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
$$a_{n} \equiv a_{2} \equiv -a_{1} \pmod{p} \text{ for any } n>1.$$
Thus $a_{n}^{2}+a_{n}+1 \equiv a_{1}^{2}-a_{1}+1 \equiv -2a_{1} \not\equiv 0 \pmod{p}$ 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. $\square$