
A Criterion for the Reducibility of a Class of Integer Polynomials over the Field of Rational Numbers
 ZHAO Shizhong, FU Hongguang, QIN XiaoLin, LIU Jing, LIU YunHao

2021, 41(12):
33513362.
DOI: 10.12341/jssms21438

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For the integer polynomials $\sum_{i=0}^m a_i\,x^i$
of degree $m$, where $a_m=1$, this paper presents a fixedpoint iterative algorithm:
%and the corresponding fixed point iterative algorithm
$$\left\{
\begin{array}{ll}
u_1=\tilde{u}_1, & \\
u_2=\tilde{u}_2, & \\
\quad\,\,\,\vdots& \\
u_{m1}=\tilde{u}_{m1}, & \\
\displaystyle{u_n=\Big{(}a_{m1}+\frac{a_{m2}}{u_{n1}}+\frac{a_{m3}}{u_{n1}u_{n2}}+\cdots+\frac{a_{0}}{u_{n1}u_{n2}\cdots u_{n(m1)}}\Big{)}\,\,(n\geq m).} &
%\displaystyle{u_n=\frac{a_{m1}}{a_m}\frac{a_{m2}}{a_mu_{n1}}\frac{a_{m3}}{a_mu_{n1}u_{n2}}\cdots\frac{a_{0}}{a_mu_{n1}u_{n2}\cdots u_{n(m1)}}\,\,(n\geq m).} &
\end{array}
\right.
$$
1) Obviously, if the iteration has a rational limit value, then the value is a zero of the polynomial, so that the polynomial is reducible over
$\mathbb{Q}$.
2) This iteration does not need to choose the initial values:
If the polynomial has $m$ rational zeros with different absolute values,
then for any $m1$ nonzero rational initial values $u_i\,(1\leq i\leq m1)$, the iteration approaches to one of the zeros. Therefore, the polynomial is reducible.
3) Assuming that $\{\zeta_i\,\big{}\,
\zeta_1\geq\zeta_{2}\geq\cdots\geq\zeta_m,\,\zeta_i\in\mathcal{C},\,1\leq
i\leq m\}$ are the distinct zeros of the above polynomial, there
exist $m$ complex numbers $\{\beta_i\,\,
\beta_i\in\mathcal{C},\,1\leq i\leq m\}$ such that $u_n$ can be
expressed in the following form
\begin{equation*}u_n=\frac{\beta_1\zeta_1^{n+1}+\beta_2\zeta_2^{n+1}+\cdots+\beta_m\zeta_m^{n+1}}{\beta_1\zeta_1^n+\beta_2\zeta_2^n+\cdots+\beta_m\zeta_m^n}%\,\,(\beta_i\in\mathcal{R})
.\end{equation*}
Among the $m$ elements of the vector $\beta$,
let $\beta_l$ be the first nonzero element and $\beta_k$ the first
nonzero element after it, that is, $\{\beta_i\,\,
\beta_i\in\mathcal{C},\,1\leq i\leq m\}=
\{\!\!\underbrace{0,0,\cdots,0}_{\mbox{All are
zero.}},\beta_l(\neq0), \underbrace{0,0,\cdots,0}_{\mbox{ All are
zero.}},\beta_k(\neq0),\cdots,\beta_m\}.$
In this case, if $\zeta_l>\zeta_k$, then the iteration converges to $\zeta_l$. Therefore, if $\zeta_l\in\mathcal{Q}$, then the polynomial is reducible.