# Conjecture that relates matrix systems with some specific functions as solution sets

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 $$\left(\begin{array}{cccc|c} 1 & x^{z}+1 & \frac{(x^{y+z} + 1)^{2} + (x^y-1)}{x^y} & \frac{(x^{y+z} + 1)^{3} + (x^y-1)}{x^y} & \frac{(x^{y+z} + 1)^{4} + (x^y-1)}{x^y} \\ 1 & x^{y+z}+1 & \frac{(x^{2y+z} + 1)^{2} + (x^y-1)}{x^y} & \frac{(x^{2y+z} + 1)^{3} + (x^y-1)}{x^y} & \frac{(x^{2y+z} + 1)^{4} + (x^y-1)}{x^y} \\ 1 & x^{2y+z}+1 & \frac{(x^{3y+z} + 1)^{2} + (x^y-1)}{x^y} & \frac{(x^{3y+z} + 1)^{3} + (x^y-1)}{x^y} & \frac{(x^{3y+z} + 1)^{4} + (x^y-1)}{x^y} \\ 1 & x^{3y+z}+1 & \frac{(x^{4y+z} + 1)^{2} + (x^y-1)}{x^y} & \frac{(x^{4y+z} + 1)^{3} + (x^y-1)}{x^y} & \frac{(x^{4y+z} + 1)^{4} + (x^y-1)}{x^y} \end{array}\right)$$

the solution set is:

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

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

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

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

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

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 .

• It's best if you learn how to use MathJax to format math here. It's a lot like TeX. x^yx^z enclosed in dollar signs becomes $x^yx^z$. To get $x^{2y+z}$, type x^{2y+z} enclosed in dollar signs. And maybe a smaller example, say with $n=1$, would be more enlightening. – Gerry Myerson Nov 4 '18 at 3:04
• If you expect anyone to take the time to read them, you should take the time to make them legible. If you don't care whether anyone reads your question or not, then take your time. – Gerry Myerson Nov 4 '18 at 3:45
• It's still very hard to read. Are $x,y,z$ parameters or unknowns? What are the unknowns? I'm not sure what a "matrix system" is supposed to be, but I would expect to find at least an equal sign in its definition... – Federico Poloni Nov 4 '18 at 16:44
• Is this a generalization of mathoverflow.net/questions/281442/…? – Mahdi Nov 4 '18 at 17:33
• Some edits have been done ,thanks goes to Gerry Myerson . – Ahmad Jamil Ahmad Masad Nov 6 '18 at 17:56

## 1 Answer

The affirmative answer and explicit solution to this question directly follows from my answer to the previous one by substituting there $$u_i:=X^{Yi+Z}+1$$ and $$x:=X^Y$$. Here I use capital letters to refer to the variables in the present question and distinguish them from those in my answer.