Baltic Way 2019

Assume that $p^2+p+1=k^3$. Then we have that $p(p+1)=k^3-1=(k-1)(k^2+k+1)$. Since $p$ is prime, either $p\mid k-1$ or $p\mid k^2+k+1$. If $p\mid k-1$, then $p\le k-1$ and $k^3\ge (p+1{)}^3>p^2+p+1=k^3$, which is impossible. Now consider the possibility that $p\mid k^2+k+1$. Since $p$ is prime, we have one of 3 cases.

1) $p=3$, which does not give a solution;

2) $p\equiv 1(\mathrm{mod}3)$. Thus, $p^2+p+1\equiv 3(\mathrm{mod}9)$, but no perfect cube is congruent to 3 (mod 9) and so we again reach a contradiction.

3) $p\equiv 2(\mathrm{mod}3)$. Then $p=2$ is not a solution. And for odd $p$ we have $4(k^2+k+1{)}^2\equiv 0(\mathrm{mod}p)$, or, equivalently, $(2k+1{)}^2\equiv -3(\mathrm{mod}p)$. This is impossible because $-3$ is not a square residue modulo $p=3\ell +2$. This detail is not quite elementary and can be checked via the properties of the Legendre symbol: $$\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)=(-1{)}^{\frac{p-1}{2}}\cdot \left((-1{)}^{\frac{3-1}{2}}\cdot \frac{p-1}{2}\left(\frac{p}{3}\right)\right)=\left(\frac{2}{3}\right)=-1.$$

Assume $p^2+p+1=(x+1{)}^3$ for a certain $x\ge 1$. This yields $p(p+1)=x(x^2+3x+3)$. Clearly $p>x$, so $p$ (prime) must divide $x^2+3x+3$; hence $$x^2+3x+3=ap\text{and}p+1=ax$$ for a certain $a\ge 1$. Elimination of $p$ leads to $$x^2-(a^2-3)x+(a+3)=0,$$ a quadratic equation whose discriminant $$D=(a^2-3{)}^2-4(a+3)$$ must be an even square. Thus $D\le (a^2-5{)}^2$, recasting into $a^2\le a+7$; and $a$ must be odd. This leaves 1 and 3 as admissible values of $a$; but neither of them yields an integer solution to the quadratic equation (displayed).