IMO Team Selection Test 1

The smallest divisor of $n$ greater than $1$ is the smallest prime divisor of $n$, hence $d$ is prime. Moreover, we have $d\mid n$, hence $d\mid d^3+m^3$, and $d\mid m^3$. This yields that $m>1$. On the other hand we have $m\mid n$, hence $m\mid d^3+m^3$, and $m\mid d^3$. Because $d$ is prime and $m>1$, we also see that $m$ equals $d$, $d^2$, or $d^3$.

In all cases the parity of $m^3$ is equal to that of $d^3$, and $n=d^3+m^3$ is even. This means that the smallest divisor of $n$ greater than $1$ equals $2$, i.e. $d=2$. In case $m=d$, we find $n=2^3+2^3=16$, in case $m=d^2$, we find $n=2^3+2^6=72$, and in case $m=d^3$, we find $n=2^3+2^9=520$. These are indeed solutions: they are even so that $d=2$, and $2\mid 16$; $4\mid 72$ and $8\mid 520$, which indeed gives $m\mid n$. $\square$