3. For a composite number $n$, let $f(n)$ be the sum of its smallest three positive divisors, and $g(n)$ be the sum of its largest two positive divisors. Find all positive composite numbers $n$ such that $g(n)$ equals a positive integer power of $f(n)$.
(Provided by He Yijie)
3. Solution 1 If $n$ is odd, then all divisors of $n$ are odd. Therefore, by the problem's condition, $f(n)$ is odd and $g(n)$ is even. Thus, $g(n)$ cannot be a positive integer power of $f(n)$.

Therefore, we only need to consider the case where $n$ is even. In this case, 1 and 2 are the smallest two positive divisors of $n$, and $n$ and $\frac{n}{2}$ are the largest two positive divisors of $n$.
Let $d$ be the smallest positive divisor of $n$ other than 1 and 2.
If there exists $k \in \mathbf{N}_{+}$ such that $g(n)=f^{k}(n)$, then
$$
\frac{3 n}{2}=(1+2+d)^{k}=(3+d)^{k} \equiv d^{k}(\bmod 3) \text {. }
$$

Since $\frac{3 n}{2}$ is a multiple of 3, we have $3 \mid d^{k}$, which implies $3 \mid d$.
By the minimality of $d$, we have $d=3$.
Therefore, $\frac{3}{2} n=6^{k} \Rightarrow n=4 \times 6^{k-1}$.
Also, $31 n$, so $k \geqslant 2$.
In summary, all possible values of $n$ are
$$
n=4 \times 6^{l}\left(l \in \mathbf{N}_{+}\right) \text {. }
$$

Solution 2 Suppose the composite number $n$ satisfies
$$
g(n)=f^{k}(n)\left(k \in \mathbf{N}_{+}\right) \text {, }
$$

and let the smallest prime divisor of $n$ be $p$. Then the second largest positive divisor of $n$ is $\frac{n}{p}$.
If the third smallest positive divisor of $n$ is $p^{2}$, then $\frac{n}{p^{2}} \in \mathbf{Z}$.
In this case, $f(n)=1+p+p^{2} \equiv 1(\bmod p)$, $g(n)=n+\frac{n}{p}=p(1+p) \frac{n}{p^{2}}(\bmod p)$, i.e., $1^{k} \equiv f^{k}(n)=g(n) \equiv 0(\bmod p)$, which is a contradiction. Therefore, the third smallest positive divisor of $n$ is not $p^{2}$.
Thus, it must be some prime $q(q>p)$.
It is easy to see that $\frac{n}{p q} \in \mathbf{Z}$.
Then $f(n)=1+p+q \equiv 1+p(\bmod q)$,
$g(n)=n+\frac{n}{p}=q(1+p) \frac{n}{p q}(\bmod q)$.
Hence $(1+p)^{k} \equiv f^{k}(n)=g(n) \equiv 0(\bmod q)$.
Since $q$ is a prime, we have $q \mid(1+p)$.
Thus, $p<q \leqslant 1+p$, which implies $p=2, q=3$.
At this point, $6^{k}=f^{k}(n)=g(n)=\frac{3}{2} n$.

Solving this, we get $n=4 \times 6^{k-1}$.
The rest is the same as Solution 1.