, 2015

Because $p q$ divides $n^{p q} + 1$, neither $p$ nor $q$ divides $n$.
Assume $p < q$. We have from Fermat's little theorem
$$0 \equiv n^{p q} + 1 \equiv n^{q} + 1 \quad (\bmod p).$$
Therefore $n^{2 q} \equiv 1 (\bmod p)$. But $n^{p-1} \equiv 1 (\bmod p)$ and $\gcd(p-1, q) = 1$, since $p-1 < q$ and $q$ is prime. We deduce that $n^{2} \equiv 1 (\bmod p)$ and hence $p$ divides $n + 1$.
Let $n + 1 = a p$ for an integer $a$. We have
$$\begin{aligned} n^{p q - 1} - n^{p q - 2} + \cdots - n + 1 &\equiv \sum_{i=0}^{p q - 1} (1 - a p)^{i} \equiv \sum_{i=0}^{p q - 1} (1 - i a p) \\ &\equiv p q - \frac{p q - 1}{2} a p^{2} q \equiv p q \not\equiv 0 \quad \left(\bmod p^{2}\right). \end{aligned}$$
Therefore $p^{2}$ does not divide $n^{p q - 1} - n^{p q - 2} + \cdots - n + 1$ and since $p^{3}$ divides
$$n^{p q} + 1 = (n + 1) \left(n^{p q - 1} - n^{p q - 2} + \cdots - n + 1\right),$$
then $p^{2}$ divides $n + 1$.