4.54 Determine all positive integers $n$ such that $n=[d(n)]^{2}$, where $d(n)$ denotes the number of positive divisors of $n$ (including 1 and itself).
(Canadian Mathematical Olympiad, 1999)
[Solution] Let the standard factorization of $n$ be
$n=p_{1}^{a_{1}} p_{2}^{a_{2} \cdots} p_{k}^{a_{k}}\left(p_{i}\right.$ is a prime, $\left.\alpha_{i} \geqslant 0\right)$.
Then, we have $\quad d(n)=\left(1+\alpha_{1}\right)\left(1+\alpha_{2}\right) \cdots\left(1+\alpha_{k}\right)$.

By the problem statement, $n=[d(n)]^{2}$, so $n$ is a perfect square, and $\alpha_{i}$ is even. Let $\alpha_{i}=2 \beta_{i}$,
$$
p_{1}^{\beta_{1}} p_{2}^{\beta_{2}} \cdots p_{k}^{\beta_{k}}=\left(1+2 \beta_{1}\right)\left(1+2 \beta_{2}\right) \cdots\left(1+2 \beta_{k}\right) .
$$

The right-hand side of the above equation is odd, so $p_{i}$ must be an odd prime, $p_{i} \geqslant 3$.
When $\beta_{i}>0$, we have $p_{i}^{\beta_{i}} \geqslant 3^{\beta_{i}}=(1+2)^{\beta_{i}}>1+2 \beta_{i}$,
$$
p_{1}^{\beta_{1}} p_{2}^{\beta_{2} \cdots} p_{k}^{\beta_{k}}>\left(1+2 \beta_{1}\right)\left(1+2 \beta_{2}\right) \cdots\left(1+2 \beta_{k}\right) .
$$
This contradicts the previous equation.
$\therefore$ For all $i, \beta_{i}=0$. Therefore, $n=1$.