Problem 1798

1. Reformulate the Problem:
Given two sets of complex numbers $A = \{a_1, a_2, \ldots, a_{2010}\}$ and $B = \{b_1, b_2, \ldots, b_{2010}\}$, we are given that:
$$\sum_{1 \leq i < j \leq 2010} (a_i + a_j)^k = \sum_{1 \leq i < j \leq 2010} (b_i + b_j)^k$$
for every $k = 1, 2, \ldots, 2010$. We need to prove that $A = B$.

2. Symmetric Polynomials:
Consider the symmetric polynomial $[e_1, e_2, \ldots, e_{2010}]$ defined as:
$$[e_1, e_2, \ldots, e_{2010}] = \sum_{\text{sym}} \prod_{i=1}^{2010} x_i^{e_i}$$
where the sum is over all permutations of $\{1, \ldots, 2010\}$.

3. Multiplication of Symmetric Polynomials:
We have the following property for symmetric polynomials:
$$[p_1, \ldots, p_{2010}] [q_1, \ldots, q_{2010}] = \sum_{\sigma} [p_1 + q_{\sigma(1)}, \ldots, p_{2010} + q_{\sigma(2010)}]$$
where the sum is over all permutations $\sigma$ of $\{1, \ldots, 2010\}$.

4. Lemma:
For a fixed $1 \leq k \leq 2010$, for every decreasing tuple of $2010$ nonnegative integers $(e_1, \ldots, e_{2010})$ with sum $k$, except when $e_1 = k$ and $e_2 = \cdots = e_{2010} = 0$, there exist two decreasing tuples $(p_1, \ldots, p_{2010})$ and $(q_1, \ldots, q_{2010})$ whose sums are positive and add up to $k$, such that when they are multiplied, the polynomial $[e_1, \ldots, e_{2010}]$ appears in the expansion.

5. Proof of Lemma:
Take the maximal $i \geq 2$ such that $e_i > 0$. Consider $(p_1, \ldots, p_{2010}) = (e_i, 0, \ldots, 0)$ and $(q_1, \ldots, q_{2010}) = (e_1, \ldots, e_{i-1}, 0, \ldots, 0)$. This construction works as required. $\blacksquare$

6. Key Claim:
We can determine the value of every polynomial $[i_1, \ldots, i_{2010}]$ where $i_1 + \cdots + i_{2010} = k$ for every $1 \leq k \leq 2010$.

7. Proof of Key Claim:
This is done by strong induction on $k$. The base case $k = 1$ is obvious since we are given a nonzero multiple of $[1, 0, \ldots, 0]$ for $k = 1$. For the inductive step, construct the set $S$ of all products $[p_1, \ldots, p_{2010}] [q_1, \ldots, q_{2010}]$ generated by the lemma as $(e_1, \ldots, e_{2010})$ varies. We know the value of every polynomial in $S$ since it is the product of lower-degree polynomials whose values we know by hypothesis.

8. Basis of Symmetric Homogeneous Polynomials:
The polynomials in $S$ and $[k, 0, \ldots, 0]$ form a basis of the vector space of symmetric homogeneous degree-$k$ polynomials in $x_1, \ldots, x_{2010}$. We need to show that the coefficient of $[k, 0, \ldots, 0]$ in the representation of $\sum_{1 \leq i < j \leq 2010} (x_i + x_j)^k$ is nonzero.

9. Coefficient Analysis:
We write:
$$\sum_{1 \leq i < j \leq 2010} (x_i + x_j)^k = 2009 \sum_{i=1}^{2010} x_i^k + \sum_{1 \leq i < j \leq 2010} \sum_{a=1}^{k-1} \binom{k}{a} x_i^a x_j^{k-a}$$
and:
$$\frac{1}{2} \sum_{a=1}^{k-1} \binom{k}{a} (x_1^a + \cdots + x_{2010}^a)(x_1^{k-a} + \cdots + x_{2010}^{k-a}) = \frac{1}{2} \sum_{a=1}^{k-1} \binom{k}{a} \left( \sum_{i=1}^{2010} x_i^k + \sum_{1 \leq i < j \leq 2010} (x_i^a x_j^{k-a} + x_i^{k-a} x_j^a) \right)$$
Since the blue parts are the same, it follows that if $\sum_{1 \leq i < j \leq 2010} (x_i + x_j)^k$ can be expressed as a linear combination of the above polynomials with the $x_1^k + \cdots + x_{2010}^k$ coefficient zero, then we must have:
$$2009 = \frac{1}{2} \sum_{a=1}^{k-1} \binom{k}{a} = \frac{1}{2} (2^k - 2) = 2^{k-1} - 1$$
but this is never true.

10. Conclusion:
Since we know the values of $\sum_{1 \leq i < j \leq 2010} (x_i + x_j)^k$ and all the polynomials in $S$, we can find the value of $[k, 0, \ldots, 0]$. Then, since $S$ and $[k, 0, \ldots, 0]$ form a basis for symmetric homogeneous degree-$k$ polynomials in $x_1, \ldots, x_{2010}$, we can find the value of every $[i_1, \ldots, i_{2010}]$ where $i_1 + \cdots + i_{2010} = k$, completing the inductive hypothesis. $\blacksquare$

11. Final Step:
By our key claim, we can find the $k$-th elementary symmetric sum of the $x_i$ for $1 \leq k \leq 2010$, which we can use to write down a unique monic degree-$2010$ polynomial whose roots are precisely $x_1, \ldots, x_{2010}$. Thus, the multiset $\{x_1, \ldots, x_{2010}\}$ is uniquely determined. $\blacksquare$