33 Find all positive integers $m$ such that $m=d(m)^{4}$.

Find all positive integers $m$ such that $m=d(m)^{4}$.
33. From the conditions, we know that \( m \) is a fourth power, so we can set
$$m=2^{4 a_{2}} \cdot 3^{4 a_{3}} \cdot 5^{4 a_{5}} \cdot 7^{4 a_{7}} \cdots \cdots$$

where \(\alpha_{2}, \alpha_{3}, \alpha_{5}, \alpha_{7}, \cdots\) are non-negative integers. And
$$d(m)=\left(4 \alpha_{2}+1\right)\left(4 \alpha_{3}+1\right) \cdots$$

is an odd number, hence \(\alpha_{2}=0\), and
$$1=\frac{4 \alpha_{3}+1}{3^{a_{3}}} \cdot \frac{4 \alpha_{5}+1}{5^{a_{5}}} \cdot \frac{4 \alpha_{7}+1}{7^{\alpha_{7}}} \cdots=x_{3} x_{5} x_{7} \cdots$$

where
$$x_{3}=\frac{4 \alpha_{3}+1}{3^{a_{3}}}, x_{5}=\frac{4 \alpha_{5}+1}{5^{a_{5}}}, \cdots$$

When \(\alpha_{3}=1\), \(x_{3}=\frac{5}{3}\); when \(\alpha_{3}=0\) or 2, \(x_{3}=1\); and when \(\alpha_{3} \geqslant 3\), \(3^{\alpha_{3}} > 4 \alpha_{3}+1\), so in this case \(x_{3}<1\). For \(p > 5\), where \(p\) is a prime, when \(\alpha_{p}=0\), \(x_{p}=1\), and when \(\alpha_{p}=1\), \(p^{\alpha_{p}}>5=4 \alpha_{p}+1\), so \(x_{p}<1\); and when \(\alpha_{p} \geqslant 2\), \(x_{p}<\frac{9}{25}\).

The above discussion shows: if \(\alpha_{3} \neq 1\), then
$$x_{3}=x_{5}=x_{7}=\cdots=1$$

Thus,
$$\alpha_{3}=0 \text{ or } 2, \alpha_{5}=0 \text{ or } 1,$$

and
$$\alpha_{7}=\alpha_{11}=\cdots=0$$

That is,
$$m=1,3^{8}, 5^{4} \text{ or } 45^{4} \text{. }$$

If \(\alpha_{3}=1\), then \(3 \mid m\), at this time, from \(m=d(m)^{4}\), we know
$$m=5^{4} \times\left(4 \alpha_{5}+1\right)^{4} \times\left(4 \alpha_{7}+1\right)^{4} \cdots$$

Thus, there exists a prime \(p \geqslant 5\) such that \(3 \mid 4 \alpha_{p}+1\), which requires \(\alpha_{p} \geqslant 2\), hence \(x_{p}<\frac{9}{25}\). This leads to
$$x_{3} x_{5} x_{7} \cdots \leqslant \frac{5}{3} \times \frac{9}{25}=\frac{3}{5}<1$$

a contradiction.
Therefore,
$$m=1,5^{4}, 3^{8}, 3^{8} \cdot 5^{4}$$
(Direct verification shows that they indeed satisfy the conditions).