Showing $x^4-24x^2+4$ is irreducible over $\mathbb{Q}$

Question

I want to show that $x^4-24x^2+4$ is irreducible over $\mathbb{Q}$.

I have tried Eisenstein's criterion using $p=2$ and substituting $x+1$ and $x-1$ but it won't work. Then what I tried was using the rational root test. I got as my possible rational roots $\pm 1, \pm 2, \pm4$. But each one does not work. From here I can consider $x^4-24x^2+4$ over $\mathbb{Z}$. So suppose that $x^4-24x^2+4$ is reducible over $\mathbb{Z}$. Then $$x^4-24x^2+4= (x^2+ax+b)(x^2+cx+d)=x^4+(a+c)x^3+(b+ac+d)x^2+(bc+ad)x+bd.$$ By matching up the coefficients we get $$a+c=0$$ $$b+ac+d=-24$$ $$bd=4$$ $$bc+ad=0$$

Then we see $a+c=0\implies a=-c$. Then $$bc+ad=0\implies bc-cd=0 \implies bc=cd\implies b=d.$$ Since $bd=4$ then $b^2=4\implies b=\pm 2$ and $d=\pm 2$. We get two cases: $b=d=2$ and $b=d=-2$. In the first case $b=d=2$ we get $b+ac+d=-24\implies 4-c^2=-24\implies c^2=28$ (I substituted $a=-c$ into $b+ac+d=-24$), a contradiction since $c$ is an integer. Similarly in the case $b=d=-2$ we have $ b+ac+d=-24\implies -4-c^2=-24\implies c^2=20$. Again a contradiction. Would this be correct?

Answer 1

$x^4-24x^2+4=t^2-24t+4=0 \Rightarrow t=12\pm 2\sqrt{35}\Rightarrow x^2-(12\pm 2\sqrt{35})=0$

I was a bit excited (Not careless ) and concluded that given polynomial is irreducible.

But It is not proper to conlcude at this stage but rather find all other roots and then check for possible combinations.

$x^2=12+2\sqrt{35}\Rightarrow x= \pm(\sqrt{7}+\sqrt{5})$ and $x^2=12-2\sqrt{35}\Rightarrow x= \pm(\sqrt{7}-\sqrt{5})$

Now we have all roots of $x^4-24x^2+4$ as $\pm(\sqrt{7}\pm\sqrt{5})$

It would be now your duty to check if :

$(x-a)(x-b)\in\mathbb{Q}[x]$ where $a,b \in \{\pm(\sqrt{7}\pm\sqrt{5})\}$ and $a\neq b$.

So, now i will leave this to you conclude this polynomial is irreducible/not in $\mathbb{Q}[x]$

Danger : This worked out only because you have only $x^4$ and $x^2$ and no $x^3$.

This is done under experts supervision(It is not me). please do not repeat this at home. :D

Answer 2

Note that you have a monic even polynomial (it is invariant under substituting $x:=-x$) of degree $4$. You can therefore consider two cases for potential factors of the polynomial: factors that are themselves even, and factors that are not; the latter come in pairs whose members are interchanged by the substitution $x:=-x$ (which flips the signs of the odd degree terms). Any non-trivial even proper factor must be of the form $x^2-r$ where $r$ satisfies $r^2-24r+4=0$, but the discriminant $560=16*35$ of that polynomial is not a square in$~\Bbb Z$, so there are no such integer$~r$ (this is the case that the initial answer by Praphulla Koushik checked).

So you are left to check a potential factorisation $x^4-24x^2+4=(x^2+ax+b)(x^2-ax+b)$. This gives equations $b^2=4$ (from the constant term) and $a^2=2b+24$ (from the coefficient of $x^2$). It is easily checked that there are no solutions $a,b\in\Bbb Z$ to this system (neither $24+4$ nor $24-4$ is a square).