然后丢了一个IMO问题进去 .....

Find all polynomials f with real coefficients such that for all reals a,b,c such that ab + bc + ca = 0 we have the following relations f(a − b) + f(b − c) + f(c − a) = 2f(a + b + c)

十秒钟以后，可爱的AI说：

To solve this problem, we can first note that if $f(x)$ is a polynomial that satisfies the given condition for all real numbers $a$, $b$, and $c$, then it must also satisfy the condition for all complex numbers $a$, $b$, and $c$. This is because the given condition is linear in $f(a - b)$, $f(b - c)$, and $f(c - a)$, and therefore if the condition holds for real $a$, $b$, and $c$, it must also hold for complex $a$, $b$, and $c$.

Next, we can use the condition $ab + bc + ca = 0$ to rewrite the given condition as follows:

$$f(a - b) + f(b - c) + f(c - a) = 2f(a + b + c)$$

$$\implies f(a - b) + f(b - a) + f(c - b) + f(b - c) + f(a - c) + f(c - a) = 2f(a + b + c) + 2f(b)$$

$$\implies 2[f(a - b) + f(b - a) + f(c - b) + f(b - c) + f(a - c) + f(c - a)] = 2f(a + b + c) + 2f(b)$$

$$\implies 2[f(a - b) + f(b - a) + f(c - b) + f(b - c) + f(a - c) + f(c - a)] = 2[f(a + b + c) + f(b)]$$