For each positive integer $n$, define the set
$$
P_{n}=\left\{n^{k} \mid k=0,1, \cdots\right\} \text {. }
$$

For positive integers $a$, $b$, and $c$, if there exists a positive integer $m$ such that the numbers $a-1$, $ab-12$, and $abc-2015$ (not necessarily distinct) all belong to the set $P_{m}$, then the positive integer triplet $(a, b, c)$ is called "lucky". Find the number of all lucky positive integer triplets.
（Supplied by He Yijie）
4. Consider the conditions that the lucky positive integer triplet $(a, b, c)$ must satisfy.
Let $m$ be a positive integer, and $\alpha, \beta, \gamma$ be non-negative integers such that
$a-1=m^{\alpha}$,
$a b-12=m^{\beta}$,
$a b c-2015=m^{\nu}$.
(1) $m$ is even.
Otherwise, from equation (1), $a$ is even. Thus, the left side of equation (2) is even, but the right side is odd, which is a contradiction.
Therefore, $m$ is even.
(2) $\gamma=0$.
Otherwise, from equation (3),
$a b c=2015+m^{\nu}$ (odd).
Hence, $a b$ is odd.
From equation (2),
$m^{\beta}=a b-12$ (odd).
Since $m$ is even, it can only be that $a b-12=1$, i.e., $a b=13$.
From equation (1), $a>1$.
Therefore, $a=13$.
Thus, $m^{\alpha}=a-1=12 \Rightarrow m=12$.
At this point, from equation (3),
$12^{\gamma}=a b c-2015=13(c-155)$,

which is impossible.
Therefore, $\gamma=0$ must hold.
Thus, $a b c=2016$.
(3) $\alpha=0$.
Otherwise, from equation (1), $a$ is an odd number greater than 1, and from equation (4), $a$ is a divisor of 2016.
Note that, $2016=2^{5} \times 3^{2} \times 7$.
Then $a$ can only be $3, 7, 9, 21, 63$.
For the cases $a=3,9,21,63$, $3 \mid a$.
Hence, $3 \mid (a b-12)$.
From equation (2), $3 \mid m$.
But from equation (1), we also have
$$
m^{\alpha}=a-1 \equiv 2(\bmod 3),
$$

which is a contradiction.
For the case $a=7$, from equation (1),
$$
m^{\alpha}=a-1=6 \Rightarrow m=6 .
$$

At this point, equation (2) becomes
$$
7 b-12=6^{\beta} \equiv \pm 1(\bmod 7),
$$

which is a contradiction.
Thus, $\alpha=0$ must hold.
Therefore, $a=2$.
From equation (4), $b c=1008$.
At this point, equation (2) becomes
$2 b-12=m^{\beta}$.
Therefore, $b>6$.
Conversely, when $b>6$, there exists a positive even number $m=2 b-12$ and a positive integer $\beta=1$ that satisfy equation (5).

The above shows that the positive integer triplet $(a, b, c)$ is lucky if and only if
$a=2, b c=1008$, and $b>6$.
Note that, the number of positive divisors of $1008=2^{4} \times 3^{2} \times 7$ is $(4+1) \times(2+1) \times(1+1)=30$, among which, the positive divisors not greater than 6 are $1,2,3,4,6$.

Thus, the number of values $b$ can take is $30-5=25$. Accordingly, the number of lucky positive integer triplets $(a, b, c)$ is 25.