The polynomial $f(x) = 2x^6 + 6x^5 + 4x^4 + 5x^3 + 3x +1 $ is irreducible in $\mathbb{Z}[x]$

Question

How can I prove that the polynomial $f(x) = 2x^6 + 6x^5 + 4x^4 + 5x^3 + 3x +1 $ is irreducible in $\mathbb{Z}[x]$?

Answer 1

Using Cohn's irreducibility criterion (see also Corollary 1 in article ON AN IRREDUCIBILITY THEOREM OF A. COHN) if we find integer $b\geq 2$ such that $0 \leq a_k \leq b-1$ and $\sum_{k=0}^{n}a_k b^k$ is a prime, then $f(x)=\sum_{k=0}^{n}a_k x^k$ is irreducible in $\mathbb{Z}[x]$.

In this case $b=14$ satisfies conditions of the criterion since $f(14)=18453443$ is a prime and so the $f(x)$ is irreducible in $\mathbb{Z}[x]$. Not as pretty as Eisenstein but works.

Answer 2

Let $f = 2x^6 + 6x^5 + 4x^4 + 5x^3 + 3x +1$ and suppose $f = gh$ for some $g,h$ in $\mathbb{Z}[x]$ with $g,h \notin \{\pm 1\}$.

Without loss of generality, we can assume the leading coefficients of $g,h$ are both positive.

It follows that one of $g,h$ is monic, and the other has leading coefficient $2$.

Also, the constant terms of $g,h$ are either both $1$ or both $-1$.

Since $f$ has no rational roots, it follows that either $g,h$ are both degree $3$, or else one is quadratic, and the other degree $4$.

Then

\begin{align*} 2x^6 + 6x^5 + 4x^4 + 5x^3 + 3x +1&= gh\;\text{in}\;\mathbb{Z}[x]\\[4pt] \implies\;x^3 + x + 1&= gh\;\text{in}\;\mathbb{Z_2}[x] \end{align*}

Note that $x^3+x+1$ has no root in $\mathbb{Z}_2$, hence, since it's cubic, it's irreducible in $\mathbb{Z_2}[x]$.

Then since $x^3 + x + 1= gh\;\text{in}\;\mathbb{Z_2}[x]$, it follows that in $\mathbb{Z_2}[x]$, one of $g,h$ equals $1$, and the other is equal to $x^3+x+1$.

Without loss of generality, assume $g = x^3 + x + 1\;\text{in}\;\mathbb{Z_2}[x]$.

Then in $\mathbb{Z}[x]$, we must have $\text{deg}(g) \ge 3$, hence either $\text{deg}(g)=3$ or $\text{deg}(g)=4$.

Suppose $\text{deg}(g)=4$.

Since $g = x^3 + x + 1\;\text{in}\;\mathbb{Z_2}[x]$, in $\;\mathbb{Z}[x]$, the leading coefficient of $g$ (i..e., the coefficient of $x^4$) must be even, hence must be $2$, and the coefficient of the $x^3$ term must be odd. Applying Vieta's formula for the sum of the roots, the sum of the roots of $g$ has a denominator of $2$ and an odd numerator, hence is not an integer. But since $g$ has leading coefficient $2$, $h$ must be monic, hence the sum of the roots of $h$ is an integer. But then the sum of the roots of $g$ plus the sum of the roots of $h$ is not an integer, contradiction, since the sum of the roots of $f$ is $-3$.

Thus, we must have $\text{deg}(g)=3$.

Since $g$ is cubic and $g = x^3 + x + 1\;\text{in}\;\mathbb{Z_2}[x]$, it follows that $g$ is monic, and

$$g = x^3 + a_2x^2 + a_1x + a_0$$

where $a_2,a_1,a_0 \in \mathbb{Z}$, with $a_2$ even, and $a_1,a_0$ odd.

Moreover, as previously, noted, either $a_0=1$ or $a_0 = -1$.

Since $g$ is cubic, so is $h$, and since $g$ is monic, the leading coefficient of $h$ is $2$.

Also, the constant terms of $g,h$ are equal, so the constant term of $h$ is $a_0$.

Since $h = 1\;\text{in}\;\mathbb{Z_2}[x]$, it follows that

$$h = 2x^3 + b_2x^2 + b_1x + a_0$$

where $b_2,b_1 \in \mathbb{Z}$, with $b_2,b_1$ even.

Expanding the product

$$(x^3 + a_2x^2 + a_1x + a_0)(2x^3 + b_2x^2 + b_1x + a_0)$$

and equating the coefficients to the corresponding coefficients of $f$, we get the equations

\begin{align*} a_0(a_1 + b_1) &= 3\tag{eq1}\\[4pt] a_0(a_2 + b_2) + a_1b_1 &=0\tag{eq2}\\[4pt] 3a_0 + a_1b_2 + a_2b_1 &= 5\tag{eq3}\\[4pt] 2a_1 + b_1 + a_2b_2 &=4\tag{eq4}\\[4pt] 2a_2 + b_2 &=6\tag{eq5}\\[12pt] \text{Using $(\text{eq}5),$}\;\,2a_2 + b_2 &=6\\[4pt] \implies\;0 + b_2 &= 2\pmod 4 &&\text{[since $a_2$ is even]}\\[4pt] \implies\; b_2 &\equiv 2 \pmod 4\\[12pt] \text{Using $(\text{eq}4),$}\;\,2a_1 + b_1 + a_2b_2 &=4\\[4pt] \implies\; 2 + b_1 + 0 &\equiv 0 \pmod 4 &&\text{[since $a_1$ is odd,}\\[-1pt] &&&\;\text{and $a_2,b_2$ are even]}\\[-1pt] \implies\; b_1 &\equiv 2 \pmod 4\\[12pt] \text{Using $(\text{eq}2),$}\;\,a_0(a_2 + b_2) + a_1b_1 &=0\\[4pt] \implies\; a_0(a_2 + 2) + 2 &\equiv 0 \pmod 4 &&\text{[since $a_1$ is odd,}\\[-1pt] &&&\;\text{and $b_2 \equiv 2\,(\text{mod}\;4)$]}\\[4pt] \implies\; a_2 &\equiv 0 \pmod 4 &&\text{[since $a_0$ is odd,}\\[-1pt] &&&\;\text{and $a_2$ is even]}\\[4pt] \text{Using $(\text{eq}3),$}\;\,3a_0 + a_1b_2 + a_2b_1 &= 5\\[4pt] \implies\; 3a_0 + 2 + 0 &\equiv 5 \pmod 4 &&\text{[since $a_1$ is odd,}\\[-1pt] &&&\;\text{$b_2 \equiv 2\,(\text{mod}\;4)$,}\\[-1pt] &&&\;\text{and $a_2,b_1$ are even]}\\[-1pt] \implies\; a_0 &\equiv 1 \pmod 4\\[4pt] \implies\; a_0 &= 1 &&\text{[since $a_0 \in \{\pm 1\}$]}\\[12pt] \text{Using $(\text{eq}1),$}\;\,a_0(a_1 + b_1) &= 3\\[4pt] \implies\; a_1 + b_1 &= 3 &&\text{[since $a_0=1$]}\\[4pt] \implies\; a_1 + 2 &\equiv 3 \pmod 4 &&\text{[since $b_1 \equiv 2\,(\text{mod}\;4)$]}\\[4pt] \implies\; a_1 &\equiv 1 \pmod 4\\[12pt] \text{Using $(\text{eq}5),$}\;\,2a_2 + b_2 &=6\\[4pt] \implies\; 0 + b_2 &\equiv 6 \pmod 8 &&\text{[since $a_2 \equiv 0\,(\text{mod}\;4)$]}\\[4pt] \implies\; b_2 &\equiv 6 \pmod 8\\[12pt] \text{Using $(\text{eq}4),$}\;\, 2a_1 + b_1 + a_2b_2 &=4\\[4pt] \implies\; 2 + b_1 &\equiv 4 \pmod 8 &&\text{[since $a_1 \equiv 1\,(\text{mod}\;4)$,}\\[-1pt] &&&\;\text{$a_2 \equiv 0\,(\text{mod}\;4)$,}\\[-1pt] &&&\;\text{and $b_2$ is even]}\\[-1pt] \implies\; b_1 &\equiv 2 \pmod 8\\[12pt] \text{Using $(\text{eq}3),$}\;\, 3a_0 + a_1b_2 + a_2b_1 &= 5\\[4pt] \implies\; a_1b_2 + a_2b_1 &= 2 &&\text{[since $a_0=1$]}\\[4pt] \implies\; a_1b_2&\equiv 2 \pmod 8 &&\text{[since $a_2 \equiv 0\,(\text{mod}\;4)$,}\\[-1pt] &&&\;\text{and $b_1$ is even]}\\[4pt] \implies\; 6a_1&\equiv 2 \pmod 8 &&\text{[since $b_2 \equiv 6\,(\text{mod}\;8)$]}\\[4pt] \implies\; 3a_1&\equiv 1 \pmod 4\\[4pt] \implies\; 3 &\equiv 1 \pmod 4 &&\text{[since $a_1 \equiv 1\,(\text{mod}\;4)$]}\\[4pt] \text{c}&\text{ontradiction} \end{align*}

Therefore $f$ is irreducible.

Answer 3

As others pointed out we can use modular factorization. I think that a quinella of modulo two and five gives us a winning betting slip with the least amount of paper-and-pencil work.

Modulo $2$ we have $$ f(x)\equiv x^3+x+1 $$ which is well known to be irreducible. Like quasi (+1) we conclude that if we have a factorization $f(x)=g(x)h(x)$ in $\Bbb{Z}[x]$ then we must have $g(x)\equiv x^3+x+1$ and $h(x)\equiv 1$. Here the leading coefficient of $h$ must be even, so $g$ must have an odd leading coefficient. Therefore $\deg g(x)=3=\deg h(x)$.

Let's look at it modulo $5$. $$ f(x)\equiv2x^6+x^5-x^4-2x+1. $$ We see that $f(-2)\equiv0\pmod5$ so in $\Bbb{Z}_5[x]$ $f$ is divisible by $(x+2)$. By polynomial division we get $$ f(x)=2(x+2)(x^5+x^4-1)\in\Bbb{Z}_5[x]. $$ That degree five factor, denote it by $m(x)$, is actually irreducible in $\Bbb{Z}_5[x]$. There are many ways to see this. The simplest may be to observe that the reciprocal of $m(x)$ is $$ \tilde{m}(x)=x^5m(\frac1x)=-(x^5-x-1). $$ This falls under the umbrella of Artin-Schreier polynomials. A standard exercise is to show that whenever $a\not\equiv 0\pmod p$, $p$ a prime, the polynomial $$ x^p-x-a $$ is irreducible in $\Bbb{Z}_p[x]$. See here for a variety of arguments with varying degrees of sophistication proving this result. With $\tilde{m}(x)$ known to be irreducible we can conclude that $m(x)$ is irreducible also.

Anyway, with the knowledge that modulo five $f$ has an irreducible factor of degree five, we have run into a contradiction with the modulo two conclusion that the only possible factorization is the product of two cubics. Therefore $f$ is irreducible.

Answer 4

Using the Berlekamp algorithm for polynomials $f\in \mathbb{F}_p[x]$, we see that $f(x)$ is irreducible for $p=29, 31, 47$, etc., see Bemte's comment. Hence $f(x)$ is irreducible in $\mathbb{Z}[x]$.

Answer 5

$f$ has degree $6$ and assumes prime values for these $13 > 2 \cdot 6$ points and so must be irreducible: $$ \begin{array}{rl} n & f(n) \\ 2 & 431 \\ 14 & 18453443 \\ 17 & 57152981 \\ 21 & 196861141 \\ 30 & 1607175091 \\ 39 & 7588383193 \\ 89 & 1027721432693 \\ 105 & 2757260165941 \\ 122 & 6757665621431 \\ 131 & 10340492467001 \\ 161 & 35484310770581 \\ 182 & 73889824920731 \\ 201 & 133863373175221 \\ \end{array} $$