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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:

enter image description here

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.

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2
  • $\begingroup$ 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. $\endgroup$ Commented Mar 20, 2021 at 19:07
  • $\begingroup$ @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. $\endgroup$
    – Red Banana
    Commented Mar 20, 2021 at 21:12

1 Answer 1

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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

enter image description here

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 *)
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