Assume $x$ is a variable belongs to $\mathbb R \setminus \{ 0,-1,+1 \}$ and consider for all $i, j \in \mathbb N$,

$$a(i,j) = \frac{(x^{i+1} + 1)^{j-1} + (x-1)}{x}$$

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 all elements in it are polynomials of integer coefficients. I put this conjecture and it was tried well; and all results were in agreement with the conjecture.

###Example

For $n=7$, the solution set will be

$$x^34 + x^33 + x^32 + x^31 + 2*x^30 + 2*x^29 + 3*x^28 + 3*x^27 + 4*x^26 + 4*x^25 + 5*x^24 + 5*x^23 + 6*x^22 + 6*x^21 + 7*x^20 + 6*x^19 + 7*x^18 + 7*x^17 + 6*x^16 + 7*x^15 + 6*x^14 + 6*x^13 + 5*x^12 + 5*x^11 + 4*x^10 + 4*x^9 + 3*x^8 + 3*x^7 + 2*x^6 + 2*x^5 + x^4 + x^3 + x^2 + 1$$

$$-x^33 - x^32 - x^31 - 3*x^30 - 3*x^29 - 5*x^28 - 5*x^27 - 9*x^26 - 9*x^25 - 12*x^24 - 13*x^23 - 16*x^22 - 18*x^21 - 21*x^20 - 20*x^19 - 24*x^18 - 25*x^17 - 22*x^16 - 28*x^15 - 24*x^14 - 26*x^13 - 22*x^12 - 23*x^11 - 19*x^10 - 19*x^9 - 16*x^8 - 16*x^7 - 11*x^6 - 11*x^5 - 6*x^4 - 6*x^3 - 6*x^2 - 7$$

$$x^30 + x^29 + 2*x^28 + 2*x^27 + 6*x^26 + 6*x^25 + 9*x^24 + 11*x^23 + 14*x^22 + 19*x^21 + 22*x^20 + 24*x^19 + 30*x^18 + 33*x^17 + 30*x^16 + 43*x^15 + 37*x^14 + 44*x^13 + 38*x^12 + 42*x^11 + 36*x^10 + 36*x^9 + 35*x^8 + 35*x^7 + 25*x^6 + 25*x^5 + 15*x^4 + 15*x^3 + 15*x^2 + 21$$

$$- x^26 - x^25 - 2*x^24 - 3*x^23 - 4*x^22 - 8*x^21 - 9*x^20 - 12*x^19 - 16*x^18 - 19*x^17 - 18*x^16 - 31*x^15 - 27*x^14 - 36*x^13 - 32*x^12 - 38*x^11 - 34*x^10 - 34*x^9 - 40*x^8 - 40*x^7 - 30*x^6 - 30*x^5 - 20*x^4 - 20*x^3 - 20*x^2 - 35$$ $$x^21 + x^20 + 2*x^19 + 3*x^18 + 4*x^17 + 4*x^16 + 10*x^15 + 9*x^14 + 14*x^13 + 13*x^12 + 17*x^11 + 16*x^10 + 16*x^9 + 25*x^8 + 25*x^7 + 20*x^6 + 20*x^5 + 15*x^4 + 15*x^3 + 15*x^2 + 35$$ $$- x^15 - x^14 - 2*x^13 - 2*x^12 - 3*x^11 - 3*x^10 - 3*x^9 - 8*x^8 - 8*x^7 - 7*x^6 - 7*x^5 - 6*x^4 - 6*x^3 - 6*x^2 - 21$$ $$x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + 7$$

Thank you.