IMC 2018 · Day 2 · P9

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 $P | Q^2 + 1$ and $Q | P^2 + 1$ then $P$ and $Q$ are coprime and the condition is equivalent with $PQ | P^2 + Q^2 + 1$.

Lemma. If $P, Q \in \mathbb{C}[x]$ are monic polynomials such that $P^2 + Q^2 + 1$ is divisible by $PQ$, then $\deg P = \deg Q$.

Proof. Assume for the sake of contradiction that there is a pair $(P, Q)$ with $\deg P \ne \deg Q$. Among all these pairs, take the one with smallest sum $\deg P + \deg Q$ and let $(P, Q)$ be such pair. Without loss of generality, suppose that $\deg P > \deg Q$. Let $S$ be the polynomial such that $$\frac{P^2 + Q^2 + 1}{PQ} = S.$$ Notice that $P$ a solution of the polynomial equation $X^2 - QSX + Q^2 + 1 = 0$, in variable $X$. By Vieta's formulas, the other solution is $R = QS - P = \dfrac{Q^2 + 1}{P}$. By $R = QS - P$, the $R$ is indeed a polynomial, and because $P, Q$ are monic, $R = \dfrac{Q^2 + 1}{P}$ is also monic. Therefore the pair $(R, Q)$ satisfies the conditions of the Lemma. Notice that $\deg R = 2 \deg Q - \deg P < \deg P$, which contradicts the minimality of $\deg P + \deg Q$. This contradiction establishes the Lemma.

By the Lemma, we have that $\deg(PQ) = \deg(P^2 + Q^2 + 1)$ and therefore $\dfrac{P^2 + Q^2 + 1}{PQ}$ is a constant polynomial. If $P$ and $Q$ are constant polynomials, we have $P = Q = 1$. Assuming that $\deg P = \deg Q \ge 1$, as $P$ and $Q$ are monic, the leading coefficient of $P^2 + Q^2 + 1$ is 2 and the leading coefficient of $PQ$ is 1, which give us $\dfrac{P^2 + Q^2 + 1}{PQ} = 2$. Finally we have that $P^2 + Q^2 + 1 = 2PQ$ 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.