SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Denote $p$ as a prime of $a_1^2+a_1+1$, note that $a_1$ is odd (since $a_1^2+a_1+1=a_1\left(a_1+1\right)+1$ is an odd number) and $p\mid a_1$. By induction, we can show that $$a_n\equiv a_2\equiv -a_1(\mathrm{mod}p)\text{ for any }n>1.$$ Thus $a_n^2+a_n+1\equiv a_1^2-a_1+1\equiv -2a_1\not\equiv 0(\mathrm{mod}p)$ so $$v_p\left(\prod _{k=1}^n\left(a_k^2+a_k+1\right)\right)=v_p\left(a_1^2+a_1+1\right).$$ Since $a_1^2<a_1^2+a_1+1<{\left(a_1+1\right)}^2$, then $a_1^2+a_1+1$ is not a perfect square. This implies that there exist some prime $p$ such that $v_p\left(a_1^2+a_1+1\right)$ is odd. This finishes the proof. $\square$