Berkeley Math Circle Monthly Contest 1

Solution:
The answer is yes. Consider the following expression:
$$f(x, y) = 1 + (x-1)(y-1) + (x-1)(y-1)(x-2)(y-2) + (x-1)(y-1)(x-2)(y-2)(x-3)(y-3) + \cdots.$$
Here, although the sum appears to be infinite, if we fix a value $y = y_{0}$, all but the first $y_{0}$ terms contain the factor $(y - y_{0})$ and therefore equal $0$. Therefore $f(x, y_{0})$ is defined and indeed is a polynomial in $x$. (The initial term $1$ is merely to ensure that $f(x, y)$ is always positive.) Symmetrically, when $x$ is fixed, $f(x, y)$ becomes a polynomial in $y$.

It remains to prove that $f$ is not a polynomial in $x$ and $y$. If so, we can expand $f$ as a finite sum of terms $c x^{a} y^{b}$; let $a_{0}$ be the largest exponent $a$ occurring. Then, for every $y_{0}$, $f(x, y_{0})$ is a polynomial in $x$ of degree at most $a_{0}$. However, we see that $f(x, y_{0})$ is a polynomial in $x$ of degree $y_{0} - 1$. Taking $y_{0} = a_{0} + 2$ yields a contradiction.