Yeah, no. $m+n$ divides $18$, and do ugly case work. Solutions are $(1,1);(1,2);(1,5)$

ZNatox wrote: Yeah, no. $m+n$ divides $18$, and do ugly case work. Solutions are $(1,1);(1,2);(1,5)$ You can also add that $m^2-mn+n^2$ divides $21$, To reduce number of cases

Can you explain how you get that?

$m^3 + n^3 = (m+n)(m^2 - mn + n^2)$, so $m^2 - mn + n^2 \mid m^2 + 20mn + n^2$, which implies $m^2 - mn + n^2 \mid m^2 + 20mn + n^2 - (m^2 - mn + n^2) = 21mn$. Now notice that $m^2 - mn + n^2$ is relatively prime to $mn$, so it must divide $21$

megarnie wrote: $m^3 + n^3 = (m+n)(m^2 - mn + n^2)$, so $m^2 - mn + n^2 \mid m^2 + 20mn + n^2$, which implies $m^2 - mn + n^2 \mid m^2 + 20mn + n^2 - (m^2 - mn + n^2) = 21mn$. Now notice that $m^2 - mn + n^2$ is relatively prime to $mn$, so it must divide $21$ Indeed but even though it is trivial to show it i think it's best to do it anyway $gcd(m^2- mn + n^2, mn)$ $=$ $gcd(m^2 + n^2, mn)=gcd(m^2, mn)=gcd(n^2, mn)=gcd(m ,n)=1$