Return to Answer

Let me denote $u_i = x^{i+1} + 1$. As I learned from the previous question, under the "matrix system" OP understands the following matrix equation (up to a factor $\frac{1}{x}$): $$My = b,$$ where $$M:=\begin{bmatrix} u_1^0 + x - 1 & u_1^1 + x - 1 & \cdots & u_1^{n-1} + x - 1 \\ \vdots & \vdots & \ddots & \vdots \\ u_n^0 + x - 1 & u_n^1 + x - 1 & \cdots & u_n^{n-1} + x - 1 \end{bmatrix} \quad\text{and}\quad b := \begin{bmatrix} u_1^n + x - 1 \\ \vdots \\ u_n^n + x - 1 \end{bmatrix}. $$

First, it is easy to see that it always has a unique solution: $$y = M^{-1} b = AA^{-1}M^{-1} b = A (MA)^{-1} b,$$ where $$A := \begin{bmatrix} \frac{1}{x} & - \frac{x-1}{x} & - \frac{x-1}{x} & \cdots & - \frac{x-1}{x} \\ 0 & 1 & 0 & \cdots & 0\\ 0 & 0 & 1 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{bmatrix}$$ so that $$MA = V = V(u_1,\dots,u_n)$$ is a Vandermonde matrix.

Second, let us find the value of $V^{-1}b$.

The Vandermonde matrix inverse $V^{-1}$ has elements $$(V^{-1})_{i,j} = [z^{i-1}]\ \frac{F(z)}{F'(u_j)(z-u_j)},$$ where $$F(z) := (z-u_1)\cdots (z-u_n)$$ and $[z^k]$ is the operator taking the coefficient of $z^k$. Hence, $V^{-1} b$ is composed of the coefficients of $$G(z):=\sum_{j=1}^n \frac{F(z)}{F'(u_j)(z-u_j)} (u_j^n+x-1)=z^n - F(z) + x-1,$$ where the latter equality holds since the right-hand and left-hand sides as polynomials in $z$ both have degree $\leq n-1$ and at $z=u_j$ evaluate to $u_j^n+x-1$ (i.e., they have equal values at $n$ distinct points).

Hence, $$(V^{-1}b)_k = [z^{k-1}]\ G(z) = \begin{cases} (-1)^{n-1} u_1\cdots u_n + x-1, & k=1;\\ (-1)^{n-k} e_{n+1-k}(u_1,\dots,u_n), & k>1; \end{cases}$$ where $e_k()$ are elementary symmetric polynomials.

Finally, from $y=A(V^{-1}b)$ we get $$y_1 = (-1)^{n+1} u_1\cdots u_n + x-1 - \frac{x-1}{x}G(1)$$ and $$y_k = (-1)^{n+k} e_{n+1-k}(u_1,\dots,u_n),\quad k>1.$$ Recalling the definition of $u_i$, we conclude that the free term of $G(1) = x -(1-u_1)\cdots(1-u_n)$ as polynomial in $x$ is zero, and thus $\frac{x-1}{x}G(1)$ is a polynomial in $x$ with integer coefficients. Then so are all $y_1,\dots,y_n$.

Let me denote $u_i = x^{i+1} + 1$. As I learned from the previous question, under the "matrix system" OP understands the following matrix equation (up to a factor $\frac{1}{x}$): $$My = b,$$ where $$M:=\begin{bmatrix} u_1^0 + x - 1 & u_1^1 + x - 1 & \cdots & u_1^{n-1} + x - 1 \\ \vdots & \vdots & \ddots & \vdots \\ u_n^0 + x - 1 & u_n^1 + x - 1 & \cdots & u_n^{n-1} + x - 1 \end{bmatrix} \quad\text{and}\quad b := \begin{bmatrix} u_1^n + x - 1 \\ \vdots \\ u_n^n + x - 1 \end{bmatrix}. $$

First, it is easy to see that it always has a unique solution: $$y = M^{-1} b = AA^{-1}M^{-1} b = A (MA)^{-1} b,$$ where $$A := \begin{bmatrix} \frac{1}{x} & - \frac{x-1}{x} & - \frac{x-1}{x} & \cdots & - \frac{x-1}{x} \\ 0 & 1 & 0 & \cdots & 0\\ 0 & 0 & 1 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{bmatrix}$$ so that $$MA = V = V(u_1,\dots,u_n)$$ is a Vandermonde matrix.

Second, let us find the value of $V^{-1}b$.

The Vandermonde matrix inverse $V^{-1}$ has elements $$(V^{-1})_{i,j} = [z^{i-1}]\ \frac{F(z)}{F'(u_j)(z-u_j)},$$ where $$F(z) := (z-u_1)\cdots (z-u_n)$$ and $[z^k]$ is the operator taking the coefficient of $z^k$. Hence, $V^{-1} b$ is composed of the coefficients of $$G(z):=\sum_{j=1}^n \frac{F(z)}{F'(u_j)(z-u_j)} (u_j^n+x-1)=z^n - F(z) + x-1,$$ where the latter equality holds since the right-hand and left-hand sides as polynomials in $z$ both have degree $\leq n-1$ and at $z=u_j$ evaluate to $u_j^n+x-1$ (i.e., they have equal values at $n$ distinct points).

Hence, $$(V^{-1}b)_k = [z^{k-1}]\ G(z) = \begin{cases} (-1)^{n-1} u_1\cdots u_n + x-1, & k=1;\\ (-1)^{n-k} e_{n+1-k}(u_1,\dots,u_n), & k>1; \end{cases}$$ where $e_k()$ are elementary symmetric polynomials.

Finally, from $y=A(V^{-1}b)$ we get $$y_1 = (-1)^{n+1} u_1\cdots u_n + x-1 - \frac{x-1}{x}G(1)$$ and $$y_k = (-1)^{n+k} e_{n+1-k}(u_1,\dots,u_n),\quad k>1.$$ Recalling the definition of $u_i$, we conclude that the free term of $G(1) = x -(1-u_1)\cdots(1-u_n)$ as polynomial in $x$ is zero, and thus $\frac{x-1}{x}G(1)$ is a polynomial in $x$ with integer coefficients. Then so are all $y_1,\dots,y_n$.

Let me denote $u_i = x^{i+1} + 1$. As I learned from the previous question, under the "matrix system" OP understands the following matrix equation (up to a factor $\frac{1}{x}$): $$My = b,$$ where $$M:=\begin{bmatrix} u_1^0 + x - 1 & u_1^1 + x - 1 & \cdots & u_1^{n-1} + x - 1 \\ \vdots & \vdots & \ddots & \vdots \\ u_n^0 + x - 1 & u_n^1 + x - 1 & \cdots & u_n^{n-1} + x - 1 \end{bmatrix} \quad\text{and}\quad b := \begin{bmatrix} u_1^n + x - 1 \\ \vdots \\ u_n^n + x - 1 \end{bmatrix}. $$

First, it is easy to see that it always has a unique solution: $$y = M^{-1} b = AA^{-1}M^{-1} b = A (MA)^{-1} b,$$ where $$A := \begin{bmatrix} \frac{1}{x} & - \frac{x-1}{x} & - \frac{x-1}{x} & \cdots & - \frac{x-1}{x} \\ 0 & 1 & 0 & \cdots & 0\\ 0 & 0 & 1 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{bmatrix}$$ so that $$MA = V = V(u_1,\dots,u_n)$$ is a Vandermonde matrix.

Second, let us find the value of $V^{-1}b$.

The Vandermonde matrix inverse $V^{-1}$ has elements $$(V^{-1})_{i,j} = [z^{i-1}]\ \frac{F(z)}{F'(u_j)(z-u_j)},$$ where $$F(z) := (z-u_1)\cdots (z-u_n)$$ and $[z^k]$ is the operator taking the coefficient of $z^k$. Hence, $V^{-1} b$ is composed of the coefficients of $$G(z):=\sum_{j=1}^n \frac{F(z)}{F'(u_j)(z-u_j)} (u_j^n+x-1)=z^n - F(z) + x-1,$$ where the latter equality can be seen from the fact that the right-hand and left-hand sides as polynomials in $z$ have degree $\leq n-1$ and at $z=u_j$ both evaluate to $u_j^n+x-1$ (i.e., they have equal values at $n$ distinct points).

Hence, $$(V^{-1}b)_k = [z^{k-1}]\ G(z) = \begin{cases} (-1)^{n-1} u_1\cdots u_n + x-1, & k=1;\\ (-1)^{n-k} e_{n+1-k}(u_1,\dots,u_n), & k>1; \end{cases}$$ where $e_k()$ are elementary symmetric polynomials.

Finally, from $y=A(V^{-1}b)$ we get $$y_1 = (-1)^{n+1} u_1\cdots u_n + x-1 - \frac{x-1}{x}G(1)$$ and $$y_k = (-1)^{n+k} e_{n+1-k}(u_1,\dots,u_n),\quad k>1.$$ Recalling the definition of $u_i$, we conclude that the free term of $G(1) = x -(1-u_1)\cdots(1-u_n)$ is zero, and thus $\frac{x-1}{x}G(1)$ is a polynomial in $x$ with integer coefficients. Then so are all $y_1,\dots,y_n$.

Let me denote $u_i = x^{i+1} + 1$. As I learned from the previous question, under the "matrix system" OP understands the following matrix equation (up to a factor $\frac{1}{x}$): $$My = b,$$ where $$M:=\begin{bmatrix} u_1^0 + x - 1 & u_1^1 + x - 1 & \cdots & u_1^{n-1} + x - 1 \\ \vdots & \vdots & \ddots & \vdots \\ u_n^0 + x - 1 & u_n^1 + x - 1 & \cdots & u_n^{n-1} + x - 1 \end{bmatrix} \quad\text{and}\quad b := \begin{bmatrix} u_1^n + x - 1 \\ \vdots \\ u_n^n + x - 1 \end{bmatrix}. $$

First, it is easy to see that it always has a unique solution: $$y = M^{-1} b = AA^{-1}M^{-1} b = A (MA)^{-1} b,$$ where $$A := \begin{bmatrix} \frac{1}{x} & - \frac{x-1}{x} & - \frac{x-1}{x} & \cdots & - \frac{x-1}{x} \\ 0 & 1 & 0 & \cdots & 0\\ 0 & 0 & 1 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{bmatrix}$$ so that $$MA = V = V(u_1,\dots,u_n)$$ is a Vandermonde matrix.

Second, let us find the value of $V^{-1}b$.

The Vandermonde matrix inverse $V^{-1}$ has elements $$(V^{-1})_{i,j} = [z^{i-1}]\ \frac{F(z)}{F'(u_j)(z-u_j)},$$ where $$F(z) := (z-u_1)\cdots (z-u_n)$$ and $[z^k]$ is the operator taking the coefficient of $z^k$. Hence, $V^{-1} b$ is composed of the coefficients of $$G(z):=\sum_{j=1}^n \frac{F(z)}{F'(u_j)(z-u_j)} (u_j^n+x-1)=z^n - F(z) + x-1,$$ where the latter equality holds since the right-hand and left-hand sides as polynomials in $z$ both have degree $\leq n-1$ and at $z=u_j$ evaluate to $u_j^n+x-1$ (i.e., they have equal values at $n$ distinct points).

Hence, $$(V^{-1}b)_k = [z^{k-1}]\ G(z) = \begin{cases} (-1)^{n-1} u_1\cdots u_n + x-1, & k=1;\\ (-1)^{n-k} e_{n+1-k}(u_1,\dots,u_n), & k>1; \end{cases}$$ where $e_k()$ are elementary symmetric polynomials.

Finally, from $y=A(V^{-1}b)$ we get $$y_1 = (-1)^{n+1} u_1\cdots u_n + x-1 - \frac{x-1}{x}G(1)$$ and $$y_k = (-1)^{n+k} e_{n+1-k}(u_1,\dots,u_n),\quad k>1.$$ Recalling the definition of $u_i$, we conclude that the free term of $G(1) = x -(1-u_1)\cdots(1-u_n)$ as polynomial in $x$ is zero, and thus $\frac{x-1}{x}G(1)$ is a polynomial in $x$ with integer coefficients. Then so are all $y_1,\dots,y_n$.

Let me denote $u_i = x^{i+1} + 1$. As I learned from the previous question, under the "matrix system" OP understands the following matrix equation (up to a factor $\frac{1}{x}$): $$My = b,$$ where $$M:=\begin{bmatrix} u_1^0 + x - 1 & u_1^1 + x - 1 & \cdots & u_1^{n-1} + x - 1 \\ \vdots & \vdots & \ddots & \vdots \\ u_n^0 + x - 1 & u_n^1 + x - 1 & \cdots & u_n^{n-1} + x - 1 \end{bmatrix} \quad\text{and}\quad b := \begin{bmatrix} u_1^n + x - 1 \\ \vdots \\ u_n^n + x - 1 \end{bmatrix}. $$

First, it is easy to see that it always has a unique solution: $$y = M^{-1} b = AA^{-1}M^{-1} b = A (MA)^{-1} b,$$ where $$A := \begin{bmatrix} \frac{1}{x} & - \frac{x-1}{x} & - \frac{x-1}{x} & \cdots & - \frac{x-1}{x} \\ 0 & 1 & 0 & \cdots & 0\\ 0 & 0 & 1 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{bmatrix}$$ so that $$MA = V = V(u_1,\dots,u_n)$$ is a Vandermonde matrix.

Second, let us find the value of $V^{-1}b$.

The Vandermonde matrix inverse $V^{-1}$ has elements $$(V^{-1})_{i,j} = [z^{i-1}]\ \frac{F(z)}{F'(u_j)(z-u_j)},$$ where $$F(z) := (z-u_1)\cdots (z-u_n)$$ and $[z^k]$ is the operator taking the coefficient of $z^k$. Hence, $V^{-1} b$ is composed of the coefficients of $$\sum_{j=1}^n \frac{F(z)}{F'(u_j)(z-u_j)} (u_j^n+x-1).$$

I claim that $$\sum_{j=1}^n \frac{F(z)}{F'(u_j)(z-u_j)} u_j^n = z^n - F(z).$$ Indeed, the polynomials (in $z$) in the right-hand and left-hand sides have degree $\leq n-1$ and have equal values at $n$ distinct points $u_1,\dots,u_n$. So, these polynomials must coincide.

Similarly, $$\sum_{j=1}^n \frac{F(z)}{F'(u_j)(z-u_j)} = 1.$$

Hence, $V^{-1} b$ is composed of the coefficients of $G(z) = z^n - F(z) + (x-1)$, that is $$(V^{-1}b)_k = [z^{k-1}]\ G(z) = \begin{cases} (-1)^{n-1} u_1\cdots u_n + x-1, & k=1;\\ (-1)^{n-k} e_{n+1-k}(u_1,\dots,u_n), & k>1; \end{cases}$$ where $e_k()$ are elementary symmetric polynomial.

Finally, from $y=A(V^{-1}b)$ we get $$y_1 = (-1)^{n+1} u_1\cdots u_n + x-1 - \frac{x-1}{x}G(1)$$ and $$y_k = (-1)^{n+k} e_{n+1-k}(u_1,\dots,u_n),\quad k>1.$$ Recalling the definition of $u_i$, we conclude that the free term of $G(1) = x -(1-u_1)\cdots(1-u_n)$ is zero, and thus $\frac{x-1}{x}G(1)$ is a polynomial in $x$ with integer coefficients. Then so are all $y_1,\dots,y_n$.

Let me denote $u_i = x^{i+1} + 1$. As I learned from the previous question, under the "matrix system" OP understands the following matrix equation (up to a factor $\frac{1}{x}$): $$My = b,$$ where $$M:=\begin{bmatrix} u_1^0 + x - 1 & u_1^1 + x - 1 & \cdots & u_1^{n-1} + x - 1 \\ \vdots & \vdots & \ddots & \vdots \\ u_n^0 + x - 1 & u_n^1 + x - 1 & \cdots & u_n^{n-1} + x - 1 \end{bmatrix} \quad\text{and}\quad b := \begin{bmatrix} u_1^n + x - 1 \\ \vdots \\ u_n^n + x - 1 \end{bmatrix}. $$

First, it is easy to see that it always has a unique solution: $$y = M^{-1} b = AA^{-1}M^{-1} b = A (MA)^{-1} b,$$ where $$A := \begin{bmatrix} \frac{1}{x} & - \frac{x-1}{x} & - \frac{x-1}{x} & \cdots & - \frac{x-1}{x} \\ 0 & 1 & 0 & \cdots & 0\\ 0 & 0 & 1 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{bmatrix}$$ so that $$MA = V = V(u_1,\dots,u_n)$$ is a Vandermonde matrix.

Second, let us find the value of $V^{-1}b$.

The Vandermonde matrix inverse $V^{-1}$ has elements $$(V^{-1})_{i,j} = [z^{i-1}]\ \frac{F(z)}{F'(u_j)(z-u_j)},$$ where $$F(z) := (z-u_1)\cdots (z-u_n)$$ and $[z^k]$ is the operator taking the coefficient of $z^k$. Hence, $V^{-1} b$ is composed of the coefficients of $$G(z):=\sum_{j=1}^n \frac{F(z)}{F'(u_j)(z-u_j)} (u_j^n+x-1)=z^n - F(z) + x-1,$$ where the latter equality can be seen from the fact that the right-hand and left-hand sides as polynomials in $z$ have degree $\leq n-1$ and at $z=u_j$ both evaluate to $u_j^n+x-1$ (i.e., they have equal values at $n$ distinct points).

Hence, $$(V^{-1}b)_k = [z^{k-1}]\ G(z) = \begin{cases} (-1)^{n-1} u_1\cdots u_n + x-1, & k=1;\\ (-1)^{n-k} e_{n+1-k}(u_1,\dots,u_n), & k>1; \end{cases}$$ where $e_k()$ are elementary symmetric polynomials.

Finally, from $y=A(V^{-1}b)$ we get $$y_1 = (-1)^{n+1} u_1\cdots u_n + x-1 - \frac{x-1}{x}G(1)$$ and $$y_k = (-1)^{n+k} e_{n+1-k}(u_1,\dots,u_n),\quad k>1.$$ Recalling the definition of $u_i$, we conclude that the free term of $G(1) = x -(1-u_1)\cdots(1-u_n)$ is zero, and thus $\frac{x-1}{x}G(1)$ is a polynomial in $x$ with integer coefficients. Then so are all $y_1,\dots,y_n$.