## Task A-4.7.

Determine all natural numbers $n$ that have exactly 12 positive divisors

$$
1=d_{1}<d_{2}<\cdots<d_{12}=n
$$

such that $d_{4}=5$ and $d_{5}^{2}+1=d_{7}$.
## Solution.

Since $d_{4}=5$, we see that exactly one of the numbers $1,2,3,4,5$ is not a divisor of the number $n$.

The numbers 1 and 5 are obviously divisors of the number $n$. If the number 2 is not a divisor of the number $n$, then the number 4 is not either, which is impossible. Therefore, the number 2 is also a divisor of the number $n$.

1 point

Now we have two possibilities. The first possibility is that $d_{1}=1, d_{2}=2, d_{3}=3$, and $d_{4}=5$. Notice that in this case, the number 6 is also a divisor of $n$, so we conclude that $d_{5}=6$ must hold.

From the second condition, we have that $d_{7}=d_{5}^{2}+1=37$, so the number $n$ has at least 4 different prime factors $(2,3,5,37)$, which means it has at least 16 divisors, which is a contradiction.

The second possibility is that $d_{1}=1, d_{2}=2, d_{3}=4$, and $d_{4}=5$. We see that in this case, the number 10 is also a divisor of $n$, which means $d_{5} \leqslant 10$.

Since 3 is not a divisor of $n, d_{5}$ cannot be 6 or 9 either.

We need to check the possibilities that $d_{5}$ is 7, 8, or 10.

If $d_{5}=7$, then $d_{7}=7^{2}+1=50$, so there should be other divisors of 50, such as 10 and 25, between $d_{5}$ and $d_{7}$, which is impossible. If $d_{5}=8$, then $d_{7}=8^{2}+1=65$, and similarly, we get a contradiction because there should be the number 13 and the number 26 between $d_{5}$ and $d_{7}$.

If $d_{5}=10$, then $d_{7}=10^{2}+1=101$, which is a prime number. Since $n$ is divisible by 4 and 5, the only possibility would be $d_{6}=20$.

The divisors of the number $n$ can be paired so that their product is equal to $n$, i.e., $d_{i} \cdot d_{13-i}=n$, for each $i \in\{1,2, \ldots, 12\}$. In particular, it must be $n=d_{6} \cdot d_{7}=2020$.

It remains to verify that the number $n=2020$ indeed satisfies all the conditions of the problem, i.e., that there are no other divisors between 20 and 101 and that it has exactly 12 divisors in total. Both conditions can be easily verified by listing all the divisors of 2020. They are, in order,

$$
1,2,4,5,10,20,101,202,404,505,1010,2020
$$

Note: Once it is concluded that $d_{5} \leqslant 10$, it is also possible to conclude in the following way. Notice that the number 20 is also a divisor of the number $n$. If we assume that $d_{5}d_{4}=5$. Therefore, it must be $d_{5}=10$. (1 point)