I suppose the system is supposed to have $ n$ equations, and the last one is $ x_1+2^{n-1}x_2+3^{n-1}x_3+\cdots+n^{n-1}x_n=n+2^{n-1}+n^{n-1}$ There's a simple solution

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Now try and find a reason why there's no other solution

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lordWings wrote: I suppose the system is supposed to have $ n$ equations, and the last one is $ x_1 + 2^{n - 1}x_2 + 3^{n - 1}x_3 + \cdots + n^{n - 1}x_n = n + 2^{n - 1} + n^{n - 1}$ There are only 4 equations for this problem in the official book. So I don't understand what you mean.

Let substitute $ x_2\rightarrow x_2 - 1$ and $ x_n\rightarrow x_n - 1$. Now, the new system becomes: $ \{\begin{array}{c} \displaystyle x_1 + x_2 + x_3 + \cdots + x_n = n \\ \\ x_1 + 2x_2 + 3x_3 + \cdots + nx_n = n \\ \\ x_1 + 2^2x_2 + 3^2 x_3 + \cdots + n^2x_n = n \\ \\ x_1 + 2^3x_2 + 3^3x_3 + \cdots + n^3x_n = n \end{array} $ where $ x_2,x_n\ge - 1$ and $ x_1,x_3,...,x_{n - 1}\ge0$. $ \Rightarrow (2 - 1)x_2 + ... + (n - 1)x_n = 0$ $ (2 - 1)^2x_2 + ... + (n - 1)^2x_n = 0$ $ \Leftrightarrow (2^2x_2 + ... + n^2x_n) - 2(2x_2 + ... + nx_n) + (x_2 + ... + x_n)$ $ = (n - x_1) - 2(n - x_1) + (n - x_1) = 0$. $ (2 - 1)^3x_2 + ... + (n - 1)^3x_n = 0$ $ \Leftrightarrow (2^3x_2 + ... + n^3x_n) - 3(2^2x_2 + ... + n^2x_n) + 3(2x_2 + ... + nx_n) - (x_2 + ... + x_n)$ $ = (n - x_1) - 3(n - x_1) + 3(n - x_1) - (n - x_1) = 0$. $ \Rightarrow \sum_{k = 2}^{n}(k - 1)x_k((k - 1)^2 - 2(k - 1) + 1) = 0$ $ \Leftrightarrow \sum_{k = 3}^{n}(k - 1)x_k(k - 2)^2 = 0$, but obviously we also have $ \sum_{k = 3}^{n}(k - 2)(k - 1)^2x_k = 0$. So $ (n - 1)\sum_{k = 3}^{n}(k - 1)x_k(k - 2)^2 - (n - 2)\sum_{k = 3}^{n}(k - 2)(k - 1)^2x_k = 0$ $ \Rightarrow \sum_{k = 3}^{n - 1}(k - 2)(k - 1)(k - n)x_k = 0$ $ \Rightarrow x_i = 0$ for every $ i\in \{3,4,...,n - 1\}$. Now, we can obtain easily that $ x_1 = n$ and $ x_2 = x_n = 0$ (or for the initial $ x_2 = x_n = 1$).

denote $ a_i = \sum_{k = 3, k\neq i}^{n - 1}x_ k$, $ b_i = \sum_{k = 3, k\neq i}^{n - 1}kx_ k$, $ c_i = \sum_{k = 3, k\neq i}^{n - 1}k^2x_ k$, $ d_i\sum_{k = 3, k\neq i}^{n - 1}k^3x_ k$... and consider the systems (for all $ 3\leq i\leq n - 1$) $ \{\begin{array}{c} \displaystyle x_1 + x_2 + x_i + x_n = n + 2 - a_i \\ \\ x_1 + 2x_2 + ix_i + nx_n = 2n + 2 - b_i \\ \\ x_1 + 2^2x_2 + i^2 x_i + n^2x_n = n^2 + n + 4 - c_i \\ \\ x_1 + 2^3x_2 + i^3x_i + n^3x_n = n^3 + n + 8 - d_i \end{array} $ after solving this system we have that $ x_i = \dfrac{n( - 2a_i + 3b_i - c_i) - ( - 2b_i + 3c_i - d_i)}{(n - i)(i - 1)(i - 2)}$, for all $ 3\leq i\leq n - 1$... so we have that $ n( - 2a_i + 3b_i - c_i) - ( - 2b_i + 3c_i - d_i)\geq 0$... this implies that $ \sum_{k = 3, k\neq i}^{n - 1}x_k( - 2n + 3kn - k^2n + 2k - 3k^2 + k^3) = \sum_{k = 3, k\neq i}^{n - 1}(k - 1)(n - k)(2 - k)x_k\geq 0$ clearly this implies that $ x_k = 0$, since $ (k - 1)(n - k)(2 - k) < 0$ for all $ 3\leq k\leq n - 1$... taking two distinct indexes, say $ i = 3, i = 4$, it follows that $ x_3 = x_4 = ... = x_{n - 1} = 0$... from this it's easy to conclude that $ x_1 = n, x_2 = 1, x_n = 1$, and we're done

We have $(n-2)(n^2+n+4)<n^3+n+8$. Therefore, $x_{n-1}+x_n>0$. But if $x_{n-1}$ or $x_n$ are greater than $1$, we immediately arrive at a contradiction. Thus, there are two cases: If $x_n=1$, it follows that $x_1=n$ and $x_2=1$ (subtract the second equation from the first). If $x_{n-1}=1$, it follows that $x_3=1$, $x_1=n$ or $x_2=2$, $x_1=n-1$. Both cases clearly lead to a contradiction. Thus, $(x_1$, $\dots$, $x_n)=(n$, $1$, $0$, $\dots$, $0$, $1)$.

I just used Gaussian elimination method to get the solution.