China Hong Kong Mathematical Olympiad

If $p^2 \mid n$ for some prime $p$, then $p \mid \varphi(p^2) \mid \varphi(n)$. This implies $p \mid \varphi(n)\sigma(n) + 1$, and hence $p \mid n + \varphi(n)\sigma(n) + 1$. Therefore, $n$ must be squarefree.

If $n = pq$ for some distinct primes $p$ and $q$, then
$$\varphi(pq)\sigma(pq) + 1 = (p-1)(q-1)(p+1)(q+1) + 1 = p^2q^2 - p^2 - q^2 + 2.$$
This is a multiple of $pq$ if and only if $pq \mid p^2 + q^2 - 2$. We now prove that $ab \mid a^2 + b^2 - 2$ has no solution in distinct integers $a, b > 1$ using Vieta's jumping.

Suppose on the contrary that the divisibility is solvable. Let $(a, b)$ be a solution such that $a > b > 1$ and $b$ is the smallest possible. Let $a^2 + b^2 - 2 = kab$ for some positive integer $k$. Rewrite this equation as
$$a^2 - kba + (b^2 - 2) = 0.$$
This is a quadratic equation in $a$. Let $a'$ be another solution to this equation. Since $a + a' = kb$, $a'$ is an integer. Since $aa' = b^2 - 2 > 0$, $a'$ is positive. Also, we have
$$a' = \frac{b^2 - 2}{a} < \frac{b^2}{a} < b < a.$$
This shows $(x, y) = (b, a')$ is another pair of solution to $xy \mid x^2 + y^2 - 2$ such that $x > y > 0$. As $a' < b$, by the minimality of $b$, we must have $a' = 1$. But then the divisibility becomes $b \mid 1 + b^2 - 2$, i.e. $b \mid 1$. This is impossible since $b > 1$. Thus, there is no solution.

It follows that $n$ must have at least 3 distinct prime divisors.