Exercise 18. We say that a set $A$ of polynomials with real coefficients is beautiful if, whenever $P$ and $Q$ are two distinct elements of $A$, there exist positive integers $a_{1}>\cdots>a_{2020}$ such that

$$
P Q=\sum_{i=1}^{2020} i X^{a_{i}}
$$

What is the maximal cardinality of a beautiful set?
Solution to Exercise 18 Let $A$ be a magnificent set. Suppose that $A$ has 3 distinct elements, which we denote as $P, Q, R$.

Since $\mathrm{P}^{2}=\frac{(\mathrm{PQ}) \times(\mathrm{PR})}{\mathrm{QR}}, \mathrm{P}^{2}$ is the product of two polynomials with integer coefficients, divided by a polynomial with integer coefficients. Therefore, it is a polynomial with rational coefficients that is also unitary. Indeed, the Euclidean algorithm for computing the GCD of two polynomials with rational coefficients yields a quotient and a remainder with rational coefficients, because the set of rationals is stable under division, multiplication, addition, and subtraction.

Thus, $P^{2}$ is a unitary polynomial with rational coefficients. Let $P=\sum_{n=0}^{d} a_{n} X^{n}$ with $a_{d} \neq 0$. We have $a_{d}^{2}=1$, so $a_{d}= \pm 1$. We will prove by strong downward induction that $a_{i}$ is rational for all $i$ between 0 and $n$.

The base case comes from the fact that $a_{n}= \pm 1$. For the inductive step, let $i \in\{0, n-1\}$ such that $a_{i+1}, \ldots, a_{n}$ are rational. The coefficient of degree $i+n$ of $P^{2}$ is $\sum_{j=i}^{n} a_{j} a_{n+i-j}=2 a_{i} a_{n}+\sum_{j=i+1}^{n-1} a_{j} a_{n+i-j}$.

If $i+1 \leqslant j \leqslant n-1, a_{j}$ and $a_{n+i-j}$ are rational, so $\sum_{j=i+1}^{n-1} a_{j} a_{n+i-j}$ is rational. Since the coefficient of degree $i+n$ of $P^{2}$ is rational, $2 a_{i} a_{n}$ is also rational. Since $a_{n}= \pm 1$, $a_{i}$ is rational, which concludes the induction.

Now, $P^{2}(1)=1+\cdots+2020=\frac{2020 \times 2021}{2}=1010 \times 2021$ which is divisible by 2 but not by 4. In particular, since $P_{i}$ has rational coefficients, there exists $(p, q) \in \mathbb{Z} \times \mathbb{N}^{*}$ such that $P(1)=\frac{p}{q}$. Therefore, $p^{2}=1010 \times 2021 \times q^{2}$. We obtain that $p \neq 0$, and that $V_{2}\left(p^{2}\right)=2 V_{2}(p)=V_{2}\left(1010 \times 2021 \times q^{2}\right)=$ $1+2 V_{2}(q)$. We conclude that $2 V_{2}(p)=2 V_{2}(q)+1$. Since the left term is even and the right term is odd, we obtain a contradiction.

Thus, we cannot have $|A| \geqslant 3$. However, we can have $|A|=2$ by taking $A=\left\{1, \sum_{i=1}^{2019} i X^{2019-i}\right\}$, which works well. We conclude that the maximum cardinality of a magnificent set is 2.

Alternative Solution $n^{\circ} 1$ This solution is essentially the same as the previous one but presented from a different angle. The first solution proved that the coefficients of the polynomials are rational. The additional idea is to notice that the same can be said about the polynomials divided by $\sqrt{2020}$. Suppose $A \supseteq\{P, Q, R\}$ is magnificent. Since $P^{2} \cdot Q R=P Q \cdot R P$ and $P Q, Q R, R P$ are unitary polynomials with rational coefficients, $\mathrm{P}^{2}$ is also unitary and has rational coefficients.

Let $\tilde{P}$ be the polynomial obtained by reversing the coefficients of $P$. Precisely, if $P=\sum_{i=0}^{n} p_{i} X^{i}$ with $p_{n} \neq 0$, then $\tilde{P}=\sum_{\tilde{i}=0}^{n} p_{n-i} X^{i}$. Reversing the coefficients is compatible with multiplication, i.e., $\tilde{\mathrm{PQ}}=\tilde{\mathrm{P}} \tilde{\mathrm{Q}}$.

In our case, we notice that $\frac{\mathrm{PQ}}{2020}, \frac{\text { RR }}{2020}, \frac{\tilde{\mathrm{RP}}}{2020}$ are also unitary and have rational coefficients because

$$
\frac{\tilde{P Q}}{2020}=\sum_{i=1}^{2020} \frac{i}{2020} X^{a_{1}-a_{i}}=X^{a_{1}-a_{2020}}+\frac{2019}{2020} X^{a_{1}-a 2019}+\cdots+\frac{1}{2020}
$$

Thus, repeating the argument, since $\left(\frac{\tilde{\mathrm{P}}}{\sqrt{2020}}\right)^{2} \cdot \frac{\tilde{\mathrm{QR}}}{2020}=\frac{\tilde{\mathrm{P}}}{2020} \cdot \frac{\tilde{\mathrm{RP}}}{2020}$ and $\frac{\tilde{\mathrm{PQ}}}{2020}, \frac{\tilde{\mathrm{Q} R}}{2020}, \frac{\tilde{\mathrm{RP}}}{2020}$ are unitary polynomials with rational coefficients, $\left(\frac{\tilde{\mathrm{P}}}{\sqrt{2020}}\right)^{2}$ is also unitary and has rational coefficients.

As proven in Solution 1, from $\mathrm{P}^{2} \in \mathbb{Q}[\mathrm{X}]$ and $\mathrm{P}^{2}$ being unitary, we get $\mathrm{P} \in \mathbb{Q}[\mathrm{X}]$. From $\left(\frac{\tilde{\mathrm{P}}}{\sqrt{2020}}\right)^{2} \in \mathbb{Q}[\mathrm{X}]$ and $\left(\frac{\tilde{\mathrm{P}}}{\sqrt{2020}}\right)^{2}$ being unitary, we get $\frac{\tilde{\mathrm{P}}}{\sqrt{2020}} \in \mathbb{Q}[\mathrm{X}]$. Contradiction, since $\mathrm{P} \neq 0$.

Comment from the graders: The exercise was rarely attempted. The ideas provided by the students who tried the problem are excellent and demonstrate a good understanding of the exercise. There are few actual errors, and the only remark to make is that certain concepts (degree, Viète's formulas) were doomed to fail given that one could replace $\mathrm{P} \in A$ with $XP$ and that the degrees of the elements of $A$ are poorly understood, an observation that can sometimes be salvaging.