International Mathematics Competition

The answer is all pairs $(1,1)$ and $(P, P+i),(P, P-i)$, where $P$ is a non-constant monic polynomial in $\mathbb{C}[x]$ and $i$ is the imaginary unit. Notice that if $P \mid Q^{2}+1$ and $Q \mid P^{2}+1$ then $P$ and $Q$ are coprime and the condition is equivalent with $P Q \mid P^{2}+Q^{2}+1$. Lemma. If $P, Q \in \mathbb{C}[x]$ are monic polynomials such that $P^{2}+Q^{2}+1$ is divisible by $P Q$, then $\operatorname{deg} P=\operatorname{deg} Q$. Proof. Assume for the sake of contradiction that there is a pair $(P, Q)$ with $\operatorname{deg} P \neq \operatorname{deg} Q$. Among all these pairs, take the one with smallest sum $\operatorname{deg} P+\operatorname{deg} Q$ and let $(P, Q)$ be such pair. Without loss of generality, suppose that $\operatorname{deg} P>\operatorname{deg} Q$. Let $S$ be the polynomial such that $\frac{P^{2}+Q^{2}+1}{P Q}=S$. Notice that $P$ a solution of the polynomial equation $X^{2}-Q S X+Q^{2}+1=0$, in variable $X$. By Vieta's formulas, the other solution is $R=Q S-P=\frac{Q^{2}+1}{P}$. By $R=Q S-P$, the $R$ is indeed a polynomial, and because $P, Q$ are monic, $R=\frac{Q^{2}+1}{P}$ is also monic. Therefore the pair $(R, Q)$ satisfies the conditions of the Lemma. Notice that $\operatorname{deg} R=2 \operatorname{deg} Q-\operatorname{deg} P<\operatorname{deg} P$, which contradicts the minimality of $\operatorname{deg} P+\operatorname{deg} Q$. This contradiction establishes the Lemma. By the Lemma, we have that $\operatorname{deg}(P Q)=\operatorname{deg}\left(P^{2}+Q^{2}+1\right)$ and therefore $\frac{P^{2}+Q^{2}+1}{P Q}$ is a constant polynomial. If $P$ and $Q$ are constant polynomials, we have $P=Q=1$. Assuming that $\operatorname{deg} P=\operatorname{deg} Q \geq 1$, as $P$ and $Q$ are monic, the leading coefficient of $P^{2}+Q^{2}+1$ is 2 and the leading coefficient of $P Q$ is 1 , which give us $\frac{P^{2}+Q^{2}+1}{P Q}=2$. Finally we have that $P^{2}+Q^{2}+1=2 P Q$ and therefore $(P-Q)^{2}=-1$, i.e $Q=P+i$ or $Q=P-i$. It's easy to check that these pairs are indeed solutions of the problem.