Primes and irreducible polynomials

TL;DR

This paper explores the irreducibility of polynomials derived from prime numbers' k-adic expansions, extending Murty's theorem by considering scaled primes and analyzing polynomial factors.

Contribution

It proves a stronger version of Murty's theorem for scaled primes and introduces techniques to analyze polynomial factors for larger multipliers.

Findings

Proved irreducibility for scaled primes with multiplier less than k

Developed methods to control polynomial factors with larger multipliers

Extended existing theorems on prime-based polynomials

Abstract

In 2002, M.Ram Murty showed that if p is a prime with k-adic expansion :$p = \sum_{i = 0}^n a_i k^i$ , then the polynomial $f(x) = \sum_{i = 0}^n a_ix^i$ is irreducible in $\mathbb{Z}[x]$.When $k = 10$ , it's a result of A.Cohn. I think this kind of polynomials is really interesting and worse to speak more. So I plan to find more conclusions about this kind of polynomials. In the first section of this article, author proves a stronger version of this theorem that if we multiply prime $p$ by a factor $t$ that is smaller than $k$ ,the conclusion also holds. In the second section, author further consider larger multiplier $t$ ,and gives a technique to control one of the factors of the polynomial.

Equations

$$f_{k,m}(x)=\sum _{i=0}^n{a}_i{x}^i$$

$$(i)|z|>\frac{1}{k},|z|<k$$

$$(ii)\mathrm{If}\mathrm{Re}(z)>0.Then|z|<\frac{1+\sqrt{4k-3}}{2}$$

$$(iii)\mathrm{Re}(z)<\sqrt{k}$$

$$(i)$$

$$0=\sum _{i=1}^n{a}_i{z}^i$$

$$⩾a_0-(k-1)\sum _{i=1}^n|z{|}^i$$

$$⩾1-(k-1)\frac{|z{|}^{n+1}-|z|}{|z|-1}$$

$$>1-(k-1)\frac{|z|}{1-|z|}$$

$$⩾1-(k-1)\frac{1}{k-1}=1-1=0$$

$$0$$

$$⩾a_n|z{|}^n-\sum _{i=0}^{n-1}{a}_i|z{|}^i$$

$$⩾|z{|}^n-(k-1)\sum _{i=0}^{n-1}|z{|}^i$$

$$=|z{|}^n-(k-1)\frac{|z{|}^n-1}{|z|-1}$$

$$>|z{|}^n-(k-1)\frac{|z{|}^n}{|z|-1|}$$

$$=|z{|}^n(1-\frac{k-1}{|z|-1})$$

$$>|z{|}^n(1-1)=0$$

$$(ii)$$

$$0=f_{k,m}(z)$$

$$=|a_n{z}^n+a_{n-1}{z}^{n-1}+\sum _{i=0}^{n-2}{a}_i{z}^i|$$

$$⩾\left|a_n{z}^n+a_{n-1}{z}^{n-1}\right|-\left|\sum _{i=0}^{n-2}{a}_i{z}^i\right|$$

$$⩾|z{|}^n\left|a_n+\frac{a_{n+1}}{z}\right|-\sum _{i=0}^{n-2}{a}_i|z{|}^i$$

$$⩾|z{|}^n\mathrm{Re}\left(a_n+\frac{a_{n-1}}{z}\right)-(k-1)\frac{|z{|}^{n-1}-1}{|z|-1}$$

$$\Rightarrow 0$$

$$=\frac{|z{|}^{n-1}}{|z|-1}\left(|z{|}^2+|z|-(k-1)\right)⩾0$$

$$(iii)$$

$$\mathrm{Re}\left(z^2\right)=2{\mathrm{Re}}^2(z)-|z{|}^2>2(\sqrt{k}{)}^2-{\left(\frac{1+\sqrt{4k-3}}{2}\right)}^2=\frac{2k+1-\sqrt{4k-3}}{2}⩾0$$

$$\mathrm{Re}\left(z^3\right)={\mathrm{Re}}^3(z)-3\mathrm{Re}(z)\cdot {\mathrm{Im}}^2(z)=4{\mathrm{Re}}^3(z)-3\mathrm{Re}(z)|z{|}^2$$

$$0$$

$$⩾\left|a_n{z}^n+a_{n-1}{z}^{n-1}+a_{n-2}{z}^{n-2}+a_{n-3}{z}^{n-3}\right|-\left|\sum _{i=0}^{n-4}{a}_i{z}^i\right|$$

$$⩾|z{|}^n\left|a_n+\frac{a_{n-1}}{z}+\frac{a_{n-2}}{z^2}+\frac{a_{n-3}}{z^3}\right|-(k-1)\sum _{i=0}^{n-4}|z{|}^i$$

$$⩾|z{|}^n\mathrm{Re}\left(a_n+\frac{a_{n-1}}{z}+\frac{a_{n-2}}{z^2}+\frac{a_{n-3}}{z^3}\right)-(k-1)\frac{|z{|}^{n-3}-1}{|z|-1}$$

$$\Rightarrow 0>$$

$$|z|>\sqrt{k}\Rightarrow |z{|}^4-|z{|}^3-(k-1)>k^2-k\sqrt{k}-(k-1)=(\sqrt{k}-1)[\sqrt{k}(k-1)-1]$$

$$Weonlyneedtocheckthecase:\mathrm{deg}{f}_{k,m}⩽3andk⩽3.$$

$$|f(x)|=|a|\prod _{i=1}^n\left|x-a_i\right|\prod _{j=1}^m(x^2-2\mathrm{Re}(z_j)x+|z_j{|}^2)$$

$$\mathrm{Re}\left(z_j\right)<n+\frac{1}{2}$$

$$\Rightarrow {\left(n+1-\mathrm{Re}\left(z_j\right)\right)}^2+{\mathrm{Im}}^2(z_j)>{\left(n-\mathrm{Re}\left(z_j\right)\right)}^2+{\mathrm{Im}}^2(z_j)$$

$$\Rightarrow |f(k+1)|$$

$$>|a|\prod _{i=1}^n|k-a_i|\prod _{j=1}^m[k^2-2\mathrm{Re}(z_j)k+|z{|}^2]=|f(k)|$$

$$\mathrm{Re}(z)<\sqrt{k}<\left[\sqrt{k}-\frac{1}{2}\right]+\frac{3}{2}<n+\frac{1}{2}$$

$$\Rightarrow g(n)<g(n+1)\Rightarrow g(n+1)⩾g(n)+1$$

$$g(k)$$

Full text

In the first section of this article,author proves a stronger version of this theorem that if we multiply prime $p$ by a factor $t$ that is smaller than $k$ ,the conclusion also holds.In the second section, author further consider larger multiplier $t$ ,and gives a technique to control one of the factors of the polynomial.

1. Basic Proof of the Theorem

Definition 1.1**.**

[TABLE]

, where $0\leqslant a_{i}\leqslant k-1,a_{i}\in Z,\quad f_{k,m}(k)=m,m>0$

It’s obvious that $a_{i}$ are the coefficients of $m$’s k-adic expansion.So it can be named as the polynomial generated by k-adic expansion of $m$

Lemma 1.2**.**

If $z$ is a non-zero root of $f_{k,m}$, it has the following properties:

[TABLE]

Proof.

Assume $f_{k,m}=x^{t}\sum_{i=0}^{n}a_{i}x^{i}$, because we just consider non-zero roots, $W.L.O.G$, $t=0,\quad f_{k,m}=\sum_{i=1}^{n}a_{i}x^{i},a_{0}\neq 0$

[TABLE]

If $|z|<\frac{1}{k},f(z)=0$,then

[TABLE]

Contradict. Thus $|z|>\frac{1}{k}$

Similarly, If $|z|>k,f(z)=0\Rightarrow$

[TABLE]

Contradict, thus we have proved (i)

[TABLE]

If $\operatorname{Re}(z)>0$, then $1/z=\bar{z}/|z|^{2}\Rightarrow\operatorname{Re}(1/z)=\frac{\operatorname{Re}(z)}{|z|^{2}}>0$

If $|z|\geqslant\frac{1+\sqrt{4k-3}}{2}$, we have:

[TABLE]

Because $a_{n}\geqslant 1,\quad a_{n-1}\geqslant 0\quad\operatorname{Re}\left(\frac{1}{z}\right)>0$ and $|z|\geqslant\frac{1+\sqrt{4k-3}}{2}\Rightarrow|z|>1$

[TABLE]

Contradict, thus we have proved (ii)

[TABLE]

If $\operatorname{Re}(z)>\sqrt{k}>0$, by (ii) we can know that $|z|<\frac{1+\sqrt{4k-3}}{2}$

We have the following conclusions:

[TABLE]

Let $k(x,y)=4x^{3}-3xy^{2}\quad\frac{\partial k(x,y)}{\partial x}=12x^{2}-3y^{2}\quad\frac{\partial k(x,y)}{\partial y}=-6xy$

$\operatorname{Re}(z)>\sqrt{k}>\frac{1+\sqrt{4k-3}}{4}>\frac{|z|}{2},-6|z|\operatorname{Re}(z)<0$

So $k(x,y)>k\left(\sqrt{k},\frac{1+\sqrt{4-3}}{2}\right)=\left(\frac{\sqrt{4k-3}-3}{2}\right)^{2}\geqslant 0\Rightarrow\operatorname{Re}\left(z^{3}\right)>0$

Thus $\operatorname{Re}\left(\frac{1}{z}\right)=\frac{\operatorname{Re}(z)}{|z|^{2}}>0,\quad\operatorname{Re}\left(\frac{1}{z^{2}}\right)=\frac{\operatorname{Re}\left(z^{2}\right)}{|z|^{4}}>0\quad,\operatorname{Re}\left(\frac{1}{z^{3}}\right)=\frac{\operatorname{Re}\left(z^{3}\right)}{|z|^{6}}>0$

If $\operatorname{deg}\left(f_{k,m}\right)\geqslant 4$, then we have:

[TABLE]

Because $|z|\geqslant\operatorname{Re}(z)>\sqrt{k}$ ,we have $\frac{1}{|z|-1}>0$

And $a_{n}\geqslant 1,\operatorname{Re}\left(\frac{1}{z}\right)>0,\operatorname{Re}\left(\frac{1}{z^{2}}\right)>0,\operatorname{Re}\left(\frac{1}{z^{3}}\right)>0$

[TABLE]

For $k\geqslant 2$,$(\sqrt{k}-1)[\sqrt{k}(k-1)-1]>0$,contradict.

Then we have proved that $\forall k\geqslant 2,\quad\operatorname{deg}f_{k,m}\geqslant 4,\quad\operatorname{Re}(z)<\sqrt{k}$

If $\operatorname{deg}\left(f_{k,m}\right)\geqslant 3$, by similar steps, we can get:

$0>|z|^{n-2}\cdot\frac{|z|^{3}-|z|^{2}-(k-1)}{|z|-1}$

For $k\geqslant 4,|z|^{3}-|z|^{2}-(k-1)\geqslant k\sqrt{k}-2k+1=k(\sqrt{k}-2)+1>0$, contradict.

[TABLE]

If  $\operatorname{deg}f_{k,m}=0$ there’s no root

$\operatorname{deg}f_{k,m}=1\quad a_{0}\geqslant 0\Rightarrow\operatorname{Re}(z)\leqslant 0<\sqrt{k}$,

$\operatorname{deg}f_{k,m}=2\quad\operatorname{Re}(z)=-\frac{a_{1}}{2}\leqslant 0<\sqrt{k}$,

$\operatorname{deg}f_{k,m}=3$ :

Let $f_{k,m}=a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0},0\leqslant a_{i}\leqslant 2,\ a_{3},a_{0}\geqslant 1$

W.L.O.G, $a_{0}\neq 0$ or if $a_{0}=0$, the condition is the same as $\operatorname{deg}f_{k:m}\leqslant 2$

If $z$ is a real root of $f_{k,tm}$, because $a_{i}\geqslant 0\Rightarrow z\leqslant 0\Rightarrow z=Re(z)\leqslant\sqrt{k}$

If $z$ is not real, $f_{k,m}=a_{3}\left(x-x_{0}\right)(x-z)(x-\bar{z})$

$x_{0}+z+\bar{z}=-\frac{a_{2}}{a_{3}},x_{0}\cdot|z|^{2}=-\frac{a_{0}}{a_{3}}$

Then $-x_{0}=\frac{a_{2}}{a_{3}}+2\operatorname{Re}(z)\geqslant 2\sqrt{2}$

$-x_{0}=\frac{a_{0}}{a_{3}|z|^{2}}\leqslant\frac{2}{1\cdot|z|^{2}}<\frac{2}{2}=1\Rightarrow 1>2\sqrt{2}\quad$ contradict.

In conclusion: $\forall k\geqslant 2,\operatorname{Re}(z)<\sqrt{k}$ ∎

Lemma 1.3**.**

$\forall$* polynomial $f$, if $\ \forall$ root $z$, $\operatorname{Re}(z)<n+\frac{1}{2}$, then $|f(n+1)|>|f(n)|$*

Proof.

$|f|$ could be written as:

[TABLE]

$a_{i}$ are real roots of f and $z_{j}$ are complex roots of $f$.

For single factor of degree 1: $|x-a_{i}|$,$a_{i}$ are real roots of $f$ .

$a_{i}<n+\frac{1}{2}\Rightarrow\left|n+1-a_{i}\right|>\left|n-a_{i}\right|$

For factor of degree 2 : $(x^{2}-2Re(z_{j})x+|z_{j}|^{2})$,

$x^{2}-2\operatorname{Re}(z_{j})x+|z_{j}|^{2}=\left(x-\operatorname{Re}\left(z_{j}\right)\right)^{2}+I_{m}^{2}(z_{j})$

[TABLE]

In conclusion,

[TABLE]

∎

Theorem 1.4**.**

$\forall k\geqslant 2$, $0<t<k$, $f_{k,tp}$ irreducible.

Proof.

First when $0<t\leq k-\left[\sqrt{k}-\frac{1}{2}\right]-1$

If $f_{k,tp}$ is reducible, assume $f_{k,t}=g\cdot h$, the highert coefficient of $g$ is bigger than 0 , of course so does $h$.

$tp=f_{k,tp}(k)=g(k)h(k)\Rightarrow$ p — g(k) or p — h(k) W.L.O.G ,p — h(k)

Then $g(k)=\frac{tp}{h(k)}\leqslant t\leqslant k-[\sqrt{k}-\frac{1}{2}]-1$

Because $\forall x>0,f_{k,tp}(x)>0$, then $\forall x>0,g(x)>0$

(real root of $f$ smaller than $0\Rightarrow$ real root of $g$ smaller than $0\Rightarrow\forall x>0,g(x)>0$ )

Let $g(x)=a\prod_{i=1}^{n}(x-a_{i})\sum_{j=1}^{m}(x^{2}-2Re(z_{j})x+|z_{j}|^{2})$

For $\forall n\geqslant[\sqrt{k}-\frac{1}{2}]+1,ifz$ is a root of $g$.then:

[TABLE]

Thus:

[TABLE]

However,

[TABLE]

Contradict

Thus for $t\leqslant k-[\sqrt{k}-\frac{1}{2}]-1,f_{k,tp}$ irreduable.

Especially when $k=2,2-\left[\sqrt{2}-\frac{1}{2}\right]-1=2-1\leqslant 2-1$. So all cases of $k=2$ have been solved For $k\geqslant 3,t\geqslant k-\left[\sqrt{k}-\frac{1}{2}\right]$

For other cases,

Let $f_{k},tp=g\cdot h$ , $p|h(k)$ and

[TABLE]

For factors of degree $1,a_{i}<0\Rightarrow k-a_{i}>k$

For factors of degree 2,

$\operatorname{Re}(z_{j})<\sqrt{k}\Rightarrow(k^{2}-2\operatorname{Re}(z_{j})k+|z_{j}|^{2})>(k-\operatorname{Re}(z_{j}))^{2}>(k-\sqrt{k})^{2}$

If $k\geqslant 4$ then any factor of $g$ is bigger than $k$ when $x=k$

$\Rightarrow t\geq g(k)>k$ ,contradict.

When $k=3$,

$3-a_{i}>3,(3-\sqrt{3})^{2}>1.6>1,(1.6)^{2}=2.56>3-1=k-1\geq t$

So $g$ has no factors of degree 1 and $g$ has at most 1 functor of degree 2

Let $g=b_{2}x^{2}+b_{1}x+b_{0}$,with root $z$ and $\overline{z}$. $1\leqslant b_{2}\leqslant 2,1\leqslant b_{0}\leqslant 2$,,

Then $2=g(3)=b_{2}(3-z)(3-\overline{z})>b_{2}(3-|z|)^{2}=b_{2}(3-\sqrt{\frac{b_{0}}{b_{2}}})^{2}$

So there must be $b_{2}=1$ and $b_{0}=2$ or $2>b_{2}\left(3-\sqrt{\frac{b_{0}}{b_{2}}}\right)^{2}>2$ contradict.

$g(3)=9+3b_{1}+2=2\Rightarrow b_{1}=-3\Rightarrow\quad g(x)=x^{2}-3x+2$

However $g(1)=g(2)=0\Rightarrow f_{k,tp}(1)=0$ impossible.

In conclusion, $\ \forall 0<t\leqslant k-1,f_{k,tp}$ is irreducible.

∎

Corollary 1.5**.**

If $f_{k,m}$ reducible, $g\mid f_{k,m}$ and $g(k)=n$ with highest order coefficient a¿o Then

[TABLE]

Proof.

[TABLE]

Thus $\operatorname{deg}g<\operatorname{deg}_{k-\sqrt{k}}\frac{n}{a}$

∎

Corollary 1.6**.**

$p,q$* are primes, $\forall k$, if $f_{k,pq}$ reducible, $\exists$ irreducible polynomial $g$ and $h$ such that $g(k)=p,h(k)=q$*

Proof.

If $\exists s(x)\mid f_{k,pq},\quad s(k)=1$ .

Then

[TABLE]

Thus if $f_{k,pq}$ reducible, $f_{k,pq}=gh,g>1,h>1$

\Rightarrow\left\{\begin{array}[]{l}g(k)=p\\ h(k)=q\end{array}\right. or \left\{\begin{array}[]{l}g(k)=q\\ h(k)=p\end{array}\right. ∎

2. Conditions when $t$ is bigger than k

By corollary 2, we can know that if $f_{k.pq}$ reducible, what are its irreducible factors values when $x=k$ .And by corollary 1 we can control its degree. So we can found out what one of its factor looks like. I will give some examples to show how to estimating the degree of the factor that values smaller.

Theorem 2.1**.**

When $k\geqslant 4,k\leqslant t<2k,(p,k)=1$, $f_{k,tp}$ is reducible $\Leftrightarrow$ one of its factors is $x+t-k$.

Proof.

Let $f_{k,+p}=gh$, $p|h$. If $h(k)>p\Rightarrow h(k)\geq 2p\Rightarrow g(k)<k$. By proof of Theorem 1.4 we can know that $g(k)>k$, controdict.

Hence $h(k)=p,g(k)=t$.

When $k>5$ ,by Corollary1.5 ,it’s easily to know that deg$\ g<2$

Let $g=ax+b$ , Because all real root of $f_{k,tp}$ not bigger than 0$\Rightarrow-\frac{b}{a}\leq 0$

Thus $a,b>0,g(k)=ak+b=t<2k\Rightarrow a<2\Rightarrow a=1,b=t-k$.

When $k=4,5,\operatorname{deg}g<3$

Let $g=ax^{2}+bx+c$.

Because $g(k)\equiv c\equiv t(modk)$ , $k<t<2k,c\geq 0$ $\Rightarrow t-c=k$

Thus:

[TABLE]

If $\Delta\geq 0$ , g has roots that is real and bigger than 0 , impossible.

If $g=x^{2}-3x+3$ , assume $h=\sum_{i=0}^{m}c_{m}x^{i}$

[TABLE]

If $c_{1}=1$ ,then by induce , if $\forall i<s,c_{i}=1$ ,for i = s:

[TABLE]

Thus $h(x)=\sum_{i=0}^{n}x^{i}$

However , $a_{n-1}=1*1-3*1<0$ ,contradict.

If $c_{1}=2$ , $a_{2}=c_{2}*3-2*3+1\Rightarrow a_{2}=2$ , $a_{3}=c_{3}*3-2*3+2\Rightarrow c_{3}=2$

By induction , if $\forall 1<i<s$ , $c_{i}=2$ then when $i=s$ ,we have:

$a_{s}=3*c_{s}-3*2+2\Rightarrow c_{s}=2$

However , $a_{n-1}=2*1-2*3=-4$ ,contradict.

In conclusion , deg $g$ = 1, $g=x+t-k$.

Next we need to prove $h$ is irreducible.

If $h(x)=p(x)*q(x)$ W.L.O.G $p(k)=1,\quad q(k)=P$.

$\forall$ root $z$ of $p(x)$, it is also a root of $f(x)\Rightarrow\operatorname{Re}(z)<\sqrt{k}$.

Then $p(k)>p(k-1)\Rightarrow p(k-1)\leq 0$ .That’s impossible. So $h$  is irreducible.

Thus $f_{k,tp}$ reducible $\Leftrightarrow\exists g=x+t-k$, $h$ irreducible, $f_{k,tp}=gh$. ∎

It is really hard when $k=2,3$. But we can consider $f_{3,4p},f_{3,5p},f_{2,3p}$ ,during the proof of these three conditions,except using Corollary1.5 ,we also use the technique of predict coefficients of $g(x)$ to make contradiction and to use the fact that $g(k)>g(k-[\sqrt{k}-\frac{1}{2}]-1)$

Proposition 2.2**.**

$f_{3,4p}$* reducible if and only if $x+1\mid f_{3,4p}$.*

Proof.

When $t=4,(3-\sqrt{3})^{6}>4\Rightarrow$ deg $g<6$ and $5$ is odd, $g(3)>3\times(3\sqrt{3})^{4}>4$

So $\operatorname{deg}g\leqslant 4$

If $\operatorname{deg}g=4\quad,\exists z_{1}$ and $z_{2}$ s.t

[TABLE]

If $b_{4}\geqslant 2$ ,then $g(3)>2\cdot(3-\sqrt{3})^{4}>4$

If $b_{4}=1\quad,\ g(3)=4\Rightarrow b_{0}=1$ .Thus $\left|z_{1}\right|\cdot\left|z_{2}\right|=1$.

If $\operatorname{Re}({z_{i}})\leq 0$ , $g(3)\geqslant 9$,impossible

By Lemma1.2 -(ii), $\operatorname{Re}(z_{i})>0\Rightarrow|z|<\frac{1+\sqrt{9}}{2}=2$,

$g(3)>\left(3-\operatorname{Re}(z_{1})\right)^{2}\left(3-\mathbb{R}e(z_{2})\right)^{2}>\left(3-|z_{1}|\right)^{2}\left(3-\left|z_{2}\right|^{2}\right)=\left(3-\left|z_{1}\right|\right)^{2}\left(3-\frac{1}{|z_{1}|}\right)^{2}=\left(10-3\left|z_{1}\right|-3\frac{1}{\mid z_{1}|}\right)^{2}>\left(\frac{5}{2}\right)^{2}>4$ .So $\operatorname{deg}g\leqslant 3$.

If $\operatorname{deg}g=3,$

$g$ has a real root smaller than 0. $\Rightarrow\quad g(3)>3\times(3\sqrt{3})^{2}=4.8>4$ contradict.

If $\operatorname{deg}g=2$,

$g(3)=4\Rightarrow g$ can’t have two real roots,or $g(3)>3^{2}=9>4$ contradict.

Let $g(x)=ax^{2}+bx+c\quad 1\leqslant a\leqslant 2,\quad 1\leq c\leqslant 2$

$g(3)=4\equiv 1(\bmod 3)\Rightarrow c=1$

So there are only two types of $g,g=x^{2}-2x+1$ or $g=2x^{2}-5x+1$,however,all of these polynomials has positive real roots.They can’t be factors of $f_{3,4p}$

And we also hold the conclusion that h(x) is irreducible.

So $f_{3,4p}$ reducible if and only if $x+1\mid f_{3,4p}$.

∎

Proposition 2.3**.**

$f_{3,5p}$* is reducible $\Leftrightarrow x+2\mid f_{3,5p}$*

This proposition looks like the former one at first.Indeed they are truely similar at the first step,however ,in the last step ,we have to face the factor $x^{2}-2x+2$ which its roots meet the all the properties that $f_{k.m}$ has.It need other method to do with this factor.

Proof.

[TABLE]

Let $g=\sum_{0}^{n}b_{i}x^{i}$,it obvious that $0<b_{n}<3$ and $b_{0}=2$.

If deg $g$=6:

$\operatorname{Assume}\ g=a\prod_{i=1}^{3}\left(x-z_{i}\right)\left(x-\bar{z}_{i}\right)$

$\left|z_{1}\right|^{2}\left|z_{2}\right|^{2}\left|z_{3}\right|^{2}\leqslant 2\Rightarrow\exists i\operatorname{s.t}\ \left|z_{i}\right|<\sqrt{2}$

$Thus\ g(3)>(3-\sqrt{3})^{4}(3-\sqrt{2})^{2}>5$ ,impossible.

If deg $g=4:$

Let $g=a(x^{2}-2\operatorname{Re}^{2}(z_{1})x-|z_{1}|^{2})(x^{2}-2\operatorname{Re}(z_{2})x-|z_{2}|^{2})$

$\left|z_{1}\right|^{2}\left|z_{2}\right|^{2}<2\Rightarrow a\left|z_{i}\right|<\sqrt[4]{2}$ Thus $g(3)>(3-\sqrt{3})^{2}(3-\sqrt[4]{2})^{2}>5$ ,impossible.

If $\operatorname{deg}g=3$

Let $g=b_{3}x^{3}+b_{2}x^{2}+b_{1}x+b_{0}$ $=b_{3}(x-a)(x-z)(x-\bar{z})$

Then by lemma 1: $\quad a|z|^{2}>-2,\quad|z|<2\Rightarrow\ a<-\frac{1}{2}$.

Thus $g(3)>\left(3+\frac{1}{2}\right)(3-\sqrt{3})^{2}>5$ impossible .

If $\operatorname{deg}g=2\quad g=b_{2}x^{2}+b_{1}x+2\quad b_{2}=1$ or 2 So $g=x^{2}-2x+2$ or $g=2x^{2}-5x+2$.

If $g=2x^{2}-5x+2,\quad g(\frac{1}{2})=0$ impossible.

If $g=x^{2}-2x+2$ , there’s an interesting conclusion: If $f_{3,5p}=(x^{2}-2x+2)h(x)$ ,then $2|p$,however , $f_{3,10}=x^{2}+1$ ,irreducible.So $g\nmid f_{3,5p}$.

Now let’s prove this lemma:

Lemma 2.4**.**

If $x^{2}-2x+2|f_{3,5l}$, then $2|l$

Proof.

W.L.O.G,$3\nmid l$,then let

$f_{3,5l}=\sum_{i=0}^{n}a_{i}x^{i},0\geq a_{i}\geq 2.f_{3,5l}=(x^{2}-2x+2)h(x),h(x)=\sum_{j=0}^{m}c_{j}x^{j}$ .Set $(b_{0},b_{1},b_{2})$,s.t ,$x^{2}-2x+2=b_{2}x^{2}+b_{1}x+b_{0}$

$c_{t}$ is controlled by $c_{t-1},c_{t-2},\forall t\geq 2$ by:

$0\geq a_{t}\geq 2,a_{t}=b_{0}c_{t}+b_{1}c_{t-1}+b_{2}c_{t-2}=2c_{t}-2c_{t-1}+c_{t-2}$ $c_{i}\in\mathbb{Z}$.So $c_{i}$ can only be taken at most 2 different values.To be convenient,define operation :

$G:G(c_{i+1},c_{i})=\{(c_{i+2},c_{i+1})|\operatorname{all\ possible\ pairs}(c_{i+2},c_{i+1})\}$

$G(c_{i+1},c_{i})$ has at most two elements.

Then by calculating ,we can find what adjacent pairs $(c_{i+1},c_{i})$ look like.

If $c_{1}=1$ ,

$a_{2}=2c_{2}-2+1\Rightarrow c_{2}$ can only be $1$, $G(1,1)={(1,1)}\Rightarrow c_{m}=1,c_{m-1}=1$

However, $0\leq a_{n-1}=b_{1}c_{m}+b_{2}c_{m-1}=-1<0$ ,contradict.

If $c_{1}=2$ ,there is a tree to make the relationship more clearly:

[TABLE]

We can find all possible pairs of $(c_{i},c_{i-1})$ :

$G(2,1)=\{(2,2)\},G(2,2)=\{(1,2),(2,2)\},G(1,2)=\{(0,1)\},G(0,1)=\{(0,0)\},G(0,0)=\{(0,0),(1,0)\},G(1,0)=\{(1,1),(2,1)\}$

Since we have known all possible pairs of $(c_{i},c_{i-1})$ ,and the only pair that $(c_{m},c_{m-1})$ can be taken to make $a_{n-1}\geq 0$ is $(1,2)$.

So from the tree ,we can know that the possible sequences of $c_{i}$ are all in this form:$\{1\overbrace{2\cdots 2}^{l_{1}}1\overbrace{0\cdots 0}^{l_{2}}\cdots 1\overbrace{2\cdots 2}^{l_{t}}1\}$ .It is a chain that starts from $(1,2)$ and ends with $(2,1)$ in the tree. There must be even number of 1s.

So $h(3)\equiv number\ of\ 1s\equiv 0(mod2)$,$2|n$ ∎

In conclusion, $f_{3,5p}$ reducible if and only if $x+2|f_{3,5p}$ ∎

Proposition 2.5**.**

$\quad f_{2,3p}$* is reducible if and only if $x+2|f_{2,3p}$ or $x^{2}-x+1|f_{2,3p}$*

When $k=2$, there’s a problem that for other $k,k-\sqrt{k}>1$ However, $2-\sqrt{2}<1$, so we can’t control the degree of $g$ by previous ways.

The good news is that, $g(2)>g(1)$, so if $g(2)=n$, then $\frac{g(1)}{g(2)}\leq\frac{1}{n}$ can be a good way to control degree of $g$.

Proof.

Assume $f_{2,3p}=g(x)\cdot h(x),g(2)=3,h(2)=p$

$g(2)>g(1)\Rightarrow g(1)=1$ or $2\Rightarrow\frac{g(2)}{g(1)}=3$ or $\frac{3}{2}$

Assume $g(x)=\prod_{i=1}^{n}\left(x-a_{i}\right)\prod_{j=1}^{m}\left(x^{2}-2\operatorname{Re}\left(z_{j}\right)x+|z_{j}|^{2}\right)$

Then $\frac{g(2)}{g(1)}=\prod_{i=1}^{n}\left(\frac{2-a_{i}}{1-a_{i}}\right)\prod_{j=1}^{m}\left(\frac{4+|z_{j}|^{2}-4\operatorname{Re}\left(z_{j}\right)}{\left.1+|z_{j}\right|^{2}-2\operatorname{Re}\left(z_{j}\right)}\right)$

First ,claim following inequalities :

[TABLE]

Inequality (1) is known as ”sugar water inequality”.Both of these inequalities are too easy to prove.So author will not prove them in the thesis.

By property Lemma1 and inequality(1), $|z|<2,a_{i}<0\Rightarrow\frac{2-a_{i}}{1-a_{i}}>\frac{2+2}{1+2}=\frac{4}{3}$

If $\operatorname{Re}\left(z_{j}\right)<0,$by inequality(1),

$\frac{4+\left|z_{j}\right|^{2}-4\operatorname{Re}_{2}\left(z_{j}\right)}{\left.1+|z_{j}\right|^{2}-2\operatorname{Re}\left(z_{j}\right)}>\min\left(\frac{4+\left.|z_{j}\right|^{2}}{1+\left.|z_{j}\right|^{2}},\frac{-4\operatorname{Re}\left(z_{j}\right)}{-2\operatorname{Re}\left(z_{j}\right)}\right)=\frac{8}{5}$

If $\operatorname{Re}\left(z_{j}\right)\geqslant 0$, $\frac{4-4\mathbb{R}e\left(z_{j}\right)+\left|z_{j}\right|^{2}}{1-2\mathbb{R}e\left(z_{j}\right)+\left|z_{j}\right|^{2}}$ could be written as :$1+\frac{3-2\operatorname{Re}\left(z_{j}\right)}{1-2\operatorname{Re}\left|\left.z_{j}|+|z_{j}\right|^{2}\right.}\quad$.

By lemma1.2, $\left|z_{j}\right|<\frac{1+\sqrt{5}}{2}$ and $\operatorname{Re}\left(z_{j}\right)<\sqrt{2}$.

If $\left|z_{j}\right|<\sqrt{2}$,

Then $\frac{3}{1+|z_{j}|^{2}}>1\Rightarrow\frac{3-2\operatorname{Re}(z_{j})}{1+|z_{j}|^{2}-2\operatorname{Re}(z_{j})}>1\Rightarrow\frac{4-4\operatorname{Re}\left(z_{j}\right)+\left|z_{j}\right|^{2}}{1-2\operatorname{Re}\left(z_{j}\right)+\left|z_{j}\right|^{2}}>1+1=2$

If $|Z|>\sqrt{2},$

Then $\frac{3}{1+|z_{j}|^{2}}<1\Rightarrow\frac{4-4\operatorname{Re}\left(z_{j}\right)+\left|z_{j}\right|^{2}}{1-2\operatorname{Re}\left(z_{j}\right)+\left|z_{j}\right|^{2}}>1+\frac{3-2\sqrt{2}}{1+|z_{j}|^{2}}>1+\frac{3-2\sqrt{2}}{1+(\frac{1+\sqrt{5}}{2}^{2})}>1.217$

Here we need to notice that $g(0)=1$, and the highest order coefficient is 1.Then by Vieta’s formulas, $\prod_{i=1}^{n}|a_{i}|\prod_{j=1}^{m}|z_{j}|^{2}=1$ , $\exists|a_{i}|<1\ or\ |z_{j}|<1$

For $|a_{i}|<1$ , $a_{i}<0\Rightarrow\frac{2-a}{1-a}>\frac{3}{2}$

For $|z_{j}|<1$ , it holds the inequality that:

$\begin{aligned} &\frac{4-4\operatorname{Re}\left(z_{j}\right)+\left.|z_{j}\right|^{2}}{1-2\operatorname{Re}\left(z_{j}\right)+|z_{j}|^{2}}>\frac{5-4\operatorname{Re}\left(z_{j}\right)}{2-2\operatorname{Re}\left(z_{j}\right)}\\ \Leftrightarrow&\left.\left(3-2\operatorname{Re}\left(z_{j}\right)\right)|z_{j}\right|^{2}<3-2\operatorname{Re}\left(z_{j}\right)\\ \Leftrightarrow&\left|z_{j}\right|^{2}<1\end{aligned}$

Because $|Re(z_{j})|<|z|<1$

[TABLE]

Now , we got following important constants to estimate values of factors of $\frac{g(2)}{g(1)}$ by the range of roots:

[TABLE]

Their roles are similar to the role of $k-\sqrt{k}$ in Corollary 1.5.Because in all conditions , factors of $\frac{g(2)}{g(1)}$ are bigger than 1.So we can control degree of $g$ by these constants.

Thus, there’s the estimate of degree of $g$ :

[TABLE]

If deg$g$ = 7, then $n=1,m=3,\Rightarrow|a||z_{1}|^{2}|z_{2}|^{2}|z_{3}|^{2}=1$

By lemma1 ,$|a|>\frac{1}{2}\Rightarrow|z_{1}||z_{2}||z_{3}|<\sqrt{2},\exists\ j\ \operatorname{s.t}|z_{j}|<\sqrt{2}$, so regardless what Re$(z_{j})$ is , $s(z_{j})>2$

Then $\frac{g(2)}{g(1)}>\frac{3}{2}\times 2\times(2,17)^{2}>3$,controdict.

If $\operatorname{deg}g=6$,then $\quad|a_{1}||a_{2}||z_{1}|^{2}\left|z_{2}\right|^{2}=1$

$\left|a_{1}\right|,\left|a_{2}\right|>\frac{1}{2}\Rightarrow\quad\left|z_{1}\right|^{2}\left|z_{2}\right|^{2}<4\Rightarrow\exists\left|z_{i}\right|<\sqrt{2}$

Thus $\frac{g(2)}{g(1)}>\frac{3}{2}\times 2\times 1.217\times\frac{4}{3}>3$ controdict.

$\operatorname{If\ deg}g=5$,then $|a|\left|z_{1}\right|^{2}\left|z_{2}\right|^{2}=1\Rightarrow\left|z_{1}\|z_{2}\right|<\sqrt{2}\Rightarrow\exists\left|z_{i}\right|<\sqrt{2}$

Thus $\frac{g(2)}{9(1)}>\frac{3}{2}\times 2\times 1.217>3\quad$ controdict.

If $\operatorname{deg}g=4$ :

$\left|z_{1}\right|^{2}\left|z_{2}\right|^{2}=1\Rightarrow\left|z_{1}\right|\left|z_{2}\right|=1$ WLOG $\left|z_{1}\right|<1$.

Then if $\left|z_{2}\right|\leqslant\sqrt{2}\Rightarrow\frac{g(2)}{g(1)}>2\times\frac{9}{4}>3$.Impossible.

If $\left|z_{2}\right|>\sqrt{2}\Rightarrow\left|z_{1}\right|<\frac{1}{\sqrt{2}}\Rightarrow s(z_{1})>\frac{\frac{9}{2}+2\sqrt{2}}{\frac{3}{2}+2\sqrt{2}}>2.5$

Here $s(z_{j})>1+\frac{3-2Re(z_{j})}{1+|z_{j}|^{2}-2Re(z_{j})},$  the bigger $|z_{j}|$ is and the smaller $Re(z_{j})$ is ,the smaller $s(z_{j})$ is. So $s(z_{j})>s(-\frac{1}{\sqrt{2}})=\frac{\frac{9}{2}+2\sqrt{2}}{\frac{3}{2}+2\sqrt{2}}>2.5$

Then $\frac{g(2)}{g(1)}>2.5\times 1.2=3$. Controdict.

If $\operatorname{deg}g=3$.

Case 1. $\left|a_{1}\right|\left|a_{2}\|a_{3}\right|=1,a_{1},a_{2},a_{3}<0$

Let $g=x^{3}+ax^{2}+bx+1$

[TABLE]

[TABLE]

Then $g\left(-\frac{1}{2}\right)=-\frac{1}{8}+\frac{a}{4}-a-\frac{3}{2}+1<-\frac{1}{2}-\frac{3}{4}a<0$,however, $g(0)=1>0$

So $\exists$ a real root $a_{i}$ s.t $\left|a_{i}\right|<\frac{1}{2}$, by lemma1 $\left|a_{i}\right|>\frac{1}{2}$ contradict.

Case 2: $\quad|a_{1}||z|^{2}=1$

If $\left|a_{1}\right|<1,|a_{1}|>\frac{1}{2}\Rightarrow|z|^{2}<2\Rightarrow|z|<\sqrt{2}\Rightarrow\frac{g(2)}{g(1)}>\frac{3}{2}\cdot 2=3$ contradict.

Then $\left|a_{1}\right|>1\Rightarrow|z|<1$ and $|\operatorname{Re}(z)|<1$

Because $g(2)=3$ ,then we can assume $g(x)=x^{2}+ax^{2}-(3+2a)x+1$

If $a\geqslant-1\Rightarrow g(2)=3>0,g(1)=-1-a\leq 0$ .There is a real root bigger than 0 ,impossible.

If $a<-1,g(-\frac{1}{2})=-\frac{1}{8}+\frac{a}{4}+\frac{3}{2}+a+1=\frac{5}{2}+\frac{5}{4}a-\frac{1}{8}$ , because $a\leq-2$ , $g(-\frac{1}{2})<0$

So $|a|<\frac{1}{2}$ , impossible.

If $\operatorname{deg}g=2,\quad g=x^{2}+ax+1,\quad g(2)=3\Rightarrow g=x^{2}-x+1$

If $\operatorname{deg}g=1,\quad g=x+1$

Indeed $f_{2,3\times 3}=x^{3}+1=(x^{2}-x+1)(x+1)$,so $f_{2,3p}$ can have these two factors.

In conclusion, $f_{2,3p}$ reducible $\Leftrightarrow x+1\mid f_{2,3p}$ or $x^{2}-x+1\mid f_{2,3p}$.

$\Leftrightarrow f_{2,3p}(-1)=0$ or $f_{2,3p}\left(\frac{1+\sqrt{3}i}{2}\right)=0$ ∎

Remark 2.6**.**

At last,we have done all conditions of $f_{k,tp}$ when $t<2k$.The conclusion that ′if $x^{2}-2x+2|f_{3,5l}$ ,then $2|l$′ is really interesting.The tree is generated only depend on some initial coefficient of h(x) and the constant term of g(x)!

I hope the way of proving Propositon2.5 could be improved,because while $t$ becomes bigger,the condition gets worse.Maybe ′Trees′ could help to prove some polynomials can’t be a factor of $f_{2,tp}$.

I also have some guess that I couldn’t prove now.For example, I think if $g(x)|f_{k,n}$.and $g(k)=p$ for some primes , then $g(k)|f_{k,p^{\alpha}}$ for some integer $\alpha$. Then this kind of polynomials could be a new way to represent prime numbers.

ACKNOWLEDGEMENTS.

I would like to thank Xuejun Guo who guided my former thesis and let me know this interesting question and Stonybrook University where provide my a good environment to consider this question.