Sketch of the solution I gave in contest $\newline$ Suppose by contradiction that for any $n \ge N$ we have $(a^n+a+1)(b^n+b+1)$ is a perfect square. Let $b-a=p$ $\newline$ Now fix any $r$ natural number(it can be 0). $\newline$ Take a huge prime $q $ and take $n \equiv r \pmod{q-1}$ and $n \ge N$ so we get that $(a^r+a+1)(b^r+b+1)$ is a quadratic residue mod $q$ for all huge primes $q$ so we get $(a^r+a+1)(b^r+b+1)$ is a perfect square . $\newline$ So we may take $n \in {0,1}$ and get that $(a+2)(b+2)$ and $(2a+1)(2b+1)$ are both perfect squares. Then you bash cases using the fact that gcd is $p$ or $1$ on both products

Here's a sketch with analytic tools: Suppose $b>a$. And suppose that $ (a^n+a+1)(b^n+b+1)$ is a perfect square for all large enough $n$. Let $x_n^2 = (a^n+a+1)(b^n+b+1)$, then one can prove that $\lim_{n \to \infty} \frac{x_n}{(\sqrt{ab})^n} = 1$. This inspires us to look at $y_n = ab x_{2n} - x_{2n+2}$. Then $$y_n (abx_{2n} + x_{2n+2}) = (b^3 + b^2 - b - 1) a^{2 n + 2} + (a^3 + a^2 - a - 1) b^{2 n + 2} + (a^2b^2 - 1)(a+1) (b+1),$$so $y_n$ grows like $\displaystyle c \cdot \left( \frac ba \right)^n$ for a constant $c$. Thus, we consider $z_n = ay_{n+1} - by_n$. With $z_n$, one finds success since we can actually show $z_n \to 0$ as $n \to \infty$. Since $z_n$ is a sequence of integers, this implies that $z_n = 0$ for all $n$ large enough. Thus, if you fix $N$ large, then $\displaystyle y_{N+k} = \left(\frac ba \right)^k y_N ---(*)$ for all $k$, which means that $\boxed{a \mid b}$. Now, $b-a$ is a prime $p$, so this forces $a=p$ and $b = 2p$. Now, $(*)$ forces $y_{N+k}$ to be divisible by $2^k$. But note that $a^n+a+1$ is always odd, so $x_n$ is always odd, and therefore $y_n = (2p^2)x_{2n} - x_{2n+2}$ is also always odd, a contradiction.

Can you give me more infor about this competition ?. Like the problems in 2022,2021,...