Hellenic Mathematical Olympiad ARCHIMEDES

Let $A=(m+n{)}^3$, $B=2n(3m^2+n^2)+8$. Since $(m+n{)}^3\mid 2n(3m^2+n^2)+8$ it follows that: $$(m+n{)}^3\le 2n(3m^2+n^2)+8\iff m^3+3m^{2}n+3m{n}^2+n^3\le 6m^{2}n+2n^3+8$$ $$\iff m^3-3m^{2}n+3m{n}^2-n^3\le 8\iff (m-n{)}^3\le 8\iff m-n\le 2\iff m-n\in \{0,1,2\}.$$ We distinguish the cases: $$\bullet m-n=0\iff m=n.$$ $$A\mid B\iff 8m^3\mid 8m^3+8\iff 8m^3\mid 8\iff m=1,\text{(since }m>0\text{)}\Rightarrow (m,n)=(1,1).$$ $$\bullet m-n=1.$$ $$A=(2n+1{)}^3,B=2n(3(n+1{)}^2+n^2)+8=8n^3+12n^2+6n+8=(2n+1{)}^3+7.$$ $$\text{Therefore, }A\mid B\iff (2n+1{)}^3\mid 7\Rightarrow 2n+1=1\iff n=0,m=1,(m,n)=(1,0).$$

$\bullet$ $m-n=2$. Then $A=8(n+1{)}^3=B$, and hence we have an infinity of pairs of the form $(k+2,k)$, ${\mu }_k\ge 0$.