Thread: Irreduclible polynomial

Can somebody please check if the follwoing argument is correct:

Show that $\displaystyle x^4-4x-1$ is irrdeucible over rationals.

By the rational root theorem it has no rational roots.

Assume that
$\displaystyle x^4-4x-1 = (x^2+ax+b)(x^2+cx+d)$ then

$\displaystyle a+c=0, d+ac+b=0 ad+bc=-4 bd=-1.$


$\displaystyle a=-c$ implies that $\displaystyle d+b-c^2=0$
$\displaystyle c(b-d)=-4$ $\displaystyle bd=-1$

c=-4 or b-d=-4.
Because bd=-1 it must hold that b=1,d=-1 or b=-1, d=1.

c=-4 shows that d+b=16 which is a contradiction.
if b-d=-4 we also get a contradiction. So the polynomial is irreducible.

Is it correct?

Originally Posted by peteryellow

$\displaystyle a=-c$ implies that $\displaystyle d+b-c^2=0$
$\displaystyle c(b-d)=-4$ $\displaystyle bd=-1$

c=-4 or b-d=-4.
Because bd=-1 it must hold that b=1,d=-1 or b=-1, d=1.

Since $\displaystyle bd=-1$ it means $\displaystyle b=1,d=-1$ or $\displaystyle b=-1,d=1$. In the first case it means $\displaystyle b-d = 2$ and so from $\displaystyle c(b-d)=-4\implies c=-2$. In the second case it means $\displaystyle b-d =-2$ and so from $\displaystyle c(b-d)=-4\implies c=2$. But we must have $\displaystyle c^2 = b+d$ however, $\displaystyle b+d=0$ and $\displaystyle c^2=4$ in both cases. Thus, it is impossible to find such integers.

Originally Posted by peteryellow

Can somebody please check if the follwoing argument is correct:

Show that $\displaystyle x^4-4x-1$ is irrdeucible over rationals.

By the rational root theorem it has no rational roots.

Assume that
$\displaystyle x^4-4x-1 = (x^2+ax+b)(x^2+cx+d)$ then

$\displaystyle a+c=0,\ d+ac+b=0\ ad+bc=-4\ bd=-1.$


$\displaystyle a=-c$ implies that $\displaystyle d+b-c^2=0$,
$\displaystyle c(b-d)=-4$, $\displaystyle bd=-1$ Correct as far as here!

c=-4 or b-d=-4. NO!!!
Because bd=-1 it must hold that b=1,d=-1 or b=-1, d=1.

c=-4 shows that d+b=16 which is a contradiction.
if b-d=-4 we also get a contradiction. So the polynomial is irreducible.

Is it correct?

My solution: from $\displaystyle b+d=c^2$ and $\displaystyle bd=-1$, deduce that b and d are the solutions of the quadratic equation $\displaystyle t^2 - c^2t -1=0$. But these solutions are $\displaystyle t = \tfrac12(c^2\pm\sqrt{c^2+4})$. It follows that $\displaystyle b-d = \pm\sqrt{c^2+4}$, and so $\displaystyle c\sqrt{c^2+4} = \pm4$. Therefore $\displaystyle c^2(c^2+4) = 16$. This is a quadratic equation for $\displaystyle c^2$, with irrational solutions $\displaystyle c^2 = -2\pm\sqrt{20}$. So there is no rational solution for c.