Baltic Way 2019

Assume that $b^2+b+1$ and $c^2+c+1$ are primes. Then $$a^2+a+1\equiv 0(\mathrm{mod}{b}^2+b+1).$$ This quadratic equation modulo prime has at most 2 roots. We must have either $a\equiv b(\mathrm{mod}{b}^2+b+1)$ or $a\equiv b^2(\mathrm{mod}{b}^2+b+1)$ (in the latter case $a^2+a+1\equiv b^4+b^2+1=(b^2+b+1)(b^2-b+1)\equiv 0$). We claim that we can rule out the smallest few such numbers of either of these forms. We have $a>b$, and $a\ne b^2$ since $a$ is not a perfect square. Our next possible value for $a$ is $b^2+b+1+b=(b+1{)}^2$ which is even, as is the value after that, $b^2+b+1+b^2$. Thus we have $a\ge 2(b^2+b+1)+b>2(b^2+b+1)$. By the same logic, $a>2(c^2+c+1)$. But then we have $$3(b^2+b+1)(c^2+c+1)=a^2+a+1>a^2>4(b^2+b+1)(c^2+c+1),$$ and this is a contradiction.