South-Afrika, 2011

Let $\phi(n)\sigma(n) = (n-3)(n+1)$, and let prime $p$ divide $n$ such that $n = p^a m$ for some $m$ not divisible by $p$. Then we have
$$\begin{align*} \phi(n) &= \phi(p^a)\phi(m) = p^{a-1}(p-1)\phi(m) \\ \sigma(n) &= \sigma(p^a)\sigma(m) = \left(\frac{p^{a+1}-1}{p-1}\right)\sigma(m), \end{align*}$$
so
$$\phi(n)\sigma(n) = p^{a-1}(p^{a+1} - 1)\phi(m)\sigma(m).$$
If $a \ge 2$, it follows that, modulo $p^{a-1}$,
$$0 \equiv \phi(n)\sigma(n) = (n-3)(n+1) \equiv -3.$$
Hence $a = 2$ and $p = 3$. Thus either $a = 1$ or $a = 2$ and $p = 3$.
However, if $n = 9$ does not satisfy the given equality, and if $n = 9p_1p_2\dots p_k$, then
$$\phi(n)\sigma(n) = 78(p_1^2 - 1)(p_2^2 - 1)\dots(p_k^2 - 1) \equiv_9 0,$$
since $p_1^2 \equiv_3 1$ for every prime $p_i \neq 3$, but
$$(n-3)(n+1) \equiv_9 -3.$$
Hence $n = p_1p_2 \dots p_k$ for some distinct primes $p_i$.