A2
Determine all sequences $\left(x_{1}, x_{2}, \ldots, x_{2011}\right)$ of positive integers such that for every positive integer $n$ there is an integer $a$ with
$$
x_{1}^{n}+2 x_{2}^{n}+\cdots+2011 x_{2011}^{n}=a^{n+1}+1 .
$$
Answer. The only sequence that satisfies the condition is
$$
\left(x_{1}, \ldots, x_{2011}\right)=(1, k, \ldots, k) \quad \text { with } k=2+3+\cdots+2011=2023065 .
$$

Solution. Throughout this solution, the set of positive integers will be denoted by $\mathbb{Z}_{+}$.

Put $k=2+3+\cdots+2011=2023065$. We have
$$
1^{n}+2 k^{n}+\cdots 2011 k^{n}=1+k \cdot k^{n}=k^{n+1}+1
$$
for all $n$, so $(1, k, \ldots, k)$ is a valid sequence. We shall prove that it is the only one.
Let a valid sequence $\left(x_{1}, \ldots, x_{2011}\right)$ be given. For each $n \in \mathbb{Z}_{+}$we have some $y_{n} \in \mathbb{Z}_{+}$with
$$
x_{1}^{n}+2 x_{2}^{n}+\cdots+2011 x_{2011}^{n}=y_{n}^{n+1}+1 \text {. }
$$

Note that $x_{1}^{n}+2 x_{2}^{n}+\cdots+2011 x_{2011}^{n}1$. Applying the lemma to (1) we see that $a_{m}=y=m, a_{1}=1$, and all the other $a_{i}$ are zero. This implies $\left(x_{1}, \ldots, x_{2011}\right)=(1, m, \ldots, m)$. But we also have $1+m=a_{1}+\cdots+a_{m}=1+\cdots+2011=1+k$ so $m=k$, which is what we wanted to show.