# Trouble trying to find solutions for linear equations with FindInstance?

I am trying to do the following: Given $$x=i+\sqrt{3}$$, finding a polynomial $$p$$ with rational coefficients not all zero such that $$p(x)=0$$. For this, I am doing the following:

x = I + Sqrt[3];
Table[
FindInstance[Sum[Subscript[a, i] x^i, {i, 0, n}] == 0,
Table[Subscript[a, i], {i, 0, n}], Rationals]
, {n, 1, 10}] // TableForm


This gives me only the trivial solutions with all $$a_i=0$$. I tried to ask for two solutions with:

x = I + Sqrt[3];
Table[
FindInstance[Sum[Subscript[a, i] x^i, {i, 0, n}] == 0,
Table[Subscript[a, i], {i, 0, n}], Rationals, 2]
, {n, 1, 10}] // TableForm


But then, Mathematica gives me following errors:

Observe that not all equations are going to have solutions for $$a_i$$ not all zero but I thought that in doing this, Mathematica would eventually give me one with $$a_i$$ not all zero because for some $$n$$, this solution exists.

• This not possible. If I + Sqrt[3] is a root, the the polynomial has a factor: x- I + Sqrt[3] and this factor is not rational. – Daniel Huber Mar 20 at 19:07
• @Bill I am trying some stuff on Galois theory. Consider the polynomial: $$\left(x-\sqrt{3}-i\right) \left(x-\sqrt{3}+i\right) \left(x+\sqrt{3}-i\right)\left(x+\sqrt{3}+i\right)=x^4-4 x^2+16$$ It obeys that property. I just don't know if I can do this like I am trying to do in my question. – Red Banana Mar 20 at 21:12

Clear["Global*"]

x = I + Sqrt[3];

Select[Table[vars = Array[a, n + 1, 0];
FindInstance[{vars . x^Range[0, n] == 0, Element[vars, Rationals],
Or @@ Thread[vars != 0]}, vars], {n, 1, 10}],
FreeQ[#, FindInstance] && (# =!= {}) &] // TableForm // Quiet


EDIT: Generalizing,

Assuming[Element[n, NonNegativeIntegers], -1/2 x^n - 1/128 x^(n + 6) //
Simplify]

(* 0 *)


Wash, Rinse, Repeat...

Assuming[Element[{n, m},
NonNegativeIntegers], -1/2 x^n - 1/128 x^(n + 6) - 1/2 x^m -
1/128 x^(m + 6) // Simplify]

(* 0 *)
`