Berkeley Math Circle: Monthly Contest 6

Solution:
Let $P(p, q)$ be the given condition. Expanding $P(p-q, q) + 2P(p, q) + P(p+q, q)$ gives
$$f(p+3q) - f(p-3q) = 3(f(p+q) - f(p-q)).$$
Let $F$ be the quadratic that equals $f$ at $-1, 0, 1$. Plugging in $p=0.5, q=0.5$, we get that it also intersects $f$ at $2$. Plugging in $p=1.5, q=0.5$ then gives that it also intersects $f$ at $3$, and continuing this induction gives that it is equal to $f$ for all positive integers. Continuing this reasoning in the other direction gives that it is equal to $f$ for all negative integers as well.

We now claim that $f$ is in fact equal to $F$. Let $r/s$ be a rational number. Consider the arithmetic sequence from $-r$ to $r$ with difference $r/s$. By similar reasoning, $f$ is a quadratic over this sequence. However, $F$ contains three of the terms: $-r, 0, r$. Therefore, $f$ is equal to $F$ over this sequence; in particular, $f(r/s) = F(r/s)$.

Therefore, $f$ is a quadratic, and all quadratics work, so the answer is all quadratics.