natural number, which is possible only for $n=1$. However, the expression $A$ is not defined for $n=1$, hence, $A$ cannot be a natural number for any value of the natural number $n$.
1. a) Simplify the expression

$$
A=\frac{a^{2}-1}{n^{2}+a n}\left(\frac{1}{1-\frac{1}{n}}-1\right) \frac{a-a n^{3}-n^{4}+n}{1-a^{2}}
$$

b) Is there a natural number $n$ for which the number $A$ is a natural number