IMC 2023 · Day 1 · P3

Hint: The polynomials $(x + iy)^n$ and $(x - iy)^m$ are trivial complex solutions. Suppose that $P(x, y) = (x + iy)^n (x - iy)^m Q(x, y)$, where $Q(x, y)$ is divisible neither by $x + iy$ nor $x = iy$ and consider $Q(x, y)$.

First we find all polynomials $P(x, y)$ with complex coefficients which satisfies the condition of the problem statement. The identically zero polynomial clearly satisfies the condition. Let consider other polynomials.

Let $i^2 = -1$ and $P(x, y) = (x + iy)^n (x - iy)^m Q(x, y)$, where $n$ and $m$ are non-negative integers and $Q(x, y)$ is a polynomial with complex coefficients such that it is not divisible neither by $x + iy$ nor by $x - iy$. By the problem statement we have $Q(x, y) Q(z, t) = Q(xz - yt, xt + yz)$. Note that $z = t = 0$ gives $Q(x, y) Q(0, 0) = Q(0, 0)$. If $Q(0, 0) \ne 0$, then $Q(x, y) = 1$ for all $x$ and $y$. Thus $P(x, y) = (x + iy)^n (x - iy)^m$. Now consider the case when $Q(0, 0) = 0$.

Let $x = iy$ and $z = -it$. We have $Q(iy, y) Q(-it, t) = Q(0, 0) = 0$ for all $y$ and $t$. Since $Q(x, y)$ is not divisible by $x - iy$, $Q(iy, y)$ is not identically zero and since $Q(x, y)$ is not divisible by $x + iy$, $Q(-it, t)$ is not identically zero. Thus there exist $y$ and $t$ such that $Q(iy, y) \ne 0$ and $Q(-it, t) \ne 0$ which is impossible because $Q(iy, y) Q(-it, t) = 0$ for all $y$ and $t$.

Finally, $P(x, y)$ polynomials with complex coefficients which satisfies the condition of the problem statement are $P(x, y) = 0$ and $P(x, y) = (x + iy)^n (x - iy)^m$. It is clear that if $n \ne m$, then $P(x, y) = (x + iy)^n (x - iy)^m$ cannot be polynomial with real coefficients. So we need to require $n = m$, and for this case $P(x, y) = (x + iy)^n (x - iy)^n = (x^2 + y^2)^n$.

So, the answer of the problem is $P(x, y) = 0$ and $P(x, y) = (x^2 + y^2)^n$ where $n$ is any non-negative integer.