1000. A certain natural number has only two prime divisors (in some powers), and its square has 35 different divisors. How many different divisors does the cube of this number have?
$\triangle$ Let the original number have the form:

$$
a=p_{1}^{\alpha_{1}} \cdot p_{2}^{\alpha_{2}}
$$

where $p_{1}$ and $p_{2}$ are distinct prime numbers, and $\alpha_{1}$ and $\alpha_{2}$ are natural numbers. Then

$$
a^{2}=p_{1}^{2 \alpha_{1}} \cdot p_{2}^{2 \alpha_{2}}, \quad a^{3}=p_{1}^{3 \alpha_{1}} \cdot p_{2}^{3 \alpha_{2}}
$$

Since $a^{2}$ has 35 distinct divisors, we have

$$
\left(2 \alpha_{1}+1\right)\left(2 \alpha_{2}+1\right)=35
$$

From this, we obtain:

$$
2 \alpha_{1}+1=5, \quad 2 \alpha_{2}+1=7 ; \quad \alpha_{1}=2, \quad \alpha_{2}=3
$$

Therefore,

$$
\tau\left(a^{3}\right)=\left(3 \alpha_{1}+1\right)\left(3 \alpha_{2}+1\right)=7 \cdot 10=70
$$

Answer: 70.

$1001^{\circ}$. Prove that a natural number has an odd number of distinct divisors if and only if it is a square of a natural number.

$\triangle$ Apply the statement of problem 995 (with the same notation). The product

$$
\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \ldots\left(\alpha_{n}+1\right)
$$

is odd if and only if all its factors are odd, i.e., when all numbers $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}$ are even, i.e., when the original number

$$
a=p_{1}^{\alpha_{1}} \cdot p_{2}^{\alpha_{2}} \cdot \ldots \cdot p_{n}^{\alpha_{n}}
$$

is a square of a natural number.