74th Romanian Mathematical Olympiad

a) Consider the numbers $n=72\cdot a^6$, with natural $a$. Then $2\cdot n=(12\cdot a^3{)}^2$ and $3\cdot n=(6\cdot a^2{)}^3$, which shows that every such $n$ is 'good'.

b) If $3\cdot m$ is a perfect cube, then $3\cdot m$ is a multiple of $27$. In this case $m$ is a multiple of $9$, so $m+2=M_3+2$. Since the perfect squares are $M_3$ or $M_3+1$, $m+2$ cannot be a perfect square.