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\title{An Elementary Proof of Fundamental Theorem of Algebra over Rationals}
\author{%
Liang Zhang\thanks{Corresponding author: \href{mailto:liang.zhang@unidt.com}{\texttt{liang.zhang@unidt.com}}}
\and
Xiaohua(Michael) Xuan\thanks{Email: \href{mailto:michael.xuan@unidt.com}{\texttt{michael.xuan@unidt.com}}}%
}
\date{}

\begin{document}
\maketitle

\begin{abstract}
We, with the aid of AI, give an elementary proof of the Fundamental Theorem of Algebra over Rationals, i.e., for any nonconstant polynomial $f$ with rational complex number coefficients and every positive rational number $\eps$, there exists a rational complex number $z$ such that $Q(f(z))<\eps^2$.
The proof is constructive and relies only on finite rational arithmetic, without using real numbers, infinite sets, or related concepts.


\end{abstract}

\setcounter{tocdepth}{1}
\tableofcontents
\bigskip

\section{Introduction}

The Fundamental Theorem of Algebra is a central result in mathematics.  In its
classical form, it states that every nonconstant polynomial $f$ with complex number
coefficients has a complex number $z$ such that $f(z)=0$. 

Many mathematicians contributed to the historical development of the theorem,
including d'Alembert, Euler, Lagrange, Laplace, Argand, Gauss, and
Cauchy~\cite{FineRosenberger1997,Petrova1974}.  Of particular relevance here, Gauss
attempted to prove the theorem by tracking the winding of a polynomial image
around the origin, though a topological step remained to be
justified~\cite{Gauss1799,BasuVelleman2017}.  Cauchy later gave the winding
number an integral formulation, supplying a rigorous analytic framework for
the argument~\cite{Cauchy1821,Petrova1974}.

Kuhn~\cite{Kuhn1974} gave a constructive combinatorial proof in 1974, using
vertex labelling and a combinatorial lemma on triangulated regions of the
complex plane.  His proof is an important precedent for the present use of
finite triangulations, and the augument in this paper is close in spirit to this approach.
% The emphasis here is on an approximate-zero formulation using only finite
% rational arithmetic, with quadrance bounds, explicit variation estimates for
% $\sigma(x+\ii y)\equiv|x|+|y|$, and a winding number computed directly from
% signed edge crossings.

Smale gave a short proof of the Fundamental Theorem of Algebra by
using the inverse function theorem in his study of the complexity of algorithms
for solving equations~\cite{Smale1981,Smale1985}.  He and Shub subsequently showed that
the homotopy method combined with Newton's iteration yields an efficient
algorithm with a provably good complexity bound for approximating a zero of a
polynomial~\cite{ShubSmale1985,ShubSmale1986,BCSS1998}.

While all mentioned proofs of the Fundamental Theorem of Algebra depend on the real
numbers and infinite processes, Wildberger called for a proof grounded
strictly in finite rational arithmetic and formulated two alternative versions
of the theorem in his online YouTube lectures~\cite{WildbergerFTA} in 2017.  This paper directly answers his call by
providing the requested proof.  

Using the quadrance terminology of rational trigonometry~\cite{Wildberger2005},
write $Q(a+\ii b)\equiv a^2+b^2$.  We prove the following statement:

\begin{unnumberedtheorem}[Main theorem]
For any nonconstant polynomial $f$ with rational complex number coefficients and any rational number $\eps>0$,
there exists a rational complex number $z$ such that
\[
Q\bigl(f(z)\bigr)<\eps^{2}.
\]
\end{unnumberedtheorem}

The proof uses the winding number of finite rational closed vertex lists, rational variation
bounds, and a finite triangulation to obtain a contradiction if no grid point
satisfies the conclusion.  It involves only finite objects,
rational number computation and comparisons.

\section{Notation and conventions}\label{sec:conventions}

Throughout, the symbol $\equiv$ introduces a definition, while $=$ denotes an
asserted equality.  We write
\[
t\tin T
\]
to mean that $t$ is of type $T$.  For example, $t\tin\Q$ and $t\tin\ZZ$
mean, respectively, that $t$ is a rational number and that $t$ is an integer.
When no confusion can arise, we also use the ordinary notation $t\in T$ for
this purpose.


A \emph{rational complex number} is a finite datum consisting of two rational
numbers $a$ and $b$, written as $a+\ii b$.  We write $z\tin K$ when $z$ is such
a datum, and $z\tin K^\times$ when, in addition, $z\ne0$.

A polynomial with rational complex coefficients is specified by a finite
coefficient list
\[
(a_0,a_1,\ldots,a_n),\qquad a_j\tin K,
\]
and is written as
\[
f(X)=\sum_{j=0}^{n}a_jX^j.
\]
The notation $f\tin K[X]$ asserts only that $f$ is given by such finite data.
Thus $K$, $K^\times$, and $K[X]$ are used solely as type labels.

We use the convention $u^0\equiv1$, including when $u=0$; empty sums
have value $0$, and empty products have value $1$.

We identify $a+\ii b$ with the point $[a,b]$.  For points
$A_1=[x_1,y_1]$ and $A_2=[x_2,y_2]$, define their \emph{quadrance} by
\[
Q(A_1,A_2)\equiv (x_2-x_1)^2+(y_2-y_1)^2.
\]
For $u=a+\ii b\in K$, we use the abbreviation
\[
Q(u)\equiv Q([0,0],[a,b])=a^2+b^2.
\]
Thus $Q(u)$ denotes the quadrance of $u$ from the origin.

For a positive rational number $R$, let $\Sq(R)$ denote the square with
vertices
\[
-R-\ii R,\qquad R-\ii R,\qquad R+\ii R,\qquad -R+\ii R.
\]
A point
$z=a+\ii b$ lies in $\Sq(R)$ precisely when $|a|\le R$ and $|b|\le R$,
and lies on its boundary $\partial\Sq(R)$ precisely when, in addition,
$|a|=R$ or $|b|=R$. Here $|\cdot|$ denotes the usual absolute value on $\Q$.

For any three-point list $(p_1,p_2,p_3)$, we say that $z$ is a
\emph{nonnegative rational affine combination} of the list precisely when
\[
z=\alpha p_1+\beta p_2+\gamma p_3,
\qquad
\alpha,\beta,\gamma\in\Q,
\qquad
\alpha,\beta,\gamma\ge0,
\qquad
\alpha+\beta+\gamma=1.
\]
When $p_1,p_2,p_3$ are distinct and non-collinear, this condition means that
$z$ lies on the triangle $p_1p_2p_3$.  Such a point is interior to the triangle
precisely when $\alpha,\beta,\gamma>0$, and it lies on a side precisely when
$\alpha\beta\gamma=0$.  When the three-point list is degenerate, we retain
the affine-combination language and do not call it a triangle.


\section{Rational quadrance and variation estimates}\label{sec:estimates}

\begin{definition}[Auxiliary size]\label{def:auxiliary-size}
For $u=a+\ii b\in K$, define
\[
\sigma(u)\equiv|a|+|b|.
\]
$\sigma(u)$ is called the \emph{size} of $u$.  Note that
$Q(u)=u\cdot\overline{u}$, where
$\overline{a+\ii b}\equiv a-\ii b$.
\end{definition}

\begin{lemma}[Basic inequalities]\label{lem:basic}
For all $u,v\in K$:
\[
Q(u)\le\sigma(u)^{2},\qquad
\sigma(u+v)\le\sigma(u)+\sigma(v),\qquad
\sigma(u\cdot v)\le\sigma(u)\sigma(v).\tag{2.1}\label{eq:basic-inequalities}
\]
Moreover $\sigma$ is nonnegative, $\sigma(0)=0$, $\sigma(1)=1$, and
$\sigma(u^{n})\le\sigma(u)^{n}$ for all $n\ge0$.
\end{lemma}
\begin{proof}
Set $u\equiv a+\ii b$, $v\equiv c+\ii d$.  Then
\[
\sigma(u+v)=|a+c|+|b+d|\le |a|+|c|+|b|+|d|=\sigma(u)+\sigma(v).
\]

\[
\sigma(uv)=|ac-bd|+|ad+bc|
\le |a||c|+|b||d|+|a||d|+|b||c|= (|a|+|b|)(|c|+|d|)=\sigma(u)\sigma(v).
\]

Finally,
$Q(u)=a^{2}+b^{2}\le(|a|+|b|)^{2}=\sigma(u)^{2}$. The facts $\sigma(0)=0$, $\sigma(1)=1$, $\sigma\ge0$ are immediate;
$\sigma(u^{n})\le\sigma(u)^{n}$ follows from submultiplicativity by induction.
\end{proof}

\begin{lemma}[Power variation]\label{lem:powvar}
Let $w,z\in K$, let $j\ge1$ be an integer, and let $B\in \Q$ with
$\sigma(w),\sigma(z)\le B$.  Then
\[
\sigma(w^{j}-z^{j})\le j\cdot B^{j-1}\cdot\sigma(w-z).\tag{2.2}\label{eq:power-variation}
\]
\end{lemma}
\begin{proof}
For $j\ge1$, we note that
\[
w^{j}-z^{j}=(w-z)\cdot\sum_{r=0}^{j-1}w^{j-1-r}z^{r}.
\]
The result follows from the triangle inequality and submultiplicativity of $\sigma$, and also the fact that
$\sigma(w),\sigma(z)\le B$.
\end{proof}

\begin{lemma}[Polynomial variation]\label{lem:polyvar}
Let $f(X)=\sum_{j=0}^{n}a_{j}X^{j}\in K[X]$ and let $B\in \Q$ with
$\sigma(w),\sigma(z)\le B$.  Then
\[
\sigma\bigl(f(w)-f(z)\bigr)
\le \sigma(w-z)\cdot\sum_{j=1}^{n}j\,\sigma(a_{j})\,B^{j-1}.\tag{2.3}\label{eq:polynomial-variation}
\]
\end{lemma}
\begin{proof}
By the power-variation estimate \eqref{eq:power-variation} and the
submultiplicative inequality in \eqref{eq:basic-inequalities},
\[
\sigma(a_jw^j-a_jz^j)
\le \sigma(a_j)\,jB^{j-1}\sigma(w-z).
\]
Summing this inequality over $j=1,\ldots,n$ and applying the triangle
inequality in \eqref{eq:basic-inequalities} gives
\eqref{eq:polynomial-variation}.
\end{proof}

\begin{lemma}[Affine exclusion]\label{lem:affine-exclusion}
Let $a,b,c\in K$ and $0<\eps\in \Q$.  If
$\sigma(a-b)<\eps$, $\sigma(a-c)<\eps$, and $\eps^{2}\le Q(a)$, then
the origin is not a nonnegative rational affine combination of $(a,b,c)$.
\end{lemma}
\begin{proof}
If the origin were such an affine combination, then
$a=\beta(a-b)+\gamma(a-c)$ with
$\beta,\gamma\ge0$, $\beta+\gamma\le1$, so
$\sigma(a)\le\beta\sigma(a-b)+\gamma\sigma(a-c)<\eps$, giving
$Q(a)\le\sigma(a)^{2}<\eps^{2}$, contradicting $\eps^{2}\le Q(a)$.
\end{proof}


\section{A winding number for finite closed vertex lists}\label{sec:index}

\subsection{Definitions and the half-open convention}

Throughout this section, a \emph{closed vertex list} is an oriented list
$P\equiv(p_{0},p_{1},\ldots,p_{s})$ with $p_{j}\in K$ and $p_{s}=p_{0}$.
The closed list $(a,b,c,a)$ is the \emph{three-cycle} determined by
$(a,b,c)$; its entries need not be distinct or non-collinear.
\begin{definition}[Determinant]\label{def:det}
For $u=x+\ii y$, $v=x'+\ii y'$, define
\[
\DeltaTwo{u}{v}\equiv xy'-yx'.
\]
This is twice the signed area of the oriented three-point list $(0,u,v)$. The sign decisions below depend on this
rational expression.
\end{definition}

\begin{definition}[Signed crossing, half-open convention]\label{def:cross}
For a directed edge $p\to q$, we define the \emph{signed crossing contribution}
$\chi(p,q)\in\ZZ$ by the half-open rule
\begin{equation}\label{eq:crossSign}
\chi(p,q)\equiv
\begin{cases}
1,& \operatorname{Im}(p)\le 0 < \operatorname{Im}(q)\ \text{and}\ \DeltaTwo{p}{q}>0,\\
-1,& \operatorname{Im}(q)\le 0 < \operatorname{Im}(p)\ \text{and}\ \DeltaTwo{p}{q}<0,\\
0,& \text{otherwise}.
\end{cases}
\end{equation}
Here $\operatorname{Im}(p)$ denotes the imaginary part of $p$. 
\end{definition}
\begin{remark}[Interpretation of \eqref{eq:crossSign}]\label{rmk:cross-interpret}
Let's consider the case
\[
\operatorname{Im}(p)\le0<\operatorname{Im}(q)\ \text{and}\ \DeltaTwo{p}{q}>0,
\]
Write $p=x+\ii y,\ q=x'+\ii y'$ with $\operatorname{Im}(p)=y\le0$ and
$\operatorname{Im}(q)=y'>0$, and let the intersection of $pq$ and $\operatorname{Im}(z)=0$ be $r=x_r,x_r\in \Q$.
Then $\DeltaTwo{p}{q}=x_r\,(y'-y)$.  Since $y'-y>0$, $\DeltaTwo{p}{q}>0$
iff $\operatorname{Re}(r)>0$.
\end{remark}
\begin{definition}[Winding number]\label{def:W}
For a closed vertex list $P=(p_{0},\ldots,p_{s})$, its \emph{winding number} is defined as follows:
\[
\W(P)\equiv\sum_{i=0}^{s-1}\chi(p_{i},p_{i+1})\in\ZZ.
\]
It is computable by finitely many rational number computations and comparisons.
\end{definition}
\begin{remark}[Positive scale invariance]\label{rmk:positive-scale}
Let $a\in\Q$ be positive and write
$aP\equiv(ap_{0},ap_{1},\ldots,ap_{s})$.  Then
\[
\W(aP)=\W(P).
\]
Indeed, for any edge $p_i\to p_{i+1}$,
\[
\operatorname{Im}(ap_i)=a\operatorname{Im}(p_i),
\qquad
\DeltaTwo{ap_i}{ap_{i+1}}=a^2\DeltaTwo{p_i}{p_{i+1}}.
\]
Since $a>0$, all inequalities and signs in Definition~\ref{def:cross} are
unchanged.  Hence
$\chi(ap_i,ap_{i+1})
=\chi(p_i,p_{i+1})$ for any $i$, and summing over the
edges gives $\W(aP)=\W(P)$ by Definition~\ref{def:W}.
\end{remark}


\begin{theorem}[Structural properties of $\W$]\label{thm:properties}
The winding number has the following properties, all of which can be proved by finite rational
computations.
\begin{enumerate}[label=(\arabic*)]
\item \emph{Reversal.}  $\chi(q,p)=-\chi(p,q)$;
  hence reversing all edges of a closed vertex list negates its winding number.
\item \emph{Three-cycle vanishing.}  If the origin is not a nonnegative
  rational affine combination of $(a,b,c)$, where $a,b,c\in K$, then
  $\W(a,b,c,a)=0$.
\item \emph{Additivity under common triangulation.}  
Let $P\equiv(p_0,\dots,p_n)$ and $Q\equiv(q_0,\dots,q_m)$ be closed vertex lists.
Inserting repeated vertices (equivalently, inserting zero-length edges) does not
change winding number, so we may assume $m=n$.
For each $i=0,\dots,n-1$, suppose the origin is not a nonnegative rational
affine combination of either $(p_i,p_{i+1},q_{i+1})$ or
$(p_i,q_{i+1},q_i)$.
Then
\[
\W(P)=\W(Q).
\]
\item \emph{Base value.}  For rational $R>0$, the counterclockwise boundary of
  the square $\Sq(R)$ has winding number $1$:
  \[
  \W(-R-\ii R,\,R-\ii R,\,R+\ii R,\,-R+\ii R,\,-R-\ii R)=1.
  \]
\end{enumerate}
\end{theorem}
\begin{proof}
(1) Exchanging $p$ and $q$ swaps the two branches of \eqref{eq:crossSign} and
negates $\DeltaTwo{p}{q}$, giving
$\chi(q,p)=-\chi(p,q)$.


(2) Assume that the origin is not a nonnegative rational affine combination
of $(a,b,c)$.

\medskip

\noindent\textbf{Step 1: encode the half-open convention.}
For each vertex $p$, write $y_p\equiv\operatorname{Im}(p), x_p\equiv\operatorname{Re}(p)$ and define
\[
h(p)\equiv
\begin{cases}
1,&y_p>0,\\
0,&y_p\le0.
\end{cases}
\]
For any directed edge $p\to q$ of $(a,b,c,a)$, if $h(p)=h(q)$, then
$\chi(p,q)=0$ by \eqref{eq:crossSign}.
If $h(p)\ne h(q)$, then $y_q-y_p\ne0$, so we may define the rational number
\[
r_{pq}\equiv\frac{\DeltaTwo{p}{q}}{y_q-y_p}
=\frac{y_qp-y_pq}{y_q-y_p}.
\]
Moreover,
\[
r_{pq}=(1-t)p+tq,
\qquad
t\equiv\frac{-y_p}{y_q-y_p}\in\Q,
\qquad 0\le t\le1.
\]
Thus $r_{pq}$ is a nonnegative rational affine combination of $(a,b,c)$.
Since
\[
\DeltaTwo{p}{q}=(y_q-y_p)r_{pq}
\]
One can easily check
\[
\chi(p,q)=\bigl(h(q)-h(p)\bigr)\mathbf1_{\{r_{pq}>0\}}
\qquad\text{when }h(p)\ne h(q),
\]
where $\mathbf1_{\{r_{pq}>0\}}$ is $1$ if $r_{pq}>0$ and $0$ otherwise.

\medskip

\noindent\textbf{Step 2: the intersection values have a common strict sign.}
Each of the values $r_{pq}$ just defined is nonzero, since otherwise it
would express the origin as a nonnegative rational affine combination of
$(a,b,c)$.  Nor can two such values have opposite signs: if $r_{pq}>0>r_{p'q'}$, then
\[
0=\frac{-r_{p'q'}}{r_{pq}-r_{p'q'}}\,r_{pq}+\frac{r_{pq}}{r_{pq}-r_{p'q'}}\,r_{p'q'}.
\]
The two coefficients are positive rational numbers whose sum is $1$.
Substituting the nonnegative rational affine expressions for $r_{pq}$ and $r_{p'q'}$
would therefore express the origin as a nonnegative rational affine
combination of $(a,b,c)$, again contradicting the hypothesis.
Hence all the values $r_{pq}$ for edges with $h(p)\ne h(q)$ have the same
strict sign.

\medskip

\noindent\textbf{Step 3: conclude by two cases.}

\emph{Case 1.} There exists an edge $p\to q$ with $h(p)\ne h(q)$ and
$r_{pq}>0$.

By Step 2, every edge with distinct $h$-values has a positive intersection
value.  Hence Step 1 gives
\[
\chi(p,q)=h(q)-h(p)
\]
for every such edge.  The same identity holds when $h(p)=h(q)$, since both
sides are zero.  Consequently,
\[
\begin{aligned}
\W(a,b,c,a)
&=\chi(a,b)+\chi(b,c)+\chi(c,a)\\
&=h(b)-h(a)+h(c)-h(b)+h(a)-h(c)\\
&=0.
\end{aligned}
\]

\emph{Case 2.} There is no edge with $h(p)\ne h(q)$ and $r_{pq}>0$.

Consider any edge $p\to q$.  If $h(p)=h(q)$, Step 1 gives $\chi(p,q)=0$.
Otherwise, $r_{pq}\ne0$ by Step 2, and therefore $r_{pq}<0$; Step 1 again
gives $\chi(p,q)=0$.  Thus
\[
\W(a,b,c,a)=\chi(a,b)+\chi(b,c)+\chi(c,a)=0.
\]
Both cases allow coincident vertices and collinear three-point lists.

(3) For each $i$, apply the three-cycle vanishing property established in
part~(2) above to
$(p_i,p_{i+1},q_{i+1})$ and
$(p_i,q_{i+1},q_i)$, obtaining
\[
\W(p_i,p_{i+1},q_{i+1},p_i)=0,\qquad
\W(p_i,q_{i+1},q_i,p_i)=0.
\]
Summing these $2n$ equalities, each interior edge appears twice with opposite
orientations and therefore cancels.  Only the boundary edges of $P$ and $Q$
remain, yielding $\W(P)=\W(Q)$.

(4) Direct computation: among the four directed edges, only the right side
$R-\ii R \to R+\ii R$ has an upward crossing point $r$ with
$\operatorname{Im}(r)=0$ and $\operatorname{Re}(r)>0$, and
$\DeltaTwo{R-\ii R}{R+\ii R}=2R^{2}>0$, contributing $+1$; the other three edges
contribute $0$.
\end{proof}

\section{The winding number of the power cycle}\label{sec:power-cycle}

Let $R$ be a positive rational number, and let $M\ge2$ be an even integer.
Set $h\equiv2R/M$ and subdivide the counterclockwise boundary of $\Sq(R)$ into
$4M$ segments, each of quadrance $h^2$.  Let $z_0\equiv R$, let
$z_1,\ldots,z_{4M-1}$ be the remaining successive vertices in counterclockwise
order, and set $z_{4M}\equiv z_0$.

\begin{definition}\label{def:power-cycle}
\textbf{(Power cycle):}
The \emph{power cycle} of degree $n$ for $(z_{0},z_{1},\ldots,z_{4M})$ is
  $P_{n}\equiv(z_{0}^{n},z_{1}^{n},\ldots,z_{4M}^{n}).$
\end{definition}

For a given power cycle $P_{n}\equiv(z_{0}^{n},z_{1}^{n},\ldots,z_{4M}^{n})$, let
$I_{n}\equiv\W(P_{n})$ denote the \emph{winding number} of $P_{n}$ and let
$E_{n}\equiv n\,(2R)^{n-1}\,h$.

\begin{lemma}[Parametric bounds]\label{lem:ppbounds}
For any vertex $z_{j}$ ($0\le j\le4M$):
\[
Q(z_{j})\ge R^{2},\qquad \sigma(z_{j})\le 2R,\qquad
\sigma(z_{j+1}-z_{j})=h\ \ (0\le j<4M).
\]
Consequently, for each $0\le j<4M$,
\[
\sigma\bigl(z_{j+1}^{n}-z_{j}^{n}\bigr)\le E_{n}.
\]
\end{lemma}
\begin{proof}
Since each $z_{j}$ lies on $\partial\Sq(R)$, $Q(z_{j})\ge R^{2}$ and
$\sigma(z_{j})\le 2R$.  The difference $z_{j+1}-z_{j}$ is one of $\pm h$, $\pm h \ii$ so $\sigma (z_{j+1}-z_{j})=h$.  The last bound comes from Lemma~\ref{lem:powvar} with
$B=2R$: $\sigma(z_{j+1}^{n}-z_{j}^{n})\le n(2R)^{n-1}h=E_{n}$.
\end{proof}


\begin{lemma}[Base winding number]\label{lem:I1}
$I_{1}=\W(z_{0},\ldots,z_{4M})=1$.
\end{lemma}
\begin{proof}
The edges on bottom, top, and left sides contribute $0$.
On the right side, exactly two edges have the common endpoint $R$, which
satisfies $\operatorname{Im}(R)=0$ and $\operatorname{Re}(R)=R>0$.
According to the half-open convention, the edge leaving $R$ upward contributes
$+1$, the other one contributes $0$.  Every other segment contributes $0$.  Hence the total is $1$.

\end{proof}

\begin{theorem}[Additivity of the winding number]\label{thm:Iadd}
Let $r,s\ge1$.  Assume the Fineness Condition
$4E_{r+s}^{2}<R^{2(r+s)}$. Then
\[
I_{r+s}=I_{r}+I_{s}.\tag{4.1}
\]
\end{theorem}
\begin{proof}
On the parameter triangle $0\le v\le u\le1$, consider the map
$H(u,v)\equiv\Gamma_{R}(u)^{r}\Gamma_{R}(v)^{s}$.  Here
$\Gamma_R$ is the
counterclockwise rational piecewise-affine parametrization of $\partial\Sq(R)$  starting at $R$,
given explicitly by
\[
\Gamma_R(t)\equiv
\begin{cases}
R+8R t\,\ii,
   &0\le t\le \dfrac18,\\[3pt]
(2R-8R t)+R\ii,
   &\dfrac18< t\le \dfrac38,\\[3pt]
-R+(4R-8R t)\ii,
   &\dfrac38< t\le \dfrac58,\\[3pt]
(-6R+8R t)-R\ii,
   &\dfrac58< t\le \dfrac78,\\[3pt]
R+(8R t-8R)\ii,
   &\dfrac78< t\le1.
\end{cases}
\]

The first and last branches together in the above definition of $\Gamma_R$ parametrize the right side of the square;
the other three branches parametrize the top, left, and bottom sides,
respectively.

Set $N\equiv4M$ and consider the rational parameter grid
\[
\mathcal D_N\equiv
\left\{\left(\frac{i}{N},\frac{j}{N}\right):
0\le j\le i\le N\right\}
\]
in the triangle $0\le v\le u\le1$.  Triangulate it by the oriented
three-point lists
\[
\begin{aligned}
T^+_{i,j}\equiv{}&
\left(
\left(\frac{i}{N},\frac{j}{N}\right),
\left(\frac{i+1}{N},\frac{j}{N}\right),
\left(\frac{i+1}{N},\frac{j+1}{N}\right)
\right),
&&0\le j\le i<N,\\
T^-_{i,j}\equiv{}&
\left(
\left(\frac{i}{N},\frac{j}{N}\right),
\left(\frac{i+1}{N},\frac{j+1}{N}\right),
\left(\frac{i}{N},\frac{j+1}{N}\right)
\right),
&&0\le j<i<N.
\end{aligned}
\]
For any such elementary triangle $T=(A,B,C)$, its \emph{image three-point
list} is $(H(A),H(B),H(C))$, and its \emph{image three-cycle} is
$(H(A),H(B),H(C),H(A))$.

Under $H$, the side $v=0$ maps to $z_0^s\cdot P_r$, the side $u=1$ maps
to $z_0^r\cdot P_s$, and the diagonal $u=v$ maps to $P_{r+s}$.
For every $(u,v)\in\mathcal D_N$, both $\Gamma_R(u)$ and $\Gamma_R(v)$
lie on $\partial\Sq(R)$.  Hence
\[
\sigma(\Gamma_R(u)),\sigma(\Gamma_R(v))\le2R,
\qquad
Q(\Gamma_R(u)),Q(\Gamma_R(v))\ge R^2,
\]
and therefore, by the multiplicativity of quadrance,
\begin{equation}
Q(H(u,v))
=Q(\Gamma_R(u))^rQ(\Gamma_R(v))^s
\ge R^{2(r+s)}.\tag{4.2}\label{eq:H-lower-bound}
\end{equation}

Let $(u,v)$ and $(u',v')$ be any two vertices of one elementary triangle,
and write
\[
x\equiv\Gamma_R(u),\qquad x'\equiv\Gamma_R(u'),\qquad
y\equiv\Gamma_R(v),\qquad y'\equiv\Gamma_R(v').
\]
The two parameter coordinates differ by at most one grid step, so the
corresponding boundary vertices are either equal or consecutive.  Thus
\[
\sigma(x'-x)\le h,\qquad \sigma(y'-y)\le h.
\]
Applying Lemma~\ref{lem:polyvar} to the monomials $X^r$ and $X^s$ gives
\[
\sigma(x'^r-x^r)\le r(2R)^{r-1}h,\qquad
\sigma(y'^s-y^s)\le s(2R)^{s-1}h.
\]
Using
\[
H(u',v')-H(u,v)
=(x'^r-x^r)y'^s+x^r(y'^s-y^s)
\]
together with the triangle and submultiplicative inequalities in
Lemma~\ref{lem:basic}, we obtain
\begin{align*}
\sigma\bigl(H(u',v')-H(u,v)\bigr)
&\le r(2R)^{r-1}h(2R)^s
   +(2R)^r s(2R)^{s-1}h\\
&=(r+s)(2R)^{r+s-1}h
=E_{r+s}.
\end{align*}

Now let $(a,b,c)$ be the image three-point list of any elementary triangle.
Applying the preceding estimate to its first vertex and each of the other two
gives
\[
\sigma(a-b)\le E_{r+s}<2E_{r+s},
\qquad
\sigma(a-c)\le E_{r+s}<2E_{r+s}.
\]
By the Fineness Condition and \eqref{eq:H-lower-bound},
\[
(2E_{r+s})^2<R^{2(r+s)}\le Q(a).
\]
Lemma~\ref{lem:affine-exclusion}, applied with $\eps\equiv2E_{r+s}$,
therefore shows that the origin is not a nonnegative rational affine
combination of $(a,b,c)$.  Hence every corresponding image three-cycle has
winding number $0$ by
Theorem~\ref{thm:properties}(2).  Summing over all image three-cycles and
cancelling the contribution of interior edges gives
\[
0=\W(z_0^s\cdot P_r)+\W(z_0^r\cdot P_s)-\W(P_{r+s})
  =I_r+I_s-I_{r+s}.
\]
The last equality follows from $z_0=R>0$ and
Remark~\ref{rmk:positive-scale}.
\end{proof}


\begin{theorem}[Winding number of the power cycle]\label{thm:In}
If $4E_{n}^{2}<R^{2n}$, then $I_{n}=n$.
\end{theorem}
\begin{proof}
Suppose that $4E_n^2<R^{2n}$.  We first verify the Fineness Condition for
every $1\le r<n$.
Since $h=2R/M$, for each integer $1\le k\le n$ we have
\[
E_k=k(2R)^{k-1}h=\frac{k(2R)^k}{M},
\]
and hence
\[
q_k\equiv\frac{4E_k^2}{R^{2k}}
     =\frac{4k^2\,4^k}{M^2}.
\]
Moreover, for $1\le k<n$,
\[
\frac{q_{k+1}}{q_k}
=4\left(\frac{k+1}{k}\right)^2>1,
\]
so the finite list $q_1,\ldots,q_n$ is strictly increasing.  Hence $q_n<1$ implies
$q_r<q_n<1$, or equivalently $4E_r^2<R^{2r}$, for every $1\le r<n$.

We now prove the result by induction on $n$.  For $n=1$,
Lemma~\ref{lem:I1} gives $I_1=1$.  For $n\ge2$, the induction hypothesis and
the preceding Fineness Condition give $I_{n-1}=n-1$.  Apply
Theorem~\ref{thm:Iadd} with $r=n-1$ and $s=1$.  Its required Fineness
Condition is the assumed condition for $n$, and consequently
\[
I_n=I_{n-1}+I_1=n,
\]
where Lemma~\ref{lem:I1} was used in the last equality.
\end{proof}


\begin{remark}[On the choice of the starting vertex $z_0$]
In Definition~\ref{def:power-cycle} we fixed $z_0 = R$ (the rightmost point on the boundary of the square) for concreteness. This choice is immaterial: the winding number $\W(P)$ of any closed vertex list is invariant under cyclic permutation of the list. Indeed, the sum in $\W(P) = \sum_{i} \chi(p_i,p_{i+1})$ is taken over all oriented edges exactly once; a cyclic shift merely reorders the summands, so the total is unchanged. Therefore, any other boundary vertex $z_k$ ($0 \le k < 4M$) could serve as the starting point, and the resulting power cycle would have the same winding number $I_n$. The half-open convention ensures consistent treatment of the edge that wraps around the chosen starting point, so no special handling is required.
\end{remark}

\begin{remark}[Multiplier comparison on refined boundary lists]
\label{rmk:scale-invariance-triangulation}

Let $(\zeta_0,\ldots,\zeta_N)$ be a \emph{refinement} of
$(z_0,\ldots,z_{4M})$, obtained by inserting finitely many rational points
along its boundary segments, in order, while retaining all original vertices.
Thus $\zeta_0=\zeta_N=R$.  For an integer $n\ge1$, write
\[
P_n^\ast\equiv(\zeta_0^n,\ldots,\zeta_N^n).
\]
Keep $h$ and $E_k=k(2R)^{k-1}h$ from the original subdivision.  The refined
points still satisfy
\[
\sigma(\zeta_j)\le2R,\qquad Q(\zeta_j)\ge R^2,\qquad
\sigma(\zeta_{j+1}-\zeta_j)\le h\quad(0\le j<N).
\]
In particular, the refined list need not be uniformly spaced.

Fix two consecutive vertices $u,v$ of this refined boundary list and assume
$4E_{n+1}^2<R^{2n+2}$.  For $0\le j\le N$, set
\[
A_j\equiv u\zeta_j^n,\qquad B_j\equiv v\zeta_j^n.
\]
For each $0\le j<N$, split the four-point list
$(A_j,B_j,B_{j+1},A_{j+1})$ along $A_jB_{j+1}$ into
\[
(A_j,B_j,B_{j+1}),\qquad (A_j,B_{j+1},A_{j+1}).
\]

By Lemmas~\ref{lem:powvar} and~\ref{lem:basic}, the changes in the multiplier
and in the powered boundary vertex are bounded respectively by
$h(2R)^n$ and $2RE_n=n(2R)^nh$.  Their sum is
\[
E_{n+1}=(n+1)(2R)^nh.
\]
Thus
\[
\sigma(U-A_j)\le E_{n+1}<2E_{n+1}
\qquad
\text{for }U\in\{B_j,B_{j+1},A_{j+1}\}.
\]
Since
\[
Q(A_j)=Q(u)Q(\zeta_j)^n\ge R^{2(n+1)}>(2E_{n+1})^2,
\]
Lemma~\ref{lem:affine-exclusion}, with $\eps\equiv2E_{n+1}$, excludes the
origin from both three-point lists.  Their three-cycles therefore have winding
number $0$ by Theorem~\ref{thm:properties}(2).  Summing over $j$ cancels all
diagonal and interior edges, leaving
\[
0=\W(vP_n^\ast)-\W(uP_n^\ast),
\]
so $\W(uP_n^\ast)=\W(vP_n^\ast)$.
\end{remark}

\begin{theorem}[Conditional scale invariance]\label{thm:scale2}
Let $a\in K^{\times}$, let $R>0$ be rational, let $M\ge2$ be an even
integer, and let $n\ge1$ be an integer.  Assume the fineness
condition $4E_{n+1}^{2}<R^{2n+2}$.  Then
\[
\W\bigl(a\cdot P_{n}\bigr)=I_{n}.
\]
\end{theorem}

\begin{proof}
The assumed Fineness Condition also gives $4E_n^2<R^{2n}$, by the calculation
in the proof of Theorem~\ref{thm:In}.

\medskip

\noindent\textbf{Step 1: refinement preserves winding number.}
Insert a rational point $\xi$ between consecutive boundary vertices $u,v$.
The power-variation estimate gives
\[
\sigma(\xi^n-u^n),\ \sigma(v^n-u^n)\le E_n,
\qquad Q(u^n)\ge R^{2n}>4E_n^2.
\]
Lemma~\ref{lem:affine-exclusion}, with precision $2E_n$, therefore excludes
the origin from the nonnegative rational affine combinations of
$(u^n,\xi^n,v^n)$.  For any $b\in K^\times$, the same holds for
$(bu^n,b\xi^n,bv^n)$, since an affine expression for zero could be divided
by $b$.  Theorem~\ref{thm:properties}(1)--(2) gives
\[
\chi(bu^n,b\xi^n)+\chi(b\xi^n,bv^n)=\chi(bu^n,bv^n).
\]
Thus repeating the old vertex $bu^n$ and then replacing the repeated entry
by $b\xi^n$ preserves winding number.  Every subsequent insertion satisfies
the same bounds, so any finite refinement obeys
\[
\W(bP_n^\ast)=\W(bP_n)\qquad(b\in K^\times).
\]

\medskip

\noindent\textbf{Step 2: compare multipliers on the refined list.}
Write $a=x+\ii y$ with $x,y\in\Q$, and set
\[
t\equiv\frac{R}{\max\{|x|,|y|\}},\qquad \widetilde a\equiv ta.
\]
Then $t$ is a positive rational and $\widetilde a\in\partial\Sq(R)$.
Insert $\widetilde a$ into the original boundary list if necessary, and
denote the resulting refined power cycle by $P_n^\ast$.
Remark~\ref{rmk:scale-invariance-triangulation}, applied successively along
the refined boundary from $R$ to $\widetilde a$, gives
\[
\W(\widetilde aP_n^\ast)=\W(RP_n^\ast).
\]
Together with positive scale invariance (Remark~\ref{rmk:positive-scale})
and Step 1, this yields
\[
\begin{aligned}
\W(aP_n)
&=\W(\widetilde aP_n)=\W(\widetilde aP_n^\ast)\\
&=\W(RP_n^\ast)=\W(P_n^\ast)=\W(P_n)=I_n.
\end{aligned}
\]
\end{proof}


\section{The existence of suitable radius and mesh resolution}\label{sec:radius}

Throughout, $f(X)=\sum_{j=0}^{n}a_{j}X^{j}\in K[X]$ with $n\ge1$, $a_{n}\ne0$,
and a rational $\eps>0$.  Let $A_{j}\equiv\sigma(a_{j})$ and

\[
C\equiv\sum_{j=0}^{n-1}2^{j}A_{j},\qquad
G\equiv CR^{n-1},\qquad
m\equiv Q(a_{n})\,R^{2n},\qquad
L\equiv\sum_{j=1}^{n}j\,A_{j}(2R)^{j-1}.
\]

% \begin{definition}[Radius data]\label{def:radiusdata}
% Define
% \[
% C\equiv\sum_{j=0}^{n-1}2^{j}A_{j},\qquad
% G\equiv CR^{n-1},\qquad
% m\equiv Q(a_{n})\,R^{2n},\qquad
% L\equiv\sum_{j=1}^{n}j\,A_{j}(2R)^{j-1}.
% \]

% \end{definition}

\begin{lemma}[Choice of $R$]\label{lem:existsR}
There is an integer $R\ge1$ such that
\[
Q(a_{n})\,R^{2}>2\,C^{2}.\tag{5.1}\label{eq:radius-condition}
\]
For any such $R$ one also has $2G^{2}<m$.
\end{lemma}
\begin{proof}
Set $\nu\equiv Q(a_n)>0$ and choose
\[
R\equiv1+\left\lceil\frac{2C^2}{\nu}\right\rceil.
\]
The ceiling of a rational number is computed by integer division of its
numerator by its positive denominator, with a remainder check.
Thus $R$ is an explicitly determined integer with $R\ge1$ and
$R^2\ge R>2C^2/\nu$, proving \eqref{eq:radius-condition}.
Multiplying \eqref{eq:radius-condition} by the
positive factor $R^{2n-2}$ gives $2G^{2}<m$.
\end{proof}

\begin{lemma}[Choice of $M$]\label{lem:existsM}
Let $R\ge1$ be an integer satisfying \eqref{eq:radius-condition}, whose
existence is guaranteed by Lemma~\ref{lem:existsR}.  Then there exists an
explicitly determined even integer $M\ge2$ such that, writing $h\equiv2R/M$ and
$D\equiv A_{n}E_n$, all three
strict rational number inequalities hold:
\[
2hL<\eps,\qquad (D+G)^{2}<m,\qquad
4E_{n+1}^{2}<R^{2n+2}.\tag{5.2}
\]
The last inequality also implies $4E_n^2<R^{2n}$.
\end{lemma}
\begin{proof}
For $k=n,n+1$, write
\[
e_k\equiv2Rk(2R)^{k-1},
\]
and set $d_{0}\equiv A_n e_n$.  Thus
$E_k=e_k/M$ for $k=n,n+1$, and $D=d_{0}/M$.
By Lemma~\ref{lem:existsR},
$\delta\equiv m-G^{2}>0$.  Since
\[
\frac{4E_{n+1}^2}{R^{2n+2}}
=\frac{4(n+1)^2\,4^{n+1}}{M^2},
\]
the degree-$(n+1)$ Fineness Condition is equivalent to
$M>2(n+1)2^{n+1}$.  Define
\[
\begin{aligned}
B_M&\equiv\max\left\{
1,\ \frac{4RL}{\eps},\ \frac{4Gd_0}{\delta},
\frac{2d_0^2}{\delta},\ 2(n+1)2^{n+1}
\right\},\\
M&\equiv2\left(1+\left\lfloor\frac{B_M}{2}\right\rfloor\right).
\end{aligned}
\]
These choices require only finitely many rational computations and comparisons
and an integer division.  The integer $M$ is even, $M\ge2$, and $M>B_M$.
In particular, $M^2\ge M>B_M\ge2d_0^2/\delta$, so all four inequalities
\[
\frac{4RL}{M}<\eps,\qquad
\frac{2Gd_{0}}{M}<\frac{\delta}{2},\qquad
\frac{d_{0}^{2}}{M^{2}}<\frac{\delta}{2},\qquad
\frac{4e_{n+1}^{2}}{M^{2}}<R^{2n+2}
\]
hold.  The first is exactly $2hL<\eps$, the fourth gives
$4E_{n+1}^{2}<R^{2n+2}$, and
\[
(D+G)^{2}=G^{2}+\frac{2Gd_{0}}{M}+\frac{d_{0}^{2}}{M^{2}}
<G^{2}+\frac{\delta}{2}+\frac{\delta}{2}=m.
\]
Finally, the proof of Theorem~\ref{thm:In} shows that
\[
q_k\equiv\frac{4E_k^2}{R^{2k}}
\]
is strictly increasing in $k$.  Hence $q_{n+1}<1$ implies $q_n<1$, which
is precisely $4E_n^2<R^{2n}$.
\end{proof}
% \begin{remark}\label{rmk:mesh}
% The three inequalities in (5.2) play distinct roles in \S\ref{sec:grid}: the
% first excludes $0$ from the affine combinations of image three-point lists;
% the second does the same for the boundary comparison strip; the third is
% the Fineness Condition needed for conditional scale invariance and also
% implies the degree-$n$ Fineness Condition for the power cycle.
% \end{remark}

\section{The existence of a good grid vertex}\label{sec:grid}

Choose $R,M,h$ as in \S\ref{sec:radius}; in particular,
$4E_{n+1}^2<R^{2n+2}$.  Consider the finite rational grid
\[
\mathcal{G}\equiv\{-R+ih+\ii(-R+jh):0\le i,j\le M\},
\]
listed lexicographically (the $i$-coordinate outer, $j$ inner); it has exactly
$(M+1)^{2}$ points.  As in \S\ref{sec:power-cycle}, let
$\partial\mathcal{G}$ denote the counterclockwise closed list of boundary
grid vertices, whose successive vertices subdivide $\partial\Sq(R)$ into
$4M$ segments, each of quadrance $h^2$.  Each of the $M^{2}$ cells is split by its positive
diagonal into two triangles.

\begin{theorem}[Existence of a good grid vertex]\label{thm:grid}
There exists $z\in\mathcal{G}$ with
\[
Q\bigl(f(z)\bigr)<\eps^{2}.
\]
\end{theorem}
\begin{proof}

Suppose, for contradiction, that
\[
Q(f(z))\ge\eps^{2}\qquad\text{for all }z\in\mathcal{G}.\tag{6.1}
\]

\emph{(a) Image three-point lists exclude $0$.}  For two vertices $z,w$ of one grid
triangle, $\sigma(w-z)\le2h$; since all grid points lie in $\mathcal{G}$, each grid vertex has real and imaginary parts bounded by $R$ in absolute value, so for vertices $z,w$ we have
$|\Re(z)|,|\Im(z)|,|\Re(w)|,|\Im(w)|\le R$.
Lemma~\ref{lem:polyvar} and (5.2) give, for vertices $z,w$ of the same
triangle,
\[
\sigma\bigl(f(w)-f(z)\bigr)\le \sigma(w-z)\,\sum_{j=1}^{n}j\,A_{j}(2R)^{j-1}=\sigma(w-z)\,L\le 2hL<\eps.\tag{6.2}
\]
Let $p_{0}\equiv f(z_{0}),p_{1}\equiv f(z_{1}),p_{2}\equiv f(z_{2})$ be the
three image values of a grid triangle.  If the origin were a nonnegative
rational affine combination of $(p_0,p_1,p_2)$, there would exist nonnegative
rationals $\theta_{0},\theta_{1},\theta_{2}$ with $\theta_{0}+\theta_{1}+\theta_{2}=1$
and $0=\theta_{0}p_{0}+\theta_{1}p_{1}+\theta_{2}p_{2}$.  Then
\[
p_{0}=\theta_{1}(p_{0}-p_{1})+\theta_{2}(p_{0}-p_{2}),
\]
so by (6.2) and the affine-exclusion argument (Lemma~\ref{lem:affine-exclusion})
$\sigma(p_{0})<\eps$, hence $Q(p_{0})\le\sigma(p_{0})^{2}<\eps^{2}$, contradicting
(6.1).  Thus no image three-point list admits such an affine combination of
the origin, and by Theorem~\ref{thm:properties}(2) each image three-cycle has
winding number $0$:
$\W(f(z_{0}),f(z_{1}),f(z_{2}),f(z_{0}))=0$.

\emph{(b) Edge cancellation.}  Summing the winding numbers of all $2M^{2}$
image three-cycles: every interior edge of the triangulation occurs twice, in
opposite orientations, and cancels by Theorem~\ref{thm:properties}(1).  What
remains is the image of the outer boundary:
\[
\W\bigl(f(\partial\mathcal{G})\bigr)
=\W\bigl(f((z_j)_{j=0,1,2,...,4M})\bigr)=0.\tag{6.3}
\]

\emph{(c) The boundary winding number is the leading-term winding number.}  Write
$f(z)=l(z)+g(z)$ with $l(z)\equiv a_{n}z^{n}$ and
$g(z)\equiv\sum_{j<n}a_{j}z^{j}$.  At a boundary grid vertex, $\sigma(z)\le2R$,
$Q(z)\ge R^{2}$, and $R\ge1$; hence
\[
\sigma(g(z))\le\sum_{j<n}A_{j}(2R)^{j}\le CR^{n-1}=G,\qquad
Q(l(z))\ge Q(a_{n})\,R^{2n}=m.\tag{6.4}
\]
 For
consecutive boundary vertices $z,z'$ (with $\sigma(z'-z)=h$), the power estimate
gives
\[
\sigma\bigl(l(z')-l(z)\bigr)\le D.\tag{6.5}
\] Join the leading-term closed vertex list
$(l(z_{j}))_{j}$ to the full-value closed vertex list $(l(z_{j})+g(z_{j}))_{j}$ by the
closed four-vertex list
$(l(z_{j}),\ f(z_{j}),\ f(z_{j+1}),\ l(z_{j+1}),\ l(z_j))$,
and split this list along a diagonal into the two closed three-cycles
\[
\bigl(l(z_{j}),\,f(z_{j}),\,f(z_{j+1}),\,l(z_j)\bigr),\qquad
\bigl(l(z_{j}),\,f(z_{j+1}),\,l(z_{j+1}),\,l(z_j)\bigr).
\]
Every vertex $u$ of either three-cycle satisfies
$\sigma(u-l(z_j))\le D+G$ (using (6.4)--(6.5)).  If the origin were a
nonnegative rational affine combination of either determining three-point
list, the same argument as
in Lemma~\ref{lem:affine-exclusion} would give
$Q(l(z_{j}))\le(D+G)^{2}<m$, contradicting (6.4).  Since neither list admits such a
combination, each three-cycle has winding number $0$, again finite
edge cancellation yields
\[
\W\bigl(f(\partial\mathcal{G})\bigr)
=\W\bigl((a_{n}z_{j}^{n})_{j=0,1,2,...,4M}\bigr).\tag{6.6}
\]


\emph{(d) The boundary winding number is $n$.}
Lemma~\ref{lem:existsM} also gives the degree-$n$ Fineness Condition.
Therefore, by Theorems~\ref{thm:In} and~\ref{thm:scale2},
\[
\W\bigl((a_{n}z_{j}^{n})_{j=0,1,2,...,4M}\bigr)
=\W\bigl(a_{n}\cdot P_{n}\bigr)=I_{n}=n.\tag{6.7}
\]

Equations (6.3) and (6.7) give $0=n$, contradicting $n\ge1$.  Hence (6.1) is
false, i.e.\ some grid vertex satisfies $Q(f(z))<\eps^{2}$.
\end{proof}

\section{Proof of main theorem}\label{sec:main}

\begin{theorem}[Main theorem]\label{thm:main}
For any nonconstant polynomial $f$ with rational complex number coefficients and every rational number $\eps>0$,
there exists a rational complex number $z$ such that
\[
Q\bigl(f(z)\bigr)<\eps^{2}.
\]
\end{theorem}
\begin{proof}
Write $f(X)=\sum_{j=0}^{n}a_{j}X^{j}$ with $n\ge1$, $a_{n}\ne0$.  Let
$A_{j}\equiv\sigma(a_{j})$ and $\nu\equiv Q(a_{n})>0$.  By Lemmas~\ref{lem:existsR} and
\ref{lem:existsM} we obtain an integer $R\ge1$ and an even integer $M\ge2$
satisfying the three strict inequalities (5.2).  Form the finite grid
$\mathcal{G}$ and apply Theorem~\ref{thm:grid}
: there exists some
$z\in\mathcal{G}$ satisfying $Q(f(z))<\eps^{2}$.
\end{proof}


The proof of theorem is also constructive with an algorithm for finding an approximate zero: construct
the finite grid $\mathcal{G}$ used in its proof and test its vertices one by one
until finding a vertex $z$ such that $Q(f(z))<\eps^{2}$. The theorem guarantees that the algorithm will terminate after finitely many steps. This exhaustive search is
generally slow; finding an elementary proof with more effective algorithm is left for future work.
\medskip

\begin{remark}
Given a positive integer $k$, set $k_1\equiv k+1$ and
$\eps\equiv1/k_1>0$.  Since $(k+1)^2>k$, one has
$1/k_1^2<1/k$.  By the main theorem
(Theorem~\ref{thm:main}), there exists a rational complex number $z$ such that
\[
Q(f(z))< \frac{1}{k_1^2} < \frac{1}{k}.
\]
This is one of Wildberger's formulation of the Fundamental Theorem of Algebra over
Rationals, which he named FTA I.
\end{remark}

\section{Complete approximate factorization over Rationals}\label{sec:complete}

We now establish a complete approximate factorization theorem: every
nonconstant polynomial over $K$ can be approximated, up to any prescribed
precision, by a monic (up to its leading coefficient) product of linear
factors over $K$.

\subsection{Polynomial coefficient quadrance}

For a polynomial
\[
P(X)=\sum_{j=0}^{m} c_j X^j \in K[X],
\]
define its \emph{coefficient quadrance} by
\[
\CoeffQuad{P}\equiv\max_{0\le j\le m} Q(c_j),
\qquad \CoeffQuad{0}\equiv0.
\]
For $P,H\in K[X]$, the elementary inequality
$Q(u+v)\le 2(Q(u)+Q(v))$ gives
\[
\CoeffQuad{P+H}\le 2\bigl(\CoeffQuad{P}+\CoeffQuad{H}\bigr)
\]
and, when multiplied by a linear factor,
\[
\CoeffQuad{(X-z)P} \le 2(1+Q(z))\,\CoeffQuad{P} \qquad (z\in K),
\]
because the coefficients of $(X-z)P$ are of the form $c_{j-1}-zc_j$, whose
quadrance is bounded by
$2\bigl(Q(c_{j-1})+Q(z)Q(c_j)\bigr)$.

The main goal of this section is to prove the following theorem.

\begin{theorem}[Complete approximate factorization over Rationals]\label{thm:complete}
Let $f\in K[X]$ be nonconstant with leading coefficient $a_n\neq 0$ and degree
$n\ge 1$.  For any rational $\eps>0$ there exist
$z_1,\dots,z_n\in K$ such that
\[
\CoeffQuad{f-a_n\prod_{i=1}^{n}(X-z_i)}<\eps.
\]
\end{theorem}


\subsection{Proof of Theorem~\ref{thm:complete}}

We prove the theorem by induction on $n\ge 1$.

\paragraph{Base case $n=1$.}
Write $f(X)=a_1 X+a_0$ with $a_1\neq 0$.  Take $z_1=-a_0/a_1 \in K$.  Then
\[
f(X)-a_1(X-z_1)=0,
\]
so the desired inequality holds with any $\eps>0$.

\paragraph{Inductive step.}
Assume the theorem holds for all polynomials of degree $<n$ (with $n\ge 2$).
Let $f$ be as above and fix $\eps>0$.



Set
\[
\eta\equiv\frac{\min(1,\eps)}{4},
\]
so that $\eta^2<\eps/4$.  Apply the main theorem
(Theorem~\ref{thm:main}) to $f$ with precision $\eta$: there exists
$z_1\in K$ such that
\[
Q\bigl(f(z_1)\bigr)<\eta^2<\eps/4.
\]
Set
\[
T\equiv\left\lceil Q(z_1)\right\rceil,
\]
so that $T$ is a nonnegative integer and $Q(z_1)\le T$.

Let $r\equiv f(z_1)$. By our assumption,
\[
Q(r)<\frac{\eps}{4} \tag{9.1}
\].  By polynomial division by the monic linear factor $X-z_1$, we obtain
\[
f(X)=(X-z_1)\,g(X)+r,
\]
where $g\in K[X]$ has degree $n-1$ and leading coefficient $a_n$.
Set
\[
\eps_1 \equiv \frac{\eps}{8(T+1)}.
\]
Since $n-1\ge 1$ (because $n\ge 2$), the induction hypothesis applies to $g$: there exist $z_2,\dots,z_n\in K$ such that
\[
\CoeffQuad{g-a_n\prod_{i=2}^{n}(X-z_i)}<\eps_1. \tag{9.2}
\]
Define the remainder polynomial
\[
h(X)\equiv g(X)-a_n \prod_{i=2}^{n}(X-z_i)=\sum_{j=0}^{n-1} b_j X^j,
\]
thus $Q(b_j)<\eps_1$ for every $j=0,1,\dots,n-1$.

Now compute the global error:
\[
\begin{aligned}
f(X)-a_n\prod_{i=1}^{n}(X-z_i)
&= (X-z_1)g(X)+r - (X-z_1)a_n\prod_{i=2}^{n}(X-z_i) \\
&= (X-z_1)h(X)+r.
\end{aligned}
\]
Let
\[
(X-z_1)h(X) = \sum_{j=0}^{n} c_j X^j,
\]
where
\[
c_0 = -z_1 b_0,\qquad
c_j = b_{j-1}-z_1 b_j \quad (1\le j\le n-1),\qquad
c_n = b_{n-1}.
\]
Thus the coefficients of the polynomial $f(X)-a_n\prod_{i=1}^{n}(X-z_i)$ are
\[
c_0+r,\quad c_1,\ \dots,\ c_{n-1},\quad c_n.
\]

We next estimate their quadrances.  For $1\le j\le n-1$,
\[
Q(c_j) \le 2\bigl(Q(b_{j-1})+Q(z_1)Q(b_j)\bigr)
<2(1+T)\eps_1=\frac{\eps}{4}<\eps.
\]
For the leading coefficient,
\[
Q(c_n)=Q(b_{n-1})<\eps_1<\eps.
\]
For the constant term,
\[
Q(c_0+r) \le 2\bigl(Q(r)+Q(z_1)Q(b_0)\bigr)
< \frac{\eps}{2}+2T\eps_1
= \frac{\eps}{2}+\frac{T\eps}{4(T+1)}
< \frac{3\eps}{4}<\eps,
\]
by using (9.1) and $Q(b_0)<\eps_1$.

Therefore every coefficient of the error polynomial has quadrance strictly
less than $\eps$, i.e.
\[
\CoeffQuad{f-a_n\prod_{i=1}^{n}(X-z_i)}<\eps.
\]
This completes the induction and proves the theorem.

\begin{proposition}[Wildberger's FTA II]\label{prop:wildberger-fta-ii}
Let $f\in K[X]$ be nonconstant with leading coefficient $a_n\ne0$ and degree
$n\ge1$.  For any integer $k\ge1$, there exist $z_1,\dots,z_n\in K$ such
that, if
\[
f(X)-a_n\prod_{i=1}^{n}(X-z_i)=\sum_{j=0}^{n}e_jX^j,
\]
then
\[
\sum_{j=0}^{n}Q(e_j)<\frac{1}{k}.
\]
\end{proposition}

\begin{proof}
Apply Theorem~\ref{thm:complete} with $\eps=1/(k(n+1))$.  Then
$Q(e_j)<\eps$ for every $j$, and hence
\[
\sum_{j=0}^{n}Q(e_j)<(n+1)\eps=\frac{1}{k}.
\]
\end{proof}

\begin{remark}
Theorem~\ref{thm:main} and Proposition~\ref{prop:wildberger-fta-ii} are,
respectively, Wildberger's two formulations of the Fundamental Theorem of
Algebra over Rationals.
\end{remark}

\section{Complete approximate real factorization over Rationals}
\label{sec:real-complete}

Suppose that the imaginary part of every coefficient $a_i$ of a polynomial
$f$ is zero.  Theorem~\ref{thm:complete} does not by itself give an
approximation by rational real linear and quadratic factors.  In this section
we prove such a result directly.

For
\[
P(X)=\sum_{j=0}^{m}c_jX^j\in\Q[X],
\]
define its \emph{coefficient square-sum} by
\[
\mathcal S(P)\equiv\sum_{j=0}^{m}c_j^2,
\qquad \mathcal S(0)\equiv0.
\]
Coefficients outside the displayed degree range are understood to be zero.

\begin{lemma}[Coefficient square-sum estimates]
\label{lem:real-coeff-square}
For $P,H,F\in\Q[X]$, with $\deg F=\ell$, one has
\[
\mathcal S(P+H)\le2\bigl(\mathcal S(P)+\mathcal S(H)\bigr)
\tag{10.1}\label{eq:real-sum-add}
\]
and
\[
\mathcal S(FP)\le(\ell+1)\mathcal S(F)\mathcal S(P).
\tag{10.2}\label{eq:real-sum-mul}
\]
\end{lemma}
\begin{proof}
The first inequality follows, coefficient by coefficient, from
$(x+y)^2\le2(x^2+y^2)$.  Write
$F(X)=\sum_{i=0}^{\ell}b_iX^i$ and
$P(X)=\sum_{j=0}^{m}c_jX^j$, where $m\ge0$.  For $0\le k\le m+\ell$,
the coefficient of $X^k$ in $FP$ is
$\sum_{i=0}^{\ell}b_ic_{k-i}$.  The finite rational inequality
\[
\left(\sum_{i=0}^{\ell}x_i\right)^2
\le(\ell+1)\sum_{i=0}^{\ell}x_i^2
\]
therefore gives
\[
\begin{aligned}
\mathcal S(FP)
&\le(\ell+1)\sum_{k=0}^{m+\ell}\sum_{i=0}^{\ell}b_i^2c_{k-i}^2 \\
&=(\ell+1)\left(\sum_{i=0}^{\ell}b_i^2\right)
                    \left(\sum_{j=0}^{m}c_j^2\right),
\end{aligned}
\]
which is \eqref{eq:real-sum-mul}.
\end{proof}

The next lemma is the conjugation-symmetric replacement for extracting one
approximate complex linear factor.

\begin{lemma}[Approximate real divisor]
\label{lem:approx-real-divisor}
Let
\[
f(X)=\sum_{j=0}^{n}a_jX^j\in\Q[X],
\qquad n\ge2,\qquad a_n\ne0,
\]
and let $\eps>0$ be rational.  There exist polynomials
$d,g,e\in\Q[X]$ such that
\[
f=dg+e,
\tag{10.3}\label{eq:approx-real-division}
\]
with the following properties:
\begin{enumerate}[label=\textup{(\roman*)}]
\item $d$ is monic of degree $\ell\in\{1,2\}$, while $g$ has degree
      $n-\ell$ and leading coefficient $a_n$;
\item Either $e=0$ or $\deg e<\ell$, and every coefficient of $e$ has absolute value
      strictly less than $\eps$;
\item if $\ell=1$, then $d=X-q_1$ for some $q_1\in\Q$; if $\ell=2$, then
      \[
      d=X^2-2q_1X+(q_1^2+q_2^2)
      \]
      for some $q_1,q_2\in\Q$ with $q_2\ne0$.  In the latter case the
      discriminant of $d$ is $-4q_2^2<0$.
\end{enumerate}
\end{lemma}
\begin{proof}
For $0\le j\le n$, set
\[
A_j\equiv\sigma(a_j)=|a_j|.
\]
As in Section~\ref{sec:radius}, define
\[
C\equiv\sum_{j=0}^{n-1}2^jA_j.
\]
Choose an integer $R\ge1$ satisfying
\[
Q(a_n)R^2>2C^2.
\tag{10.4}\label{eq:real-divisor-radius}
\]
This is precisely the radius condition of Lemma~\ref{lem:existsR}, applied
to $f$.  In particular, $R$ depends on $f$ but not on the precision used
below.  Define
\[
L\equiv\sum_{j=1}^{n}jA_j(2R)^{j-1}>0
\]
and the positive rationals
\[
\delta\equiv\min\left(1,\frac{\eps}{4L}\right),
\qquad
\eps_1\equiv\min\left(\frac{\eps}{8},
               \frac{\eps\delta}{4(\delta+R)}\right).
\tag{10.5}\label{eq:real-divisor-parameters}
\]
Thus
\[
L\delta\le\frac{\eps}{4},\qquad
2\eps_1\le\frac{\eps}{4},\qquad
\eps_1\frac{\delta+R}{\delta}\le\frac{\eps}{4}.
\tag{10.6}\label{eq:real-divisor-bounds}
\]

Use Lemma~\ref{lem:existsM}, with the fixed radius $R$ and precision
$\eps_1$, to construct the finite grid in Theorem~\ref{thm:grid}.  That
theorem supplies a grid vertex
\[
z=q_1+\ii q_2,\qquad q_1,q_2\in\Q,\qquad |q_1|,|q_2|\le R,
\]
such that
\[
Q(f(z))<\eps_1^2.
\tag{10.7}\label{eq:real-divisor-small-value}
\]
Write $f(z)=\alpha+\ii\beta$.  Since all quantities are rational,
\eqref{eq:real-divisor-small-value} implies directly that
\[
|\alpha|<\eps_1,\qquad |\beta|<\eps_1,
\qquad \sigma(f(z))<2\eps_1.
\tag{10.8}\label{eq:real-divisor-components}
\]

Suppose first that $|q_2|\le\delta$.  Both $z$ and $q_1$ have size at most
$2R$.  Lemma~\ref{lem:polyvar} and
\eqref{eq:real-divisor-bounds} give
\[
\begin{aligned}
|f(q_1)|
&\le \sigma\bigl(f(q_1)-f(z)\bigr)+\sigma(f(z)) \\
&\le |q_2|L+\sigma(f(z))
 <\delta L+2\eps_1
 \le\frac{\eps}{2}<\eps.
\end{aligned}
\tag{10.9}\label{eq:near-real-remainder}
\]
Division by $X-q_1$ gives
\[
f=(X-q_1)g+f(q_1),
\]
which has all the required properties with $\ell=1$.

Suppose instead that $|q_2|>\delta$.  Set
\[
d(X)\equiv(X-z)(X-\overline z)
=X^2-2q_1X+(q_1^2+q_2^2)\in\Q[X].
\]
Polynomial division in $\Q[X]$ gives
\[
f(X)=d(X)g(X)+uX+v
\tag{10.10}\label{eq:quadratic-real-division}
\]
for some $g\in\Q[X]$ of degree $n-2$ and some $u,v\in\Q$.
Since $d(z)=0$, evaluation at $z$ gives
\[
\alpha+\ii\beta=f(z)=uz+v=(uq_1+v)+\ii uq_2.
\]
Consequently, using $|q_2|>\delta$, $|q_1|\le R$, and
\eqref{eq:real-divisor-bounds},
\[
|u|=\frac{|\beta|}{|q_2|}<\frac{\eps_1}{\delta}<\eps
\tag{10.11}\label{eq:quadratic-u-bound}
\]
and
\[
|v|=|\alpha-q_1u|
<\eps_1+\frac{R\eps_1}{\delta}
=\eps_1\frac{\delta+R}{\delta}
\le\frac{\eps}{4}<\eps.
\tag{10.12}\label{eq:quadratic-v-bound}
\]
Finally,
\[
\operatorname{disc}(d)=(-2q_1)^2-4(q_1^2+q_2^2)=-4q_2^2<0.
\]
Thus \eqref{eq:quadratic-real-division} has all the required properties
with $\ell=2$.  Both cases use only a finite grid search, rational
comparisons, and rational polynomial division.
\end{proof}

\begin{theorem}[Complete approximate real factorization over Rationals]
\label{thm:complete-real}
Let
\[
f(X)=\sum_{j=0}^{n}a_jX^j\in\Q[X],
\qquad n\ge1,\qquad a_n\ne0.
\]
For every rational $\eps>0$, there exist nonnegative integers $s,t$ and
rational numbers
\[
r_1,\ldots,r_s,\qquad
q_{1,1},q_{1,2},\ldots,q_{t,1},q_{t,2}
\]
such that
\[
s+2t=n,\qquad q_{j,2}\ne0\quad(1\le j\le t),
\tag{10.13}\label{eq:real-factor-degree}
\]
and
\[
\mathcal S\left(
f-a_n\prod_{i=1}^{s}(X-r_i)
       \prod_{j=1}^{t}
       \bigl(X^2-2q_{j,1}X+(q_{j,1}^2+q_{j,2}^2)\bigr)
\right)<\eps.
\tag{10.14}\label{eq:real-factor-approximation}
\]
\end{theorem}
\begin{proof}
We prove the slightly stronger assertion allowing degree $n=0$, with both
products then empty.  The proof is by induction on $n$.

If $n=0$, the empty product is $1$ by default, and the error is zero.

If $n=1$, write
$f=a_1X+a_0$ and take $r_1=-a_0/a_1$; again the factorization is exact.

Now let $n\ge2$ and assume the assertion for all smaller degrees.  Set
\[
\eps_1\equiv\frac{\min(1,\eps)}{4},
\]
so that
\[
8\eps_1^2<\eps.
\tag{10.15}\label{eq:real-factor-eps-one}
\]
Apply Lemma~\ref{lem:approx-real-divisor} to $f$ and $\eps_1$.  We obtain
\[
f=dg+e,
\tag{10.16}\label{eq:real-factor-first-step}
\]
where $d$ has degree $\ell\in\{1,2\}$,
$g$ has degree $n-\ell$ and its leading coefficient is $a_n$, and
\[
\mathcal S(e)<2\eps_1^2.
\tag{10.17}\label{eq:real-factor-local-error}
\]
The last inequality holds because $e$ has at most two coefficients and each
has absolute value less than $\eps_1$.

Define the positive rational number
\[
C_d\equiv(\ell+1)\mathcal S(d)>0,
\qquad
\eps_2\equiv\frac{\eps}{4C_d}>0.
\]
By the induction hypothesis, there is a monic product $P$ of allowed rational
linear and negative-discriminant quadratic factors, of total degree
$n-\ell$, such that
\[
\mathcal S(g-a_nP)<\eps_2.
\tag{10.18}\label{eq:real-factor-induction-error}
\]

Set $E\equiv g-a_nP$.  Equations
\eqref{eq:real-sum-add}, \eqref{eq:real-sum-mul},
\eqref{eq:real-factor-eps-one}, \eqref{eq:real-factor-first-step},
\eqref{eq:real-factor-local-error}, and
\eqref{eq:real-factor-induction-error} give
\[
\begin{aligned}
\mathcal S(f-a_ndP)
&=\mathcal S(dE+e) \\
&\le2\bigl(\mathcal S(dE)+\mathcal S(e)\bigr) \\
&\le2\bigl(C_d\mathcal S(E)+\mathcal S(e)\bigr) \\
&<2\left(C_d\frac{\eps}{4C_d}+2\eps_1^2\right)
=\frac{\eps}{2}+4\eps_1^2
<\eps.
\end{aligned}
\]
Adjoining $d$ to the factors of $P$ preserves the required factor forms,
and their total degree is $\ell+(n-\ell)=n$.  In the quadratic case,
Lemma~\ref{lem:approx-real-divisor} gives the strict negative-discriminant
condition.  This completes the induction.
\end{proof}

\begin{corollary}[Real factorization]
\label{cor:real-factor-distance}
Under the hypotheses of Theorem~\ref{thm:complete-real}, for every integer
$k\ge1$ there is a product of the form
\[
G(X)=a_n\prod_{i=1}^{s}(X-r_i)
          \prod_{j=1}^{t}
          \bigl(X^2-2q_{j,1}X+(q_{j,1}^2+q_{j,2}^2)\bigr),
\]
with all displayed parameters rational, $s+2t=n$, and $q_{j,2}\ne0$ for
$1\le j\le t$, such that, if
\[
f(X)-G(X)=\sum_{j=0}^{n}e_jX^j,
\]
then
\[
\sum_{j=0}^{n}e_j^2<\frac1k.
\]
\end{corollary}
\begin{proof}
Apply Theorem~\ref{thm:complete-real} with $\eps=1/k$.
\end{proof}

\begin{remark}[Finite character of the construction]
The constructions in Sections~\ref{sec:complete} and~\ref{sec:real-complete}
are finite.  At each inductive step, one performs a finite grid search followed
by rational polynomial division, and the degree drops by at least one.
Therefore the induction terminates after at most $n$ steps.  For each
prescribed rational precision, Theorems~\ref{thm:complete}
and~\ref{thm:complete-real} construct finite lists of rational coefficients
using only rational number computations and comparisons.
\end{remark}

\section*{Acknowledgments}

We acknowledge the use of the Math AI agents Rethlas and Archon (developed by Bin Dong’s group at Peking University), utilizing GPT-5.6-sol and Deepseek-V4-Flash, to assist in reasoning and verification for portions of this work. All outputs were independently evaluated and refined by the authors, who retain full responsibility for the results.

The second author respectfully dedicates this paper to the memory of his former advisor, Professor Xinghua Wang, for guiding his entrance into computational mathematics~\cite{WangXuan1987} and for his lasting friendship.

We are also grateful to Norman Wildberger for proofreading the manuscript.

This work is partially supported by Smale Institute.
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\end{document}