Seventeenth Stars of Mathematics Competition

Suppose, if possible, that $(a^n+a+1)(b^n+b+1)$ is a square for all but finitely many positive integers $n$. Then $$ab\equiv \frac{(a^{p+1}+a+1)(b^{p+1}+b+1)}{(a^{p-2}+a+1)(b^{p-2}+b+1)}$$ and $$(a+2)(b+2)\equiv (a^{p-1}+a+1)(b^{p-1}+b+1)(\mathrm{mod}p)$$ are both quadratic residues modulo all but finitely many primes $p$. Consequently, $ab$ and $(a+2)(b+2)$ are both squares. (It is a fact that a positive integer which is a quadratic residue modulo all but finitely many primes is itself a square.) Let $b-a=q$, a prime, and let $d=\gcd (a,b)$. Clearly, $d=1$ or $d=q$. If $d=q$, then $a=(c-1)q$ and $b=cq$ for some integer $c>1$. Hence $c(c-1)q^2=ab$ is a square, forcing $c(c-1)$ to be a square. Coprimality of $c$ and $c-1$ forces in turn $c$ and $c-1$ to be both squares. This is impossible, since two squares of non-zero integers are at least 3 distance apart. Consequently, $a$ is coprime to $b$, so they are both squares.

Since $a$ and $b$ are squares, $a+2$ and $b+2$ are not; and since $(a+2)(b+2)$ is a square, $a+2$ is not coprime to $b+2$ (otherwise they would both be squares). Clearly, $\gcd (a+2,b+2)$ divides $(b+2)-(a+2)=b-a=q$, so $\gcd (a+2,b+2)=q$. As before, $a+2=(c^{\prime }-1)q$ and $b+2=c^{\prime }q$ for some integer $c^{\prime }>1$, and we reach a contradiction along the same lines. This ends the proof.