what is written below is a conjecture that I posed , and I ask for a proof or a disproof of it .I have checked the conjecture from $n$=$1$ up to $n$=$10$ using Matlab, and all results were in agreement with the conjecture . The conjecture is as follows : assume $x$ is a positive real parameter that does not equal $1$ , and assume $y$ and $z$ are non-zero real parameters , and consider for all $i, j \in \mathbb N$, $$a(i,j) = \frac{(x^{yi+z} + 1)^{j-1} + (x^y-1)}{x^y}$$ ; then for all $n \in \mathbb N$ , the solution set of the matrix system $[a(i,j) \mid 1 \leq i \leq n, 1 \leq j \leq (1+n)]$ exists and is unique with respect to $n$ and $x$ and $y$ and $z$ ,and each element in it is a sum of powers of $x$ with integer coefficients , and each of these powers of $x$ has the power as a linear combination of $y$ and $z$ such that the coefficients of $y$ and $z$ are non-negative integers .

###Example

For $n$=$4$,the matrix system is $$\left(\begin{array}{cccc|c} 1 & x^{z}+1 & \frac{(x^{y+z} + 1)^{2} + (x^y-1)}{x^y} & \frac{(x^{y+z} + 1)^{3} + (x^y-1)}{x^y} & \frac{(x^{y+z} + 1)^{4} + (x^y-1)}{x^y} \\ 1 & x^{y+z}+1 & \frac{(x^{2y+z} + 1)^{2} + (x^y-1)}{x^y} & \frac{(x^{2y+z} + 1)^{3} + (x^y-1)}{x^y} & \frac{(x^{2y+z} + 1)^{4} + (x^y-1)}{x^y} \\ 1 & x^{2y+z}+1 & \frac{(x^{3y+z} + 1)^{2} + (x^y-1)}{x^y} & \frac{(x^{y3+z} + 1)^{3} + (x^y-1)}{x^y} & \frac{(x^{3y+z} + 1)^{4} + (x^y-1)}{x^y} \\ 1 & x^{3y+z}+1 & \frac{(x^{4y+z} + 1)^{2} + (x^y-1)}{x^y} & \frac{(x^{4y+z} + 1)^{3} + (x^y-1)}{x^y} & \frac{(x^{4y+z} + 1)^{4} + (x^y-1)}{x^y} \end{array}\right)$$

the solution set is:

$s_1=- x^{y+z} - x^{z+2y} - x^{z+3y} - x^{z+4y} - x^{3y+2z} - x^{4y+2z} - 2x^{5y+2z} - x^{6y+2z} - x^{6y+3z} - x^{7y+2z}$ $\qquad- x^{7y+3z} - x^{8y+3z} - x^{9y+3z} - x^{9y+4z} - 1$

$s2$=$3$$x^{y+z}$ + $3$$x^{z+2y}$ + $3$$x^{z+3y}$ + $3$$x^{z+4y}$ + $2$$x^{3y+2z}$ +

$\qquad 2x^{4y+2z}$ + $4$$x^{5y+2z}$ + $2$$x^{6y+2z}$ + $x^{6y+3z}$ + $2$$x^{7y+2z}$ + $x^{7y+3z}$ + $x^{8y+3z}$ + $x^{9y+3z}$ + $4$

$s3$=-$3$$x^{y+z}$ - $3$$x^{z+2y}$ - $3$$x^{z+3y}$ - $3$$x^{z+4y}$ - $x^{3y+2z}$ - $x^{4y+2z}$ - $2$$x^{5y+2z}$ - $x^{6y+2z}$ - $x^{7y+2z}$ - $6$

$s4$=$x^{y+z}$ + $x^{z+2y}$ + $x^{z+3y}$ + $x^{z+4y}$ + $4$

Another example that explains the previous example by assuming $x$=$2$,$y$=$1$,and $z$=$1$: we will have for all $i, j \in \mathbb N$, $a(i,j)$ = ($(2^{i+1} + 1)^{j-1}$ + $1$)/$2$

since $n$=$4$,we will find the solution set of the matrix system $[a(i,j) \mid 1 \leq i \leq 4, 1 \leq j \leq 5]$

The matrix system is $$\left(\begin{array}{cccc|c} 1 & 3 & 13 & 63 & 313 \\ 1 & 5 & 41 & 365 & 3281 \\ 1 & 9 & 145 & 2457 & 41761 \\ 1 & 17 & 545 & 17969 & 592961 \end{array}\right)$$

The solution set is : $s1$=$-17053$ , $s2$=$10104$ , $s3$=$-1306$ , $s4$=$64$ . Notice that $s1$ , $s2$ , $s3$ , $s4$ can be found by substituting $x$=$2$ , $y$=$1$ , $z$=$1$ in the formulas of $s1$ , $s2$ , $s3$ , $s4$ in the first example .

Thank you .