Problem 1132

Solved by Matthieu Vogel

If $m(x)=x^{2020}$, Albert has a winning strategy. He starts by setting $a_{0}=1$. Thus, at the end of the game, $f(0)=1$. It is therefore impossible for $f$ to be divisible by $m$, otherwise, since $m(0)=0$, we would have $f(0)=0$.

If $m(x)=x^{2}+1$, Homer has a winning strategy. He pairs the $a_{i}$ (which is possible since 4 divides 2020) in the form $\left(a_{4 k}, a_{4 k+2}\right),\left(a_{4 k+1}, a_{4 k+3}\right)$.

Except for the pair $\left(a_{0}, a_{2}\right)$, Homer adopts the following strategy: he plays in the same pair as Albert and chooses the same real number. We then obtain that for any real number $a, a\left(X^{4 k}+X^{4 k+2}\right)=$ $a X^{4 k}\left(X^{2}+1\right), a\left(X^{4 k+1}+X^{4 k+3}\right)=a X^{4 k+1}\left(X^{2}+1\right)$, so we will indeed get a sum of polynomials divisible by $X^{2}+1$.

However, if Homer proceeds in this way, at the end, he will obtain a polynomial congruent to $X^{2020} \equiv(-1)^{1010} \equiv 1$ modulo $X^{2}+1$. For this, Homer would like the polynomial obtained with $a_{2}$ and $a_{0}$ to be congruent to -1 modulo 2020. In this case, we would indeed have $X^{2020}+a_{2} X^{2}+a_{0} \equiv$ $X^{2020}-1 \equiv 0$ modulo $X^{2}+1$. It is therefore sufficient to have $a_{0}=a_{2}-1$.

Thus, Homer applies the following strategy:

- If Albert plays $a_{0}$, Homer sets $a_{2}$ to $a_{0}+1$.
- If Albert plays $a_{2}$, Homer sets $a_{0}$ to $a_{2}-1$.

In summary, the polynomial obtained at the end is divisible by $X^{2}+1$, so Homer wins!