4. Find all natural $n$ for which both statements are true:
1) the greatest prime divisor of the number $n$ is equal to $\sqrt{n}$,
2) the greatest prime divisor of the number $(n+72)$ is equal to $\sqrt{n+72}$.
Answer: 49 and 289.

Solution. Let the mentioned prime divisors be $p_{1}$ and $p_{2}$. Then $n=p_{1}^{2}, n+72=p_{2}^{2}$, from which $\left(p_{2}+p_{1}\right)\left(p_{2}-p_{1}\right)=72$. Consider the factorizations of the number 72 into natural factors: $72=m \cdot k$ $(m>k)$. Then $p_{1}=\frac{m-k}{2}, p_{2}=\frac{m+k}{2}$.

1) $m=72, k=1: p_{1}$ is not an integer, contradiction.
2) $m=36, k=2: p_{1}=17, p_{2}=19$.
3) $m=24, k=3: p_{1}$ is not an integer, contradiction.
4) $m=18, k=4: p_{1}=7, p_{2}=11$.
5) $m=12, k=6: p_{1}=3, p_{2}=9, p_{2}$ is not a prime, contradiction.
6) $m=9, k=8: p_{1}$ is not an integer, contradiction.

The pairs that fit are $p_{1}=17, p_{2}=19$ and $p_{1}=7, p_{2}=11$. Since $n=p_{1}^{2}$, the required numbers are 49 and 289.

Comment. The complete answer is found by trial and error, and it is not proven that there are no other suitable natural numbers - 3 points. Only one answer is found by trial and error - 1 point. With correct reasoning, some cases are overlooked - up to 5 points.