59th Ukrainian National Mathematical Olympiad

We rewrite the equation as follows. $$n^3=p^3+2p^{2}q+2p{q}^2+q^3=(p+q)(p^2+pq+q^2)$$ Clearly, $p\ne q$, because, otherwise, the equation $n^3=6p^3$ would hold, which is not possible for positive integers. Suppose there exists such $s>1$, which is a factor of both $p+q$ and $(p^2+pq+q^2)$. But then $s\mid (p+q{)}^2-(p^2+pq+q^2)=pq$. If, for example, $s\mid p$, then from $s\mid (p+q)$ follows $s\mid q\Rightarrow$ from $p,q$ being prime, it must be that $p=q$, which leads to contradiction. If, on the other hand, $((p+q),(p^2+pq+q^2))=1$, then each factor must be a cube of a positive integer. Suppose $p+q=x^3$, $p^2+pq+q^2=y^3$, and $n=xy$. Then, $$pq=(p+q{)}^2-(p^2+pq+q^2)=x^6-y^3=(x^2-y)(x^4+x^{2}y+y^2).$$ Since $x^4+x^{2}y+y^2>x^3=p+q$, we obtain $x^4+x^{2}y+y^2=pq$ and $x^2-y=1\Rightarrow pq=x^4+x^2(x^2-1)+(x^2-1{)}^2=3x^4-3x^2+1=3x^2(x-1)(x+1)+1\equiv 1(\mathrm{mod}9)$, since $x(x-1)(x+1)\not\equiv 3(\mathrm{mod}9)$. What is left is to search through the cases modulo 9. Cube of an integer modulo 9 can be equal to 0, \pm 1. Let us list all possible cases for remainders modulo 9 of primes $p,q$ so that condition $pq\equiv 1(\mathrm{mod}9)$ is satisfied. $$p\equiv 1(\mathrm{mod}9)\Rightarrow q\equiv 1(\mathrm{mod}9)\Rightarrow x^3=p+q\equiv 2(\mathrm{mod}9)\{\text{ – contradiction.}\}$$ $$p\equiv 2(\mathrm{mod}9)\Rightarrow q\equiv 5(\mathrm{mod}9)\Rightarrow x^3=p+q\equiv 7(\mathrm{mod}9)\{\text{ – contradiction.}\}$$ $$p\equiv 4(\mathrm{mod}9)\Rightarrow q\equiv 7(\mathrm{mod}9)\Rightarrow x^3=p+q\equiv 2(\mathrm{mod}9)\{\text{ – contradiction.}\}$$ $$p\equiv 8(\mathrm{mod}9)\Rightarrow q\equiv 8(\mathrm{mod}9)\Rightarrow x^3=p+q\equiv 7(\mathrm{mod}9)\{\text{ – contradiction.}\}$$ All the cases were checked, which concludes the proof that such numbers do not exist.