# If $\gamma\in \Bbb A$ then there exists a $\pm$quadratic coefficiented polynomial for which $\gamma$ is a root.

$\mathbf{1.\space Proposition}$

Let $\gamma$ be a solution to the equation:

$$\sum_{i=0}^n \rm a_i\rm x^i=0, \rm a_i\in\Bbb Z, a_n=1.$$

Then, there exists a polynomial $p\in \Bbb Z[x]$ such that: $$p=\sum _{i=0}^m\rm s_ix^i$$

Where $\rm s_i=(-1)^{k_i}\tau_i^2 \space\space\forall \rm i\in \{0,..,m\}, k_i\in \Bbb Z, \tau_i\in \Bbb Z$, for which $\gamma$ is a root.

$\mathbf{1.1\space Generalization}$.

There exist a monic polynomial satisfying the above conditions.

$\mathbf{1.2\space Generalization}$

Same conditions as $\rm1$, but $\rm s_i=(-1)^{k_i}\tau_i^r$, for all natural $\rm r$.

$\mathbf{1.3\space Generalization}$

There exists a monic polynomial satisfying $\rm1.2$.

$\mathbf{Important \space implication \space of \space 1.1 }$

If $\rm 1.1$ is true, it means that the condition of $\gamma$ being the root of a monic polynomial $p\in \Bbb Z[x]$ is actually equivalent to the 'stronger' condition of being a solution to a $\pm$quadratic coefficiented polynomial $q\in \Bbb Z[x]$.

• Can you find such an equation for $\sqrt3$ ? – lhf Sep 30 '15 at 16:55
• $(x^4-9)$ works for $\sqrt{3}$. – Aravind Sep 30 '15 at 17:50
• You can always do it for $\sqrt[n]{m}$ since the polynomial $x^{2n}-m^2=(x^n-m)(x^n+m)$ will work...but for more complicated things, I have no idea. – Ben Sheller Sep 30 '15 at 19:23
• Could you write $1+\sqrt5$ in this manner ? – Lucian Sep 30 '15 at 19:34
• @BenS.: Then the answer to the OP's question is yes. Try $x^4-4x^3+4x^2-9$. – Lucian Sep 30 '15 at 20:48

Proposition: Let $\gamma$ be an algebraic integer, $r$ an odd positive integer. Then there exists a non-zero polynomial $f\in\mathbb{Z}[x]$ with $r$-power coefficients such that $f(\gamma)=0$.
Theorem (Birch): Let $k$ and $r$ be postive integers, with $r$ odd. Then there exists a positive integer $n$ such that, if $f_1,\ldots,f_k$ are homogeneous polynomials with integer coefficients in $n$ variables, there exists $\vec{x}=(x_1,\ldots,x_n)\neq (0,\ldots,0)\in\mathbb{Z}^n$ such that $f_1(\vec{x})=\ldots=f_k(\vec{x})=0$.
Proof of proposition: We are given $\gamma$ and odd $r$. Let $k=\deg(\gamma)$, and let $n$ be as in the conclusion of Birch's Theorem. Choose an integral basis $\alpha_1,\ldots,\alpha_k$ of $\mathbb{Z}[\gamma]$. For each $m\geq 0$, there are unique integers $a_{m,1},\ldots,a_{m,k}$ such that $\gamma^m=a_{m,1}\alpha_1+\ldots+a_{m,k}\alpha_k$. For $i=1,\ldots,k$, define $$f_i(x_1,\ldots,x_n)=a_{1,i}x_1^r+\ldots+a_{n,i}x_n^r.$$ The $f_i$ are $k$ homogeneous polynomial of degree $r$ in $n$ variables, so by Birch's theorem, they have a common non-zero solution $(x_1,\ldots,x_n)$. Now $$x_1^r\gamma+x_2^r\gamma^2+\ldots+x_n^r\gamma^n=0,$$ as desired.