66th Belarusian Mathematical Olympiad

Answer: $(756;1008)$, $(600;960)$.

Let $d$ denote the greatest common divisor of $a$ and $b$, i.e., $a=d{a}_1$, $b=d{b}_1$, where $\gcd (a_1,b_1)=1$. Then the given equality can be presented in the form $$d(a_1^3+b_1^3)=1911a_1{b}_1.$$ It follows that $d{a}_1^3:b_1$, and, since $a_1$ and $b_1$ are coprime, $d:b_1$. Similarly, $d:a_1$. Hence, $d:a_1{b}_1$ because $a_1$ and $b_1$ are coprime. Let $d=d_1{a}_1{b}_1$, then the equation can be presented in the form $$d_1(a_1^3+b_1^3)=1911,a_1\le b_1.(1)$$ Note that $a_1^3+b_1^3\ge (a_1+b_1{)}^3/4$, whence $4\cdot 1911\ge d_1\cdot (a_1+b_1{)}^3$; therefore, $8000>(a_1+b_1{)}^3$, or $a_1+b_1<20$.

Further, $a_1^3+b_1^3=(a_1+b_1{)}^3-3a_1{b}_1(a_1+b_1)$, and (1) can be rewritten as $$d_1(a_1+b_1)((a_1+b_1{)}^2-3a_1{b}_1)=1911.(2)$$ We see that $a_1+b_1$ is a divisor (less than 20) of $1911=3\cdot 7^2\cdot 13$. Hence exactly three cases are possible.

1. $a_1+b_1=3$, $a_1=1$, $b_1=2$. Then $d_1\cdot 3\cdot (9-6)=1911$, but $9$ is not a divisor of $1911$.

2. $a_1+b_1=7$. If $a_1=1$, $b_1=6$, then $d_1\cdot 7\cdot (49-18)=1911$, but $49-18=31$ is not a divisor of $1911$. If $a_1=2$, $b_1=5$, we have $d_1\cdot 7\cdot (49-30)=1911$, but $49-30=19$ is not a divisor of $1911$. Finally, if $a_1=3$, $b_1=4$, then $d_1\cdot 7\cdot (49-36)=1911$, which gives $d_1=21$, whence $d=21\cdot 3\cdot 4=252$, i.e., the pair $a=756$, $b=1008$ is a solution.

3. $a_1+b_1=13$. If $a_1=1$, $b_1=12$, then $d_1\cdot 13\cdot (169-36)=1911$, but $169-36=133$ does not divide $1911$. If $a_1=2$, $b_1=11$, then $d_1\cdot 13\cdot (169-66)=13d_1\cdot 103$ does not divide $1911$. If $a_1=3$, $b_1=10$, then $d_1\cdot 13\cdot (169-90)=13d_1\cdot 79$ does not divide $1911$. If $a_1=5$, $b_1=8$, then $d_1\cdot 13\cdot (169-120)=d_1\cdot 13\cdot 49=1911$, which gives $d_1=3$, whence $d=3\cdot 5\cdot 8=120$, i.e., the pair $a=600$, $b=960$ is a solution. Finally, if $a_1=6$, $b_1=7$, then $d_1\cdot 13\cdot (169-126)=13d_1\cdot 43$ does not divide $1911$.

Thus, the equation has two solutions mentioned above.