1. ## Irreduclible polynomial

Can somebody please check if the follwoing argument is correct:

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

By the rational root theorem it has no rational roots.

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

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

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

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?

2. Originally Posted by peteryellow
$a=-c$ implies that $d+b-c^2=0$
$c(b-d)=-4$ $bd=-1$

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

3. Originally Posted by peteryellow
Can somebody please check if the follwoing argument is correct:

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

By the rational root theorem it has no rational roots.

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

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

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