Belarusian Mathematical Olympiad

From the definition of $\circ$ it is clear that $m\circ n\ge 1+n^2$, so $n\le \sqrt{496}<23$. Note that the minimal number that has at least 7 divisors is 24, hence it is sufficient to consider only $k$ from 1 to 6.

Exactly two divisors have only prime numbers, so for $k=2$ the numbers $m$ and $n$ are prime. From $m\circ n=497$ it follows that $mn=496$. But $496=2^4\cdot 31$ doesn't equal to the product of two prime numbers.

Exactly three divisors have only squares of prime numbers. Let $m=p^2$, $n=q^2$, then $m\cdot n=1+pq+(pq{)}^2=497$ and $pq(pq+1)=496=2^4\cdot 31$, hence $p=31$, $q=2$ which doesn't suit.

Consider the case $k=4$. Since $1<m_2<m_3<m$ are all divisors of $m$, then $m=m_2{m}_3$, similarly, $n_2{n}_3=n$. Factor the number $m\circ n$: $$m\circ n=1+m_2{n}_2+m_3{n}_3+(m_2{m}_3)(n_2{n}_3)=(1+m_2{n}_2)(1+m_3{n}_3).$$ Since $497=7\cdot 71$, then $m_2{n}_2=6$ and $m_3{n}_3=70$. Numbers $m_2$ and $n_2$ are prime, therefore one of them is equal to 2, the other is equal to 3, and each of the numbers $m_3$ and $n_3$ is either prime or a square of $m_2$ and respectively $n_2$. It is easy to check that for $70=2\cdot 5\cdot 7$ none of the options fits.

Exactly five divisors have only the fourth powers of prime numbers. Up to 23 there is exactly one such number $-n=2^4$. Note that $$2^4\circ 2^4=(4^5-1)/3=341<497,\text{a}{3}^4\circ 2^4=(6^5-1)/5=1555>497.$$ For larger values of $m$ there certainly is no solutions.

Exactly six divisors have only numbers $p^5$ and $p{q}^2$, where $p$ and $q$ are prime. Up to 23 there are only three such numbers: 12, 18 and 20, consider them one by one.

Let $n=12$. Among the divisors of 12 there are exactly two odd: $n_1=1$ and $n_3=3$. The number 497 is odd too, hence $m_2=2$ and $m_3=4$. Therefore, $m$ has the form $2^5$ or $4p$, where $p$ is prime and greater than 3. $m=2^5$ is impossible since $12\cdot 32=529$. In the case $m=4p$ we get the equation $17+64p=497$, which has no integer solutions.

Let $n=18$. From $18\circ m>18\cdot m+9\cdot 5+6\cdot 4+3\cdot 3+2\cdot 2+1\cdot 1$, follows the inequality $18\le m<414/18=23$. Considering the two options $m=18$ and $m=20$, we obtain the only solution (20, 18).

Suppose $n=20$. Form $20\circ m>20m+10+5+4+2+1$ follows the inequality $20\le m<475/20=23.75$. But $20\circ 20>500$, hence there is no solutions.