58th Ukrainian National Mathematical Olympiad

Notice, that from conditions of the task, it follows, that $d_1\cdot d_l=d_2\cdot d_{l-1}=n$. Using this equality and first equality from conditions, it is possible to rewrite the second equality from the task: $$N^b=d_{l-1}+d_l=\frac{n}{d_2}+\frac{n}{d_l}=\frac{n(d_1+d_2)}{d_1{d}_2}=\frac{n{N}^a}{d_1{d}_2}\text{ or }n=d_1{d}_2{N}^{b-a}.$$

Apart from that, if $b=a$, then $d_1+d_2=N^a=N^b=d_{l-1}+d_l$, so $d_1=d_{l-1}$; $d_2=d_l$. Then the number needed has four divisors, or exactly two proper divisors. There are two forms of such numbers: $n=pq$, where $p,q$ are prime numbers, or $n=p^3$, where $p$ is prime. So, we can consider only case $b>a$.

Consider two cases.

1. Let $N$ be odd number. Then from equality $d_1+d_2=N^a$ we have, that sum $d_1+d_2$ is odd number, then from these divisors there is exactly one even, another is odd. So it is obvious, that $d_1=2$, as it is the least even divisor. Then $d_2=N^a-2$ must be prime number. Let $d_2=N^a-2=p$, then $N^a=p+2$. So, $n=2\cdot p\cdot N^{b-a}$.

a) If $N$ is not prime number, then as $N\le p+2$, it has prime divisor $q\le p+2$. Obviously, $q\ne p$, and so $q<p$. But this means, that for $b>a$ number $n=2\cdot p\cdot N^{b-a}$ is divisible by $q<p=d_2$, and then number $p$ can not be $d_2$. We get contradiction.

b) If $N$ is prime number, then for $a>1$ we get, that it follows from equality $N^a=p+2$, that $N<p+2$, then it is proper divisor of number $n$, less than $p$, that is impossible. So, $a=1$, i.e. numbers $p$ and $N=p+2$ is pair of twin primes. And then $\forall b\ge 1$ we have, that number $n=2\cdot p\cdot (p+2{)}^{b-1}$ is solution. Really, $d_{l-1}=2(p+2{)}^{b-1}$, $d_l=p(p+2{)}^{b-1}$ and $d_{l-1}+d_l=2(p+2{)}^{b-1}+p(p+2{)}^{b-1}=(p+2{)}^b=N^b$.

2. Let $N$ be even number. As $n=d_1{d}_2{N}^{a-b}$, then $n$ is also even, i.e. $d_1=2$. Then as $d_1=2$, then $d_2=4$, as sum of two the least divisors is even. Then $N^a=6$. Then $N=6$, $a=1$, thus $n=2\cdot 4\cdot 6^{b-1}$. If $b>a=1$, then number 3 is divisor of number $n$, then $d_2=3$ – contradiction. Thus $b=a=1$.