China National Team Selection Test

Proof Let $S_k = \sum_{i=1}^{2010} a_i^k$ and $\tilde{S}_k = \sum_{i=1}^{2010} b_i^k$. We first show by induction that $S_k = \tilde{S}_k$ for $k = 1, 2, \ldots, 2010$.

Setting $n = 1$ in the given equality, we have $2009S_1 = 2009\tilde{S}_1$, and hence $S_1 = \tilde{S}_1$. Assume that $S_j = \tilde{S}_j$ for $j = 1, 2, \ldots, k-1$, where $2 \le k \le 2010$; we are going to show that $S_k = \tilde{S}_k$.

By the binomial expansion theorem, we have
$$\begin{align*} \sum_{1 \le i < j \le 2010} (a_i + a_j)^k &= \sum_{1 \le i < j \le 2010} \sum_{l=0}^{k} \binom{k}{l} a_i^l a_j^{k-l} \\ &= 2009S_k + \sum_{1 \le i < j \le 2010} \sum_{l=0}^{k-1} \binom{k}{l} a_i^l a_j^{k-l} \\ &= 2009S_k + \frac{1}{2} \sum_{1 \le i < j \le 2010} \sum_{l=1}^{k-1} \binom{k}{l} a_i^l a_j^{k-l} \\ &= 2009S_k + \frac{1}{2} \sum_{l=1}^{k-1} \sum_{i=1}^{2010} \binom{k}{l} a_i^l (S_{k-l} - a_i^{k-l}) \\ &= 2009S_k + \frac{1}{2} \sum_{l=1}^{k-1} \left( \binom{k}{l} S_{k-l} \sum_{i=1}^{2010} a_i^l - \sum_{i=1}^{2010} \binom{k}{l} a_i^k \right) \\ &= 2009S_k + \frac{1}{2} \sum_{l=1}^{k-1} \left( \binom{k}{l} S_{k-l} S_l - \binom{k}{l} S_k \right) \\ &= \frac{1}{2} \sum_{l=1}^{k-1} \binom{k}{l} S_{k-l} S_l + (2010 - 2^{k-1}) S_k. \quad \textcircled{1} \end{align*}$$

Similarly, we have
$$\sum_{1 \le i < j \le 2010} (b_i + b_j)^k = \frac{1}{2} \sum_{l=1}^{k-1} \binom{k}{l} \tilde{S}_{k-l} \tilde{S}_l + (2010 - 2^{k-1}) \tilde{S}_k. \quad \textcircled{2}$$

Since
$$\sum_{1 \le i < j \le 2010} (a_i + a_j)^k = \sum_{1 \le i < j \le 2010} (b_i + b_j)^k,$$
by ①, ② and inductive hypothesis $S_i = \tilde{S}_i$ for $i = 1, 2, \dots, k-1$, we have $S_k = \tilde{S}_k$ (it is worth noting that $2010$ is not a power of $2$, i.e. $2010 - 2^{k-1} \neq 0$). This completes the inductive proof that $S_k = \tilde{S}_k$ for all $k = 1, 2, \dots, 2010$.

Set
$$(x - a_1) \cdots (x - a_{2010}) = x^{2010} + A_1 x^{2009} + \cdots + A_{2010}, \quad \textcircled{3}$$
$$(x - b_1) \cdots (x - b_{2010}) = x^{2010} + B_1 x^{2009} + \cdots + B_{2010}. \quad \textcircled{4}$$
By Newton's formula, we have
$$S_k + A_1 S_{k-1} + \cdots + A_{k-1} S_1 + k A_k = 0, \quad \textcircled{5}$$
$$\tilde{S}_k + B_1 \tilde{S}_{k-1} + \cdots + B_{k-1} \tilde{S}_1 + k B_k = 0, \quad \textcircled{6}$$
for $k = 1, 2, \dots, 2010$.

It follows from ⑤, ⑥ and $S_k = \tilde{S}_k$, $k = 1, 2, \dots, 2010$, by the easy inductive argument, we have
$$A_k = B_k, \quad k = 1, 2, \dots, 2010.$$
The right hand sides of equations ③ and ④ are equal, and so are their left hand sides, i.e. $A = B$.