smc

The equation is equivalent to $m(m+1)(m+5)=27n^3+27n^2+9n+1$. Taking mod 3, we get $m(m+1)(m+2)=1(\mathrm{mod}3)$. However, $m(m+1)(m+2)$ is always divisible by $3$ for any integer $m$. Thus, the answer is $\boxed{0}$