Find all functions $f: \mathbb{Q} \rightarrow \mathbb{R}$ from rational to real numbers such that for all rational $p, q$,

$$
f(p+2 q)-f(p-2 q)=2(f(p+q)-f(p-q)) .
$$
Let $P(p, q)$ be the given condition. Expanding $P(p-q, q)+2 P(p, q)+$ $P(p+q, q)$ gives

$$
f(p+3 q)-f(p-3 q)=3(f(p+q)-f(p-q)) \text {. }
$$

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.