Saudi Arabia Mathematical Competitions

For any integer $a$ we have $a^3\equiv 0,1,8(\mathrm{mod}9)$, hence for any integers $x$ and $y$, we have $$x^3+y^3\equiv 0,1,2,7,8(\mathrm{mod}9).$$ For $n\ge 6$, we have $n!+4\equiv 4(\mathrm{mod}9)$, that is there are no solutions in this case.

If $n=1$, then the equation becomes $x^3+y^3=5$, which has no solutions because of the possible residues above.

If $n=2$, then we get the equation $x^3+y^3=6$, which has no solutions because of the possible residues above.

If $n=3$, then the equation becomes $x^3+y^3=10$. For any integer $a$ we have $a^3\equiv 0,\pm 1(\mathrm{mod}7)$, hence for any integers $x$ and $y$, we have $$x^3+y^3\equiv 0,\pm 1,\pm 2(\mathrm{mod}7).$$ The equation has no solutions because of the possible residues above.

If $n=4$, then the equation becomes $x^3+y^3=28$, with solution $(x,y)=(3,1)$.

If $n=5$, then we get the equation $x^3+y^3=124$, with solution $(x,y)=(5,-1)$.

The equation has solutions in integers if and only if $n=4$ and $n=5$.