### International Mathematics Competition for University Students

July 22 – 28 2018, Blagoevgrad, Bulgaria

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### Problem 9

Problem 9. Determine all pairs $\displaystyle P(x)$, $\displaystyle Q(x)$ of complex polynomials with leading coefficient $\displaystyle 1$ such that $\displaystyle P(x)$ divides $\displaystyle Q(x)^2+1$ and $\displaystyle Q(x)$ divides $\displaystyle P(x)^2+1$.

(Proposed by Rodrigo Angelo, Princeton University and Matheus Secco, PUC, Rio de Janeiro)

Solution. The answer is all pairs $\displaystyle (1,1)$ and $\displaystyle (P,P+i)$, $\displaystyle (P,P-i)$, where $\displaystyle P$ is a non-constant monic polynomial in $\displaystyle \mathbb{C}[x]$ and $\displaystyle i$ is the imaginary unit.

Notice that if $\displaystyle P | Q^2 + 1$ and $\displaystyle Q | P^2 + 1$ then $\displaystyle P$ and $\displaystyle Q$ are coprime and the condition is equivalent with $\displaystyle PQ | P^2 + Q^2 + 1$.

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

Proof. Assume for the sake of contradiction that there is a pair $\displaystyle (P,Q)$ with $\displaystyle \deg P \neq \deg Q$. Among all these pairs, take the one with smallest sum $\displaystyle \deg P + \deg Q$ and let $\displaystyle (P,Q)$ be such pair. Without loss of generality, suppose that $\displaystyle \deg P > \deg Q$. Let $\displaystyle S$ be the polynomial such that

$\displaystyle \frac{P^2+Q^2 + 1}{PQ} = S.$

Notice that $\displaystyle P$ a solution of the polynomial equation $\displaystyle X^2-QS X +Q^2 + 1 = 0$, in variable $\displaystyle X$. By Vieta's formulas, the other solution is $\displaystyle R = QS-P = \dfrac{Q^2 + 1}{P}$. By $\displaystyle R=QS-P$, the $\displaystyle R$ is indeed a polynomial, and because $\displaystyle P, Q$ are monic, $\displaystyle R=\dfrac{Q^2 + 1}{P}$ is also monic. Therefore the pair $\displaystyle (R,Q)$ satisfies the conditions of the Lemma. Notice that $\displaystyle \deg R = 2\deg Q - \deg P < \deg P$, which contradicts the minimality of $\displaystyle \deg P+ \deg Q$. This contradiction establishes the Lemma.

By the Lemma, we have that $\displaystyle \deg (PQ) = \deg (P^2+Q^2+1)$ and therefore $\displaystyle \dfrac{P^2+Q^2+1}{PQ}$ is a constant polynomial. If $\displaystyle P$ and $\displaystyle Q$ are constant polynomials, we have $\displaystyle P = Q = 1$. Assuming that $\displaystyle \deg P = \deg Q \geq 1$, as $\displaystyle P$ and $\displaystyle Q$ are monic, the leading coefficient of $\displaystyle P^2+Q^2+1$ is $\displaystyle 2$ and the leading coefficient of $\displaystyle PQ$ is $\displaystyle 1$, which give us $\displaystyle \dfrac{P^2+Q^2+1}{PQ} = 2$. Finally we have that $\displaystyle P^2+Q^2 + 1 = 2PQ$ and therefore $\displaystyle (P-Q)^2 = -1$, i.e $\displaystyle Q = P + i$ or $\displaystyle Q = P - i$. It's easy to check that these pairs are indeed solutions of the problem.