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 $k=1,2,\dots ,2010$.

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 $j=1,2,\dots ,k-1$, where $2\le k\le 2010$; we are going to show that $S_k={\tilde{S}}_k$.

By the binomial expansion theorem, we have $$\begin{array}{rl}\sum _{1\le i<j\le 2010}(a_i+a_j{)}^k& =\sum _{1\le i<j\le 2010}\sum _{l=0}^k\left(\genfrac{}{}{0ex}{}{k}{l}\right)a_i^l{a}_j^{k-l}\\ & =2009S_k+\sum _{1\le i<j\le 2010}\sum _{l=0}^{k-1}\left(\genfrac{}{}{0ex}{}{k}{l}\right)a_i^l{a}_j^{k-l}\\ & =2009S_k+\frac{1}{2}\sum _{1\le i<j\le 2010}\sum _{l=1}^{k-1}\left(\genfrac{}{}{0ex}{}{k}{l}\right)a_i^l{a}_j^{k-l}\\ & =2009S_k+\frac{1}{2}\sum _{l=1}^{k-1}\sum _{i=1}^{2010}\left(\genfrac{}{}{0ex}{}{k}{l}\right)a_i^l(S_{k-l}-a_i^{k-l})\\ & =2009S_k+\frac{1}{2}\sum _{l=1}^{k-1}\left(\left(\genfrac{}{}{0ex}{}{k}{l}\right)S_{k-l}\sum _{i=1}^{2010}{a}_i^l-\sum _{i=1}^{2010}\left(\genfrac{}{}{0ex}{}{k}{l}\right)a_i^k\right)\\ & =2009S_k+\frac{1}{2}\sum _{l=1}^{k-1}\left(\left(\genfrac{}{}{0ex}{}{k}{l}\right)S_{k-l}{S}_l-\left(\genfrac{}{}{0ex}{}{k}{l}\right)S_k\right)\\ & =\frac{1}{2}\sum _{l=1}^{k-1}\left(\genfrac{}{}{0ex}{}{k}{l}\right)S_{k-l}{S}_l+(2010-2^{k-1})S_k.\stackrel{◯}{1}\end{array}$$

Similarly, we have $$\sum _{1\le i<j\le 2010}(b_i+b_j{)}^k=\frac{1}{2}\sum _{l=1}^{k-1}(\genfrac{}{}{0ex}{}{k}{l}){\tilde{S}}_{k-l}{\tilde{S}}_l+(2010-2^{k-1}){\tilde{S}}_k.\stackrel{◯}{2}$$

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. $2010-2^{k-1}\ne 0$). This completes the inductive proof that $S_k={\tilde{S}}_k$ for all $k=1,2,\dots ,2010$.

Set $$(x-a_1)\cdots (x-a_{2010})=x^{2010}+A_1{x}^{2009}+\cdots +A_{2010},\stackrel{◯}{3}$$ $$(x-b_1)\cdots (x-b_{2010})=x^{2010}+B_1{x}^{2009}+\cdots +B_{2010}.\stackrel{◯}{4}$$ By Newton's formula, we have $$S_k+A_1{S}_{k-1}+\cdots +A_{k-1}{S}_1+k{A}_k=0,\stackrel{◯}{5}$$ $${\tilde{S}}_k+B_1{\tilde{S}}_{k-1}+\cdots +B_{k-1}{\tilde{S}}_1+k{B}_k=0,\stackrel{◯}{6}$$ 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 $$A_k=B_k,k=1,2,\dots ,2010.$$ The right hand sides of equations ③ and ④ are equal, and so are their left hand sides, i.e. $A=B$.