### 2011 IMO SL #A2

Determine all sequences $(x_1,x_2,\ldots,x_{2011})$ of positive integers, such that for every positive integer $n$ there exists an integer $a$ with$\sum^{2011}_{j=1} j x^n_j = a^{n+1} + 1$

The answer is  $\boxed{(1,K,K,\cdots,K) \text{ for } K=2+3+\cdots+2011}$.

Denote the corresponding sequence value for place $n$ be $a_n$.

We claim that the sequence $(a_n)$ is bounded, or in other words that there exists a constant $C$ such that $(a_n)$ is always less than it. Note that $$a_n = (x_1^n+2x_2^n + \cdots + 2011x_{2011}^n-1)^{\tfrac{1}{n+1}}<\big[x_1+2x_2+\cdots+2011x_{2011}+100\big]^{\tfrac{n}{n+1}}.$$ Let $$C=\big[x_1+2x_2+\cdots+2011x_{2011}+100\big],$$ and then we have $$\big[x_1+2x_2+\cdots+2011x_{2011}+100\big]^{\tfrac{n}{n+1}}=C^{\tfrac{n}{n+1}}<C,$$ so we have proven our claim.

For sufficiently large primes $p$, $a_{p-1}=K$ holds true. Denote $n=p-1$. Consider the given condition $\bmod{p}$ to obtain $$1+2+\cdots+2011\equiv a_{p-1}+1\pmod p,$$ and therefore for large $p>\max(a_n)$, we get $a_{p-1}=K$, as desired by the claim.

Hence, for sufficiently large primes $p$, by dividing by $K$, we obtain $\left(\frac{x_1}K\right)^{p-1}+\cdots+2011\left(\frac{x_{2011}}K\right)^{p-1}=K+\frac1{K^{p-1}}.$ From this we get $x_i \le K$ for all defined $i$. Considering $p \to \infty$, we get $(x_i/K)^{p-1}\in\{0,1\}$, where equality holds true when $x_i=K$. The RHS approaches $K$, and hence we must have $x_1<K$ and $x_2=x_3 \cdots = x_{2011}=K$.

By $n=1$, we have that $(x_1-1)+K^2$ is a perfect square. However, $x_1-1<K$, so we know that $x_1=1$.

We are done. $\square$

1. schukkayapally1/15/2022

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