orl wrote: 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 \[\sum^{2011}_{j=1} j x^n_j = a^{n+1} + 1\] I am pleased to see a problem of mine on the IMO Shortlist. I went to IMO as a contestant not too long ago. Here I reproduce the solution given in the Shortlist booklet. Answer: The only sequence that satisfies the condition is \[ (x_1,x_2,\ldots,x_{2011})=(1,k,\ldots,k) \text{ with } k=2+3+\cdots+2011=2023065.\] Solution: Throughout this solution, the set of positive integers will be denoted by $\mathbb{Z}_+$. Put $k=2+3+\cdots+2011=2023065$. We have \[ 1^n+2k^n+\cdots+2011k^n = 1+k\cdot k^n = k^{n+1}+1\] 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 $(x_1,x_2,\ldots,x_{2011})$ be given. For each $n\in\mathbb{Z}_+$ we have some $y_n\in\mathbb{Z}_+$ with \[ x_1^n+2x_2^n+\cdots+2011x_{2011}^n = y_n^{n+1}+1.\] Note that $x_1^n+2x_2^n+\cdots+2011x_{2011}^n < (x_1+2x_2+\cdots+2011x_{2011})^{n+1}$, which implies that the sequence $(y_n)$ is bounded. In particular, there is some $y\in\mathbb{Z}_+$ with $y_n=y$ for infinitely many $n$. Let $m$ be the maximum of all the $x_i$. Grouping terms with equal $x_i$ together, the sum $x_1^n+2x_2^n+\cdots+2011x_{2011}^n$ can be written as \[x_1^n+2x_2^n+\cdots+2011x_{2011}^n=a_mm^n+a_{m-1}(m-1)^n+\cdots+a_1\] with $a_i\geq 0$ for all $i$ and $a_1+\cdots+a_{2011}=1+2+\cdots+2011$. So there exist arbitrarily large values of $n$, for which \[a_mm^n+\cdots+a_1-1-y\cdot y^n=0.\] The following lemma will help us determine the $a_i$ and $y$: Lemma: Let integers $b_1,\ldots,b_N$ be given and assume that there are arbitrarily large positive integers $n$ with $b_1+b_22^n+\cdots+b_NN^n=0$. Then $b_i=0$ for all $i$. Proof: Suppose that not all $b_i$ are zero. We may assume without loss of generality that $b_N\neq 0$. Dividing through by $N^n$ gives \[|b_N| = \left|b_{N-1}\left(\dfrac{N-1}{N}\right)^n+\cdots+b_n\left(\dfrac{1}{N}\right)^n\right|\leq (|b_{N-1}|+\cdots+|b_1|)\left(\dfrac{N-1}{N}\right)^n.\] The expression $\left(\frac{N-1}{N}\right)^n$ can be made arbitrarily small for $n$ large enough, contradicting the assumption that $b_N$ be non-zero. The lemma is proved. $\square$ We obviously have $y>1$. Applying the lemma we see that $a_m=y=m,a_1=1$, and all the other $a_i$ are zero. This implies $(x_1,\ldots,x_{2011})=(1,m,\ldots,m)$. 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.

Really nice problem

eaglet wrote: orl wrote: 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 \[\sum^{2011}_{j=1} j x^n_j = a^{n+1} + 1\] I am pleased to see a problem of mine on the IMO Shortlist. I went to IMO as a contestant not too long ago. Here I reproduce the solution given in the Shortlist booklet. Answer: The only sequence that satisfies the condition is \[ (x_1,x_2,\ldots,x_{2011})=(1,k,\ldots,k) \text{ with } k=2+3+\cdots+2011=2023065.\] Solution: Throughout this solution, the set of positive integers will be denoted by $\mathbb{Z}_+$. Put $k=2+3+\cdots+2011=2023065$. We have \[ 1^n+2k^n+\cdots+2011k^n = 1+k\cdot k^n = k^{n+1}+1\] 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 $(x_1,x_2,\ldots,x_{2011})$ be given. For each $n\in\mathbb{Z}_+$ we have some $y_n\in\mathbb{Z}_+$ with \[ x_1^n+2x_2^n+\cdots+2011x_{2011}^n = y_n^{n+1}+1.\] Note that $x_1^n+2x_2^n+\cdots+2011x_{2011}^n < (x_1+2x_2+\cdots+2011x_{2011})^{n+1}$, which implies that the sequence $(y_n)$ is bounded. In particular, there is some $y\in\mathbb{Z}_+$ with $y_n=y$ for infinitely many $n$. Let $m$ be the maximum of all the $x_i$. Grouping terms with equal $x_i$ together, the sum $x_1^n+2x_2^n+\cdots+2011x_{2011}^n$ can be written as \[x_1^n+2x_2^n+\cdots+2011x_{2011}^n=a_mm^n+a_{m-1}(m-1)^n+\cdots+a_1\] with $a_i\geq 0$ for all $i$ and $a_1+\cdots+a_{2011}=1+2+\cdots+2011$. So there exist arbitrarily large values of $n$, for which \[a_mm^n+\cdots+a_1-1-y\cdot y^n=0.\] The following lemma will help us determine the $a_i$ and $y$: Lemma: Let integers $b_1,\ldots,b_N$ be given and assume that there are arbitrarily large positive integers $n$ with $b_1+b_22^n+\cdots+b_NN^n=0$. Then $b_i=0$ for all $i$. Proof: Suppose that not all $b_i$ are zero. We may assume without loss of generality that $b_N\neq 0$. Dividing through by $N^n$ gives \[|b_N| = \left|b_{N-1}\left(\dfrac{N-1}{N}\right)^n+\cdots+b_n\left(\dfrac{1}{N}\right)^n\right|\leq (|b_{N-1}|+\cdots+|b_1|)\left(\dfrac{N-1}{N}\right)^n.\] The expression $\left(\frac{N-1}{N}\right)^n$ can be made arbitrarily small for $n$ large enough, contradicting the assumption that $b_N$ be non-zero. The lemma is proved. $\square$ We obviously have $y>1$. Applying the lemma we see that $a_m=y=m,a_1=1$, and all the other $a_i$ are zero. This implies $(x_1,\ldots,x_{2011})=(1,m,\ldots,m)$. 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. I have a question. How was the lemma applied on the $(1)$ expression? please show me it, thanks in advance

Here's a similar solution to above, phrased slightly differently, Once again, we note that $a^{n+1}+1 = \sum^{2011}_{j=1}j x^n_j < (x_1+2x_2+3x_3 + \ldots +2011x_{2011})^{n+1}+1$, so $a < (x_1+2x_2+3x_3 + \ldots +2011x_{2011})$, so $a$ is bounded. So $a$ takes some value infinitely many times. Call this value $d$. Now look only at the terms that take this value $d$. Set $\sum^{2011}_{j=1}j x^n_j = d^{n+1}+1$ and divide both sides by $d^n$ to get $\sum^{2011}_{j=1}j \left(\frac{x_j}{d}\right)^n = d+d^{-n}$. Since $d$ is used infinitely many times, take the limit of both sides as $n$ goes to infinity. The RHS goes to $d$. Note that $x_j \le d$ always, or else the LHS goes to infinity. On the LHS, for each of the $x_j$, either $x_j = d$, or $x_j < d$. When we take the limit, the terms with $x_j < d$ go to 0. Let $S$ be the set of $j$ with $x_j = d$. Then it's clear that $d = \sum_{j \in S}j$. So we rewrite the sum on the LHS as $\sum^{2011}_{j=1}j x^n_j = \sum_{j \in S}j x^n_j + \sum_{j \not\in S}j x^n_j$. But note that for $j \in S$, $x_j = d$, so $\sum_{j \in S}j x^n_j = d^{n+1}$. So the remaining things must sum to $1$, so $\sum_{j \not\in S}j x^n_j = 1$. Since all the $x_i$ are positive integers, this trivially implies that $x_1 = 1$ and $S = \{2, 3, \ldots, 2011\}$, which characterizes all sequences.

Here's a slightly diffrent proof. First let $k=2011*1006$ Lemma: All sequences that are good(meaning they satisfy the condition) satisfy $x_{i}< k$. Proof: Assume the opposite. Then if $x_{m}=r$ is the maximum of all $x_{i}$ than the left side is not greater than $x_{m}^(n+1)$ so $a\le x_{m}-1$ but the left side is no less than $x_{m}^n+1$ so it's easy to see that we can choose large enough $n$ such that $x_{m}^n>(x_{m}-1)^{n+1}$ which gives a contradiction. Back to the problem. For a large prime $p$ consider $n=p-1$. Then the left side is congruent to $k$ modulo $p$ but the right side is congruent to $a+1$. So $a\ge k-1$. Now using the lemma it's easy to get that $x_{2}=x_{3}=....=x_{n}=k-1$ and $x_{1}=1$(just consider how many times $(k-1)^n$ appears on the $LHS$.

@mathocean: absolutely incredible solution. You made the problem seem trivial.

Let the $a$ for a given $n$ be $a_n$. $X$ be the largest of the $x_i$ and say that $X=x_k=x_{\ell}=...=x_j$ are all the values that are $X$. Then as $n$ goes to infinity, $\frac{ax_a^n}{X^n}$ goes to $0$ if $x_a \neq X$ and so $\frac{\sum ax_a^n}{(k+\ell+...+j)X^n}$ goes to $1$. Therefore if $k+\ell+...+j=A$ we then have $\frac{(a_n^{n+1}+1)}{AX^n}$ goes to $1$ and therefore $\frac{a_n}{A^{1/(n+1)}X^{n/(n+1)}}$ goes to $1$, but notice that $\frac{A^{1/(n+1)}}{X^{1/(n+1)}}$ goes to $1$, and therefore $\frac{a_n}{X}$ goes to $1$ and since they are integers, $a_n=X$ for sufficiently big $n$ and this implies that $A=a_n=X$ for big enough $n$. For such an $n$ we get that $AX^n+\sum_{a : x_a \neq X}ax_a^n = a^{n+1}+1$ and so $\sum_{a : x_a \neq X}ax_a^n$ is $1$ and this implies that there is only one such $x_a$, namely $x_1$. Therefore $(x_1.,,,.x_{2011})=(1,X,...,X)$ and notice $X=k+\ell+...+j=2+3+...+2011$ and this is the only solution.

Wow, really nice to see an analytic flavored problem on an IMO Shortlist. Here's an idea of a different sort to nail it with the lemma in the first post. As done earlier we get that $a$ is bounded. Now, choose a prime $p > \text max [x_1,...,x_{2011},a]$ and set $n \equiv 0 \bmod (p-1)$ and so by FLT, we have $a \equiv (2+3+...+2011)=2023065 \bmod p$ for all very huge primes $p$ which implies $a=2023065$. Now, we argue that no $x_i$ can be larger than $a$, for otherwise, we divide by $a^n$ both the sides and let $n$ tend to $\infty$ to get a contadiction to the fact that $a=2023065$ is bounded. Now, it is evident that not less than $2010$ terms can be equal to $a$ and not all terms can be less than $a$ and none greater than $a$, so $x_2=...=x_{2011}=a=2023065$ and $x_1=1$.

anantmudgal09 wrote: Wow, really nice to see an analytic flavored problem on an IMO Shortlist. Here's an idea of a different sort to nail it with the lemma in the first post. Now, it is evident that not less than $2010$ terms can be equal to $a$ and not all terms can be less than $a$ and none greater than $a$, so $x_2=...=x_{2011}=a=2023065$ and $x_1=1$. How can you prove the last claim?

Who can help to prove the last claim of anantmudgal09?

If not, then the LHS is at most $(a-1)a^n+\mathcal{O}(b^n)$ for some $b<a$ which fails for large $n$.

The answer is $(x_1, x_2, \dots, x_{2011}) = (1, k, \dots, k)$ where $k = 2+\dots+2011$. Clearly this works. First, let $m = \max (x_i)$. Then \[ m^n < a_n^{n+1} + 1 < Cm^n \implies \sqrt[n+1]{m^n-1} < a_n < \sqrt[n+1]{Cm^n-1}. \]where $C = 1 + \dots + 2011$. As $n \to \infty$, both sides approach $m$, and since $a_n$ is an integer this means for large enough $n$ we conclude $a_n = m$. Then, dividing through gives that for large $n$ we have \[ \left( \frac{x_1}{m} \right)^n + 2 \left( \frac{x_2}{m} \right)^n + 3 \left( \frac{x_3}{m} \right)^n + \dots + 2011 \left( \frac{x_{2011}}{m} \right)^n = m + m^{-n}. \]Now, of the $2011$ terms on the left, we say the $k$th term is small if $x_k < m$. If $n$ is large enough, then the sum of all small terms is a positive real number less than $10^{-5}$. That means the fractional part of the left-hand side will be exactly the sum of the small terms; but it is also equal to $m^{-n}$, and so we conclude there is exactly one small term, at $k=1$. The claimed answer then follows.

I was thinking of solution based on a generating function. Yea that one is the only solution lol

https://imgur.com/a/dA3rYdw Small typo at the end, the solution should be (1,p,p, ... ,p)

For each \(n\), let \(a_n\) be the corresponding value of \(a\). Let \(K=2+3+\cdots+2011\). The only answer is \((1,K,K,\ldots,K)\), which clearly works. Now we prove it is the only solution. I contend: Claim: The sequence \((a_n)\) is bounded. Proof. This is more-or-less obvious. For rigor, write \[a_n^{n+1}=-1+\left(x_1^n+\cdots+2011x_{2011}^n\right)\le\left(x_1+\cdots+2011x_{2011}\right)^{n+1},\]and the claim follows. \(\blacksquare\) Claim: For sufficiently large primes \(p\), we have \(a_{p-1}=K\). Proof. Let \(n=p-1\). Taking the given equation modulo \(p\) gives \[1+2+\cdots+2011\equiv a_{p-1}+1\pmod p,\]so by taking \(p>\max(a_n)\), we have \(a_{p-1}=K\). \(\blacksquare\) Finally, for sufficiently large \(p\) we have \[\left(\frac{x_1}K\right)^{p-1}+\cdots+2011\left(\frac{x_{2011}}K\right)^{p-1}=K+\frac1{K^{p-1}}.\]From the above, \(x_i\le K\) for each \(i\). Taking \(p\to\infty\), we have \((x_i/K)^{p-1}\in\{0,1\}\), with equality if and only if \(x_i=K\). The right-hand expression approaches \(K\), so we must have \(x_1<K\) and \(x_2=\cdots=x_{2011}=K\). By \(n=1\) in the problem statement, \((x_1-1)+K^2\) is a perfect square, but \(x_1-1<K\), so \(x_1=1\).

Let the value for $a$ for each $n$ be $a_n$. Claim: $(a_n)$ is bounded. Proof: We have \begin{align*} 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}} = C^{\tfrac{n}{n+1}} < C \end{align*}for a constant $C$, as claimed. $\square$ Claim: Let $x_m = \max \{x_1,\ldots,x_{2011} \}$. Then $a_n=x_m$ for sufficiently large $n$. Proof: Since $(a_n)$ is bounded, there exists some $a$ such that infinitely many $n$ satisfy $a_n=a$. Suppose $a_n=a_{n+N}=a$ for some $n$ and sufficiently large $N$. Then we have \begin{align*} x_1^n + 2x_2^n + \cdots + 2011x_{2011}^n &= a^{n+1}+1 \\ x_1^{n+N}+2x_2^{n+N} + \cdots + 2011x_{2011}^{n+N} &= a^{n+N+1}+1. \end{align*}For sufficiently large $N$, the LHS multiplies by $x_m^N$, while the RHS multiplies by $a^N$; this implies $x_m=a$. Since $a_n=a$ for infinitely many $n$, and since $(a_n)$ is a nondecreasing sequence, we conclude $a_n=x_m$ for all $n\ge M$ for some constant $M$. $\square$ Now, for large enough $n$, \[ \left(\frac{x_1}{x_m}\right)^n + \cdots + 2011 \left(\frac{x_{2011}}{x_m}\right)^n = \frac{x_m^{n+1}+1}{x_m^n} = x_m + \frac{1}{x_m^n}. \]We know the $m$'th term on the LHS in an integer, in particular it is $m\cdot 1^n=m$. However, taking $n$ sufficiently large will make all the other terms at most $10^{-100}$, and summing them will make the fractional part of the LHS less than 1. But the fractional part of the RHS is less than that of the LHS, since $\tfrac{1}{x_m^n}$ is smaller than all the other small terms on the LHS. This is a contradiction unless there is exactly 1 non-integer term in the LHS, call it $x_i$, and $x_i=1$. So all of $x_1,\ldots,x_{2011}$ are equal to $x_m$, except for some $x_i=1$. Now it is easy to see that $x_1=1, x_2=\cdots=x_{2011}=2+\cdots+2011$ is the only sequence that works.

Let $k = \sum_{i = 2}^{2011} i = 2011 \cdot 1006 - 1$. We claim the only $2011$-tuple satisfying the desired is $(1, k, k, \ldots, k)$. It is easy to check that this indeed works (we may take $a = k$ for each $n$). We now show this is the only such $2011$-tuple. Let $m = \max(x_1, x_2, \ldots, x_{2011})$, and let $a_n$ be the $a$ corresponding to $n$ (in the given equation). We have \begin{align*} a_n^{n + 1} + 1 &= \sum_{j = 1}^{2011} jx_j^n \\ a_n^{n + 1} &< (k + 1)m^n. \end{align*}This is enough to imply that $a_n \leq m$. Now, pick $n$ very large, and suppose $a_n < m$. In this case, \begin{align*} \sum_{i = 1}^{2011} jx_i^n > m^n > a_n^{n + 1}, \end{align*}contradiction. Hence, we must have $a_n = m$. Therefore, for all sufficiently large $m$, we have \begin{align*} m \cdot m^n + 1 &= \sum_{i = 1}^{2011} jx_i^n. \end{align*}Let $S$ be the subset of $\{1, \ldots, 2011\}$ for which $x_s = m$ iff $s \in S$. Let $t = \sum_{s \in S} s$, and let $\ell_n = \sum_{s \not\in S} sx_s^n$. The above equation rewrites as \begin{align*} m \cdot m^n + 1 &= t \cdot m^n + \ell_n \\ (m - t) \cdot m^n &= \ell_n - 1. \end{align*}If $m > t$, for large $n$ we have \begin{align*} (m - t) \cdot m^n > m^n > \ell_n. \end{align*}If $m < t$, for large $n$ the left side will be negative, while the right side will be positive. Therefore, we must have $m = t$, which implies that $\ell_n = 1$. This implies that $S = \{2, 3, \ldots, 2011\}$: if there exists $s > 1$ which is not in $S$, we have $\ell_n \geq s \cdot x_s^n \geq s > 1$. Therefore, $\ell_n = x_1^n = 1$, implying that $x_1 = 1$. Finally, we have $m = t = \sum_{i = 2}^{2011} i = k$, implying that $x_i = m$ for all $i > 1$. Thus, the tuple described initially is the only solution, and we are done.

The only tuple which works is $(1,k,k, \dots, k)$ where $k = 2 + 3 + \cdots + 2011$. It is easy to see that taking $a = k$ works in this instance. Let $f(n)$ denote the left hand side and $m = \max (x_1, x_2, \dots x_{2011}).$ Let $C$ be the "coefficient" of $m$ in $f(n)$. Then it is easy to see that \[ \lim_{n \to \infty} \frac{f(n)}{m^n} = C. \]Now we split into cases based on the size of $C$. If $C > m$, our goal is to show that \[ m^{n+1} + 1 < f(n) < (m+1)^{n+1} +1 \tag{$\dagger$}. \]This obviously implies the problem because it shows that the suitable value of $a$ such that $f(n) = a^{n+1} + 1$ is not an integer. However, this is easy to see by dividing by $m^n$ and taking the limit: we have \[ \lim_{n \to \infty} \frac{m^{n+1}+1}{m^n} = m < C < \lim_{n \to \infty} \frac{ (m+1)^{n+1} + 1}{m^n} = \lim_{n \to \infty} m + n +1 = \infty. \]Thus taking $n$ large enough establishes the inequality in $(\dagger)$. If $C < m$, then we have \[ \lim_{n \to \infty} \frac{(m-1)^{n+1} + 1}{m^n} = \lim_{n \to \infty} m \left( \frac{m-1}{m} \right)^{n+1} = 0 < C < \lim_{n \to \infty} \frac{m^{n+1} + 1}{m^n} = m. \]Hence, for large enough $n$, \[ (m-1)^{n+1} + 1 < f(n) < m^{n+1} + 1 \] If $C = m$, then we can obtain in the same way as the first case that \[ m^{n+1} + 1 \leq f(n) < (m+1)^{n+1} +1. \]Thus we must have equality in the lower bound which is clearly possible only if $(x_1, x_2, \dots, x_{2011}) = (1, k, k, \dots, k)$. Having exhausted all possible cases, we conclude $ (1, k, k, \dots, k)$ is the only possible solution.

The only sequence is $(1, 2023065, 2023065, \ldots 2023065)$, with all $a_{n} = 2023065$. It's easy to see that this works. Let $x_{k}$ be the maximum of all $x_{i}$. Then, observe that \[x_{k}(a_{n}^{n+1} + 1) = x_{k}(x_{1}^{n} + 2x_{2}^{n} + \ldots 2011x_{2011}^{n}) \geq x_{1}^{n+1} + \ldots 2011x_{2011}^{n+1} = a_{n+1}^{n+2} + 1\]so $a_{n+1} < \sqrt[n+2]{x_{k}a_{n}^{n+1} + x_{k}}$. If $a_{n} > x_{k}$, then $a_{n+1} < \sqrt[n+2]{a_{n}^{n+2}} = a_{n}$. If $a_{n} < x_{k}$, then $a_{n+1} < \sqrt[n+2]{x_{k}^{n+2}} = x_{k}$. This means that if $a_{i} < x_{k}$, it will always be less than $x_{k}$, and the sequence is decreasing until $a_{i} < x_{k}$. Then, our two cases are if the sequence becomes constant at $a_{n} = x_{k}$, or if $a_{n} < x_{k}$ for large enough $n$. Case 1: $a_{n} < x_{k}$ for large enough $n$. This implies $a_{n}$ is periodic, let the period be $T$, and $a_{n} = M$. Then, we will have \[M^{T}x_{1}^{n} + 2M^{T}x_{2}^{n} + \ldots 2011M^{T}x_{2011}^{n} - M^{T} + 1 = a_{n+T} + 1 = x_{1}^{n+T} + 2x_{2}^{n+T} + \ldots 2011x_{2011}^{n+T}\]\[x_{1}^{n}(M^{T} - x_{1}^{T}) + 2x_{2}^{n}(M^{T} - x_{2}^{T}) + \ldots 2011x_{2011}^{n}(M^{T} - x_{2011}^{T}) = M^{T} - 1\]We can let $c_{i} = i(M^{T} - x_{i}^{T})$ and $k = M^{T} - 1$, both of which are constants. Then, we have \[x_{1}^{n}c_{1} + x_{2}^{n}c_{2} + \ldots x_{2011}^{n}c_{2011} = k\]Observe that if $x_{i} = x_{j}$, then either $c_{i}, c_{j} > 0, c_{i} = c_{j} = 0,$ or $c_{i} , c_{j} < 0$. If they weren't all equal to $0$, then as $n$ goes to infinity, the LHS also goes to either infinity or negative infinity, which means it can not be constant. If they were all $0$, this implies all $x_{i} = M$, which gives \[\frac{2011\cdot 2012}{2} M^{n} = a_{n}^{n+1} + 1\]No solutions exist to this. Case 2: $a_{n} = x_{k}$ for large enough $n$. Then, if we let $a_{1}, a_{2}, \ldots a_{l}$ such that $x_{a_{i}} = x_{k}$, then \[(a_{1} + a_{2} + \ldots + a_{l})x_{k}^{n} + \ldots = x_{k}^{n+1} + 1\]If $x_{k} < a_{1} + a_{2} + \ldots a_{l}$, then the LHS is greater than the RHS. If $x_{k} > a_{1} + a_{2} + \ldots a_{l}$, then the RHS grows faster than the LHS, so for large enough $n$ there won't be a solution. Thus, $x_{k} = a_{1} + a_{2} + \ldots a_{l}$, which implies the sum of the other stuff is equal to $1$. The only way this is possible is if $a_{i}$ include all indices besides $1$, and $x_{1} = 1$, so our only solution is $(1, \frac{2012\cdot 2011}{2} - 1, \frac{2012\cdot 2011}{2} - 1, \ldots \frac{2012\cdot 2011}{2} - 1)$

We claim the only solution is $(1, N, N, \cdots ,N)$, where $N = 2 + 3 + \cdots + 2011$. Assume the set $x_{n_1}, x_{n_2}, \cdots, x_{n_m}$ are the largest $x_i$, let's say they're all equal to $X$. Then, our new equation is \begin{align*} X^n(n_1 + n_2 + \cdots + n_m) + \text{[tiny stuff]} = a_n \cdot a_n^n + 1. \end{align*}Now taking $n$ gigantic forces $a_n = X$ for all large $n$. Now, let's say that $S = n_1 + n_2 + \cdots + n_m$. We rewrite the equation as follows: \begin{align*} \text{[tiny stuff]} + X^n(S - X) = 1. \end{align*} If $S \neq X$, then the left hand side explodes and we win. If $S = X$, then observe that the sum of all the tiny stuff has to be $1$, which forces $x_1 = 1$ which is the only element in the set of not maximums. Thus everything else is $n_1 = 2, n_2 = 3, \cdots, n_{m = 2010} = 2011$, giving us our solution set.

Quite a nice problem in my opinion$.$ The only solution to the problem is $$(x_1, x_2, \cdots, x_{2011})=(1, a, a, \cdots, a), \text{where } a=2+3+\cdots 2011$$It is evident that for this $2011-$tuple$,$ the given sum is exactly $a^{n+1}+1$ for all $n.$ Let us now prove it is the only one$.$ Suppose that for all sufficiently large $n,$ the corresponding $a=a(n)$ is (strictly) bigger than $\max(x_1, x_2, \cdots, x_{2011}).$ Then $$a+\frac{1}{a^n}=\sum_{j=1}^{2011}\bigg(j\bigg(\frac{x_j}{a}\bigg)^{n}\bigg)\rightarrow 0,$$a clear contradiction$.$ Therefore for all sufficiently large $n,$ say $n\geq M,$ we must have $a\leq \max(x_1, x_2, \cdots, x_{2011}).$ But similarly if $a=a(n)<\max(x_1, x_2, \cdots, x_{2011})$ for all sufficiently large $n,$ then $$a+\frac{1}{a^n}=\sum_{j=1}^{2011}\bigg(j\bigg(\frac{x_j}{a}\bigg)^{n}\bigg)\rightarrow \infty,$$again a contradiction $.$ We conclude that for infinitely many sufficiently large $n,$ it must be the case that $a=a(n)=\max(x_1, x_2, \cdots, x_{2011}).$ Let us now look at the indices $i_1, \cdots, i_k,$ for which $$x_{i_1}=\cdots =x_{i_k}=\max(x_1, x_2, \cdots, x_{2011})$$and let the other numbers be (if they exist) $$x_{i_{k+1}}, \cdots x_{i_{2011}}<\max(x_1, x_2, \cdots, x_{2011}).$$Actually if all of $x_1, x_2, \cdots x_{2011}$ are equal$,$ then for the constant $a$ that we found$,$ we would have that for infinitely many $n,$ $$a+\frac{1}{a^n}=\sum_{j=1}^{2011}\bigg(j\bigg(\frac{x_j}{a}\bigg)^{n}\bigg)=\sum_{j=1}^{2011}j,$$is fixed$,$ while the left-hand side varies$,$ a clear contradiction$.$ Therefore $k\leq 2010.$ Let us rewrite the equality as $$a+\frac{1}{a^n}=\sum_{j=1}^{2011}\bigg(j\bigg(\frac{x_j}{a}\bigg)^{n}\bigg)=\sum_{t=1}^{k}\bigg(i_{t}\bigg(\frac{x_{i_t}}{a}\bigg)^{n}\bigg)+\sum_{t=k+1}^{2011}\bigg(i_{t}\bigg(\frac{x_{i_t}}{a}\bigg)^{n}\bigg)$$$$\iff a-\sum_{t=1}^{k}i_{t}=\sum_{t=k+1}^{2011}\bigg(i_{t}\bigg(\frac{x_{i_t}}{a}\bigg)^{n}\bigg)-\frac{1}{a^n}$$Note that $k\leq 2010$ implies that the right-hand side in non-negative$,$ as there is at least one number in the sum$,$ namely $x_{i_{k+1}},$ and so the sum is at least $1\cdot \frac{1}{a^n}.$ Assume that the sum is greater than zero$.$ This means that $a-\sum_{t=1}^{k}i_{t}$ is fixed but not zero (thus positive)$.$ But then $n\rightarrow \infty$ gives that the right-hand side goes arbitrarily close to $0$ (since $x_{i_t}<a$ for all $k+1\leq t\leq 2011$) $,$ whereas the left-hand side is fixed$,$ cotradiction$.$ We conclude that both sides must be zero and that is possible if and only if $k=2010, i_{2011}=1$ and $x_{i_{2011}}=1.$ It follows that $$(x_1, x_2, \cdots, x_{2011})=(1, a, \cdots, a),$$where $a=2+\cdots +2011$ and we already established that it is a solution$.$ $\blacksquare$

lol Suppose that $x_{i_1},x_{i_2}\ldots x_{i_m}$ are the greatest numbers in the sequence which share a value $x$. Then, as $n$ approaches infinity, the LHS is approximately $(\sum i_j) x^n$, and because $\frac{x^{n+1}}{(x+1)^{n+1}},\frac{(x-1)^{n+1}}{x^{n+1}}$ both approach $0$ as $n$ approaches infinity, it's necessary that $a=x$. Now we can obviously solve that $(x_1,x_2,\ldots ,x_{2011})=(1,2023065,\ldots ,2023065)$ where $2023065=2+3+\ldots +2011$.

Very nice problem. My answer is: \((1,k,k,\ldots,k)\), where $k= 2 + 3 + \dots + 2011$ My first step was: Claim: $a$ is bounded. Proof: Let $x= max(x_i)$. Then: $$a^{n+1} + 1 < (1+2+\ldots+2011)x^n \implies a< \sqrt[n+1]{(1+2+\ldots+2011)x^n} \implies a<(1+2+\ldots+2011)x $$So the claim is true. $\square$ Claim: $a= 2+\dots+2011$ for infinites $n$ Proof: Let $n= p-1$ with $p$ being a large prime. Then by FLT we have $a \equiv (2+3+...+2011) \bmod p$ but $a$ is bounded and $p$ is a large prime so the claim is true. $\square$ Now for this infinites $n$ we have that: \[\left(\frac{x_1}a\right)^{n}+\cdots+2011\left(\frac{x_{2011}}a\right)^{n}=a+\frac1{a^{n}}.\]So RHS is bounded then $x_i \leq a$. Taking $n$ sufficiently large we have $(x_i/a)^{n}\to 0$ if $x_i < a$ otherwise is $1$. But RHS tends to $a$, so we have \(x_1<a\) and \(x_2=\cdots=x_{2011}=a\) because $a=2+\dots+2011$ . It's easy to complete it from here taking $n=1$, so $x_1=1$.

The answer is $\boxed{(1,k,k,k,\ldots,k)}$, where $k=2+3+\cdots+2011$. Indeed, we can just set $a=k$. For any $n$ define $a$ to be $a_n$. It suffices to show that nothing else works. Let $m=\max(x_1,x_2,\ldots,x_{2011})$. We note that $m^n<a_n^{n+1}+1<(1+2+\ldots+2011)m^n$. This implies $m^n-1<a_n^{n+1}<(1+2+\ldots+2011)m^n$, so \[\sqrt[n+1]{m^n-1}<a_n<\sqrt[n+1]{(1+2+\ldots+2011)m^n}.\] Both sides approach $m$, so for sufficiently large $n$, we must have $a_n=m$. So we get $x_1^n+2x_2^n+3x_3^n+\ldots+2011x_{2011}^n=m^{n+1}+1$. Dividing both sides by $m^n$ gives \[\left(\frac{x_1}{m}\right)^n+2\left(\frac{x_2}{m}\right)^n+\cdots+2011\left(\frac{x_{2011}}{m}\right)^n=m+\frac{1}{m^n}.\] Note that for sufficiently large $n$, the fractional part of the LHS is the sum of all the terms with $x_i<m$. But it's also $\frac{1}{m^n}$. This implies $x_1=1$, and $x_2=x_3=\cdots=x_{2011}=m$. Checking gives $m=2+3+\cdots+2011$.

Let $M = \text{max} (x_1, x_2, \ldots, x_n)$, and let $I = \{i | x_i = M\}$, $S=\sum_{i \in I} i$ (in other words the coefficient of $M^n$ when $\sum^{2011}_{j=1} j x^n_j$ is simplified). Notice that for large enough $n$, the $SM^n$ extracted from terms in the summation becomes much larger than the other terms. Moreover, this term is much larger than $(M-1)^{n+1}$ and much smaller than $(M+1)^{n+1}$ (for sufficiently large $n$). Thus for those $n$ we must have $a=M$. In addition, for sufficiently large $n$, if $S<M$, then the sum will be much smaller than $M^{n+1}$, and if $S>M$, the sum will be much larger than $M^{n+1}$. Therefore $S=M$ as well. Therefore, $\sum^{2011}_{i\notin I} i x^n_i =1$ for all $n$ large. However, for all $i>1$ the corresponding term is always greater than $1$ and thus cannot be in the sum. (Therefore, they are in $I$.) This leaves $i=1$, for which clearly only $x_1=1$ works. Thus a solution must have $x_1 = 1, S = 2 + 3 + \ldots + 2011 =2023065\Rightarrow M = 2023065$, which implies $x_2, x_3, \ldots, x_2011 = 2023065$. It is readily checked that this works.

Let $x_M$ be the maximum $x_i$, and $c = \frac{2011*2012}{2}$. We claim the answer is $x_1 = 1$ and $x_i = c-1$ for all $i>1$, with $a$ constant and equal to $c-1$, which works. Firstly, if $(a_n)_n$ is the sequence of the $a$'s that satisfy the statement, take the $n$-th root on both sides and divide by $a_n$. Taking the limit with $n\to \infty$, we get that \[\lim_{n\to \infty} \frac{x_M}{a_n}\sqrt[n]{\sum^{2011}_{j=1} j \frac{x^n_j}{x^n_M} } = 1.\]However, \[\lim_{n\to \infty}\sqrt[n]{\sum^{2011}_{j=1} j \frac{x^n_j}{x^n_M} } = 1,\]\[\implies \lim_{n\to \infty} a_n = x_M.\]Since $a_n$ is a sequence of integers, this implies that it equals $x_M$ for large $n$. We now prove $x_M = c-1$. Take $n+1 = p$ to be a large prime. Then, taking both sides modulo $p$, we conclude that \[c-1 \equiv x_M \pmod p,\]so $x_M = c-1$. Now we claim $a_i = i$ for all $i>1$. Suppose this is false for some $i$. Then, \[x_M^n(c-1)+1 \le x_1^n + ix_i^n + (c-1-i)x_M^n\]\[\implies ix_M^n + 1\le x_1^n + ix_i^n \le x_{M}^n + i(x_M-1)^n,\]a contradiction for large $n$. Therefore, $x_i=c-1$ for all $i>1$, so $x_1 = 1$. $\blacksquare$

We claim that the only sequence $(x_1,x_2,...,x_{2011})$ that works is the sequence $x_1=1,x_2=x_3=...=x_{2011}=M$, where $M=2011.1006-1$, which obviously works. Assume that $\sum^{2011}_{j=1} j x^n_j = a_n^{n+1} + 1$ for all $n \in \mathbb{Z}_{>0}$. We claim that $a_n$ is bounded above. Indeed, observe that $a_n^{n+1} \leq T^n(1+2+...+2011)=T^n2011.1006$ for all $n \in \mathbb{Z}_{>0}$. So, if $a_n$ was unbounded, the LHS would become much greater than RHS for large values of $n$, a contradiction. Hence, $a_n$ is bounded by some $K$. Moreover, observe that $a_{n+1}^{n+2} \geq a_n^{n+1} \implies a_{n+1} \geq a_n^{\frac{n+1}{n+2}}$, so for large $n$, $a_n$ is non decreasing. However, since $a_n$ is bounded by $K$, it is eventually constant. Therefore, there is an $N$ such that $a_n=K$ for all $n \geq N$. Thus, $$K^{n+1}(K-1)=\sum^{2011}_{j=1} j x^n_j(x_j-1) $$$\implies \sum^{2011}_{j=1} j x^n_j(K-1)=\sum^{2011}_{j=1} j x^n_j(x_j-1)]+(K-1) \implies \sum_{j=1}^{2011} jx_j^n(K-x_j)=K-1$. Observe that since RHS is fixed, for all $x_j \neq K$, $x_j=1$, otherwise $x_j^n$ would explode for large $n$. Thus, since the expression is equals to $K-1>0$, then there is a $j$ such that $x_j=1$, so it has contribution $j(K-1)$ in the sum, so $j=1$. Therefore, $x_1=1$ and $x_2=x_3=...=x_{2011}=K$. Now, notice that $1+2011.1006K^n=K^{n+1}+1 \implies K=2011.1006$, so $(x_1,x_2,x_3,...,x_{2011})=(1,K,K,...,K)$ for $K=2011.1006$ is the only solution, as desired. $\blacksquare$

Why is analysis so hard for me.I had to look at the answer which is $(1,n,...n)$,where $n=2+3...2011$ as a hint .Still ok... Ok so let $a_n$ be the integer sequence of the values of $a$,for each $n$. Then let $x$ be the maximum over all $x_i$.Obv $x>1$.Then,we have that $(x^n)^{\frac {1}{n+1}}<a_n<(Ax^n)^{\frac {1}{n+1}}$,where $A=1+..2011$. Now,taking limit as $n$ tending to inf,we get that $a_n=x$ for large $n$,as both sides approach $x$ by squeeze and noting that $a_i$ is an integer sequnce.Write the LHS as $Cx^n+B$,where $C$ is weight of all $x_i=x$.Now for some $x_i$ not $x$,we have$\frac{x_i}{x}<1$,so dividing by $x^n$ and again taking $n$ large we see $A=x$, and $B=1$,which is possible only when $x_1=1$,and all other $x_i=B=2+3...2011$.

Suppose that such a sequence exists. Let $a_n$ be the integer $a$ satisfying the condition, where $a_n$ is positive. It's easy to see that this sequence is nondecreasing. Suppose FTSOC that this sequence will never be constant for sufficiently large $n$. Then the sequence is clearly unbounded. Now let $x_{\text{min}}$ and $x_{\text{max}}$ be the minimum and maximum of the sequence $x_1,x_2,\dots,x_{2011}$. Then we have \[x_{\text{max}}\ge\frac{\displaystyle{\sum_{j=1}^{2011}jx_j^{n+1}}}{\displaystyle{\sum_{j=1}^{2011}jx_j^n}}=\frac{a_{n+1}^{n+2}+1}{a_n^{n+1}+1}\ge \frac{a_n^{n+2}+1}{a_n^{n+1}+1}\ge a_n-1\]which is a contradiction. Now let this constant value be $a_0$. We have that \[x_{\text{max}}=\lim_{n\to \infty}\frac{\displaystyle{\sum_{j=1}^{2011}jx_j^{n+1}}}{\displaystyle{\sum_{j=1}^{2011}jx_j^n}}=\lim_{n\to \infty}\frac{a_{n+1}^{n+2}+1}{a_n^{n+1}+1}=a_0.\]For sufficiently large $n$ the terms in the LHS not equal to the maximum become negligible. Taking $\bmod\text{ }a_0^n$ gives that \[\sum jx_j^n=1\]where we sum over all $x_j$ not equal to $x_{\text{max}}$. Then clearly we must have $x_1=1$ while $x_2,x_3,\dots,x_{2011}=a_0=2+3+\dots+2011$. This is a solution, so we are done. $\blacksquare$

Clearly, $x_1^n+2x_2^n+\dots+2011x_{2011}^n<(x_1+2x_2+\dots+2011x_n)^{n+1}$ which implies $a\le x_1+2x_2+\dots+2011x_n.$ By pigeonhole principle, there exists $y$ such that $\sum^{2011}_{j=1} j x^n_j=y^{n+1}+1$ for infinitely many $n$. $~$ Now, if any $x_i> y$ then $ix_i^n$ will increase faster than $y^{n+1}$, so for increasingly large $n$, the LHS will be larger than RHS. Thus, $x_i\le y.$ $~$ Now, suppose we have $x_{i_1}=x_{i_2}=\dots=x_{i_m}=y.$ We have \[\sum^{2011}_{j=1} j \left (\frac{x}{y}\right)^n=y+y^{-n}\]and note that the LHS is at least $i_1+i_2+\dots+i_m$. When $n$ gets large, the other terms will approach zero, so $i_1+i_2+\dots+i_m=y$ and the sum of $jx^j$ for $x$ not in $x_{i_r}$ will be one, which implies $x_1=1$ and $x_2=x_3=\dots=x_n=2+3+\dots+2011$ which works.

The only sequence that works is \[\boxed{x_j=\begin{cases} 1 & j=1 \\ \binom{2012}{2}-1 & j>1 \end{cases}}.\]This clearly works, so we will prove that this is the only solution. Let $x_i$ be the maximum of the sequence. Obviously $x_i>1$. For large $n$, \[x_i^n\le a^{n+1}\le 2011^2 x_i^n.\]Therefore, by taking large enough $n$, $x_i\le a\le x_i$, so $a=x_i$. This means that $x_i\le 2011^2$. Now divide both sides of the original equation by $x_i^n$. Since $n$ is large enough, whenever $x_j<x_i$, $x_j$'s term in the sum is arbitrarily small(if we let $n$ be arbitrarily large). Therefore, equating the fractional part of both sides, we get \[\sum_{j,x_j<x_i}jx_j^n=1.\]Therefore, there is exactly one $j$ that $x_j<x_i$, that one $j$ is $j=1$, and $x_1=1$. Everything else should be equal to $x_i$, so if we let $C=\sum_{j=2}^{2011}j=\binom{2012}{2}-1$, \[1+Cx_i^n=x_i^{n+1}+1.\]Therefore, $x_i=C\;\blacksquare$

Very nice analytic number theory problem. Let $M = 2+3+\cdots + 2011$. I claim the only sequence that works is $(1, M, M, \dots, M)$. Throughout this proof, $N$ is taken to be a sufficiently large integer. Let $a_n$ be the corresponding number $a$ for each value of $n$; furthermore, suppose that $x_k$ is the (perhaps non-unique) maximal term among the $x_i$. Claim. $a_N = x_k$. Proof. Suppose otherwise. Let $S_N$ denote the left-hand sum, and notice that $kx_k^N < S_N < Mx_k^N$. Suppose that $S_N = (a_k+i)^{N+1} + 1$, where $i \geq 1$ (the case where $i$ is negative is analogous). Then $$S_n > a_k^{N+1} + i(N+1) a_k^N = (a_k+iN+i)a_k^N > Mx_k^N.$$Hence contradiction. $\blacksquare$ Claim. $x_i = x_k$ for all $k \geq 2$. Again assume otherwise. Let $S$ be the set of all $x_i$ not equal to $x_k$. Then $$\ell x_k^N + 1 < S_N = \ell x_k^N + \sum_{x_i \in S} ix_i^N < \ell x_k^N + 2011|S| (x_k-1)^N < (\ell+1)x_k^N$$which is an obvious contradiction. $\blacksquare$ It follows that $1 + x_k^{N+1} = S_N = x_1 + Mx_k^N$ if and only if $x_k = M$ and $x_1 = 1$, which yields the desired characterization.

The answer is $x_1=1$ and $x_i=2011\triangle -1$ for all $i\neq 1$, which clearly works. Let $$M=\max(x_1,x_2\dots ,x_{2011}).$$Then, consider $$\sqrt[n+1]{x_1^n+2x_2^n\dots +2011x_{2011}^n-1}$$which the problem asserts is an integer for all $n$. Since constant factors are negligible when $n+1$th rooting, and the exponent of $\frac{n}{n+1}$ converges to 1, the limit of this expression for sufficently large $n$ is $M$. Hence, in order for it to always be an integer, it must equal exactly $M$ for all sufficently large $n$. Thus, for all sufficently large $n$, we have $$x_1^n+2x_2^n\dots +2011x_{2011}^n-1=M^{n+1}.$$Taking modulo $M^n$, we have $$\sum_{x_i\neq M} ix_i^n\equiv 1\pmod{M^n}.$$However, as long as $n$ is sufficently large, the left hand side is smaller than $M^n$, so we must actually have $$\sum_{x_i\neq M} ix_i^n=1.$$Since each summand is a positive integer, the sum is only allowed to consist of one term, and that term must be the $i=1$ term. Thus, $x_1=1$ and $x_i=M$ for all $i\neq 1$. Thus, the equation simply reduces to $$(2011\triangle -1)M^n=M^{n+1}$$$$M=2011\triangle -1.$$ remark: kind of silly in the fact that it only matters that one of the coefficents is 1. in general, using the same method the solution is when a term with coefficent 1 equals 1 while other $x_i$ equal the sum of all other coefficents.

this was a great problem,

The only answer is $(x_1,x_2,...,x_{2011})=(1,2023065,2023065,...,2023065)$. It's easy to check that this sequence satisfies the problem condition. Suppose $x_k$ is the maximum number in the sequence $(x_1,x_2,...,x_{2011})$. Clearly, $x_k\ge 2$. Let \[ U_n= \sum^{2011}_{j=1} j x^n_j = y_{n}^{n+1} + 1\]for every natural number $n$. Note that $$\sum^{2011}_{j=1} j x^n_j = y_{n}^{n+1} + 1 \le 2023066x_{k}^{n} $$Hence, $y_n\le \sqrt[n+1]{2023066x_k^{n}-1}$ for all positive integer $n$. Note that $f(n)=\sqrt[n+1]{2023066x_k^{n}-1}< \max(2023066,x_k)$ for all natural number $n$. Hence, the sequence $y_n$ is bounded above. Now, by applying Van der Waerden's Theorem, we have constants $a,b,c\in \mathbb{N}$ such that \[\sum^{2011}_{j=1} j x^{a+bz}_{j} = c^{a+ bz+1} + 1\]for all $z\in \mathbb{N}_0$. Now, by observing that $c^b=\frac{U_{a+zb+b}-1}{U_{a+bz}-1}$, taking limit to infinity, we get $$\lim_{z\to \infty} \frac{x_1^{a+bz+b}+2x_2^{a+bz+b}+...+2011x_{2011}^{a+bz+b}-1}{x_1^{a+bz}+2x_2^{a+bz}+...+2011x_{2011}^{a+bz}-1}=\frac{\sum_{x_j=x_k} jx_j^b}{\sum_{x_j=x_k} j}=\frac{x_k^b \sum_{x_j=x_k} j}{\sum_{x_j=x_k} j}=x_k^b=c^b$$Hence, we conclude that $c=x_k$. We have $x_i\le c$ for all $i\in \{1,2,...,2011 \}$. By rewriting the equation $c^b=\frac{U_{a+zb+b}-1}{U_{a+bz}-1}$, we get that $$x_1^{a+bz}(c^b-x_1^b)+2x_2^{a+bz}(c^b-x_2^b)+...+2011x_{2011}^{a+bz}(c^b-x_{2011}^b)=c^b-1$$for all $z\in \mathbb{N}_0$. Note that the right hand side is always constant, so as $z$ grows, we must have the left hand side always constant, too. If there is an index $l$ such that $2\le x_l \le c-1$, we have $$c^b-1\ge lx_l^{a+bz}(c^b-x_l^b)\ge l(2^{a+bz})(c^b-(c-1)^b)$$So, as $z$ grows, at some point, we have the left hand side is bigger than the right hand side of the equation. Hence, we must have either $x_i=1$ or $x_i=c$ for all $i=1,2,...,2011$. If there is an index $l\ge 2$ such that $x_l=1$, then we have $c^b-1\ge l(c^b-1)\ge 2c^b-2\implies c^b\le 1$, which is a contradiction (as $c=x_k\ge 2$). Notice also we cannot have $x_i=c$ for all $i\in \{1,2,...,2011 \}$. Hence, we must have $x_1=1$, and $x_2=x_3=...=x_{2011}=c$. Now, the beginning equation becomes $1+2023065c^n=1+y_n^{n+1} \iff 2023065c^n=y_n^{n+1}$. Now, for every prime $p$, we must have $n+1\mid v_p(2023065)+nv_p(c) \implies n+1\mid v_p(2023065)-v_p(c)$. Take $n$ large enough to get $v_p(c)=v_p(2023065)$ for all primes $p$. This implies $c=2023065$. Hence, the only answer is $(x_1,x_2,...,x_{2011})=(1,2023065,2023065,...,2023065)$ and it's easy to check that this sequence satisfies the problem condition.

a graceful and beautiful problem

Throughout this proof, let $X$ stand as a placeholder for $2+ 3 + \dots + 2011$. We claim that the only sequence satisfying the problem condition is $(1, X, X, \dots, X)$. (This sequence obviously works.) For each $n$, let $S_n$ denote $\sum_{j=1}^{2011} jx_j^n$ and let $a_n$ denote the value of $a$ in the equation above. Claim: $a_n = \max (x_i)$ for sufficiently large $n$. Proof: Observe that \[\max (x_i) = \lim_{n \to \infty} \sqrt[n]{S_{n}} = \lim_{n \to \infty} \sqrt[n+1]{S_n -1} = \lim_{n \to \infty} a_n.\]Since $a_n$ must be an integer, it follows that $a_n = \max (x_i)$ for sufficiently large $n$. $\square$ Claim: $a_n = \max (x_i) = X$ for sufficiently large $n$.

prime. Then, we set $n=p-1$ and take both sides of the given equation modulo $p$. We can simplify using Fermat's little theorem, which will give us \[1 + 2 + \dots + 2011 \equiv a_{p-1}+1 \pmod{p}, \]from which it follows that $\max ( x_i) = a_{p-1} = 2 + 3 + \dots + 2011$. $\square$ Claim: $a_n = \max (x_i) = X$ for all $n$. Proof: Let $p$ be a sufficiently large prime. Then, we set $n = p-1+k$ for any integer $k$ which gives us \[\sum_{j=1}^{2011} jx_j^{k} \equiv X^{k+1} + 1 \pmod{p},\]and because $p$ is sufficiently large, this equivalence becomes an equality. $\square$ So, we have shown that $S_n = X^{n+1}+1$. It follows that our initially claimed sequence must be unique, since it is the unique way to characterize the linear recurrence $S_n$ using $n$ terms.

orl wrote: 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 \[\sum^{2011}_{j=1} j x^n_j = a^{n+1} + 1\] Proposed by Warut Suksompong, Thailand Let $a_n^{n+1}+1==\sum_{i=1}^{2011}ix_i^n\Rightarrow a_n=\sqrt[n+1]{\sum_{i=1}^{2011}ix_i^n-1}$ Let $x_k=max{x_i}$ then we get that: $a_n a_n=\sqrt[n+1]{x_k^n[\sum_{i=1}^{2011}i\frac{x_i^n}{x_i^k}-\frac{1}{x_k^n}}]\Rightarrow \lim_{n\rightarrow +\infty }a_n=\lim_{n\rightarrow +\infty }\sqrt[n+1]{x_k^n[\sum_{i=1}^{2011}i\frac{x_i^n}{x_i^k}-\frac{1}{x_k^n}}]=\lim_{n\rightarrow +\infty }\sqrt[n+1]{x_k^n*c}$ Where $c=\sum_{1\leqslant i\leqslant 2011}^{x_i= x_k}i$ Since $a_n$ is a interger for every $n$ we get that $x_k=c=a_n$ for every $n>N$ Then for $n>N$ we have that: $a_n^{n+1}+1=\sum_{i=1}^{2011}ix_i^{n}\Rightarrow 1=\sum_{1\leqslant i\leqslant 2011}^{x_i\neq x_k}ix_i^n$ So we get that $x_1=1$ and that $x_i=x_k$ for every $i>1$ so The answer is $(x_1, x_2, \dots, x_{2011}) = (1, k, \dots, k)$ where $k = 2+\dots+2011$.