<s>[INST] $x^4 -10x^2 +1 $ is irreducible over $\mathbb Q$ I have seen the thread Show that $x^4-10x^2+1$ is irreducible over $\mathbb{Q}$ but this didn't really have a full solution.
Is it true that if it is reducible then it can be factored into a linear factor or quadratic factor in the form $x^2 - a$. Is $a$ in the rationals? And what exactly is a linear factor.
What I did was the following:
Let $y=x^2$. Then $y^2 - 10y +1=0 \iff y^2 -10y = -1$, add $25$ to both sides:
$$y^2 - 10y + 25 = -1 + 25 \iff  (y - 5)^2 = 24 \iff y = 5 \pm 2\sqrt6$$
So $x^2=5 \pm 2 \sqrt6= (\sqrt2 \pm \sqrt3 )^2$
So $x^4 -10x^2 +1 =(x^2 -(\sqrt2 + \sqrt3 )^2)(x^2 - (\sqrt2 - \sqrt3)^2)$
Then difference of two squares:
$ (x-(\sqrt2 + \sqrt3))(x+(\sqrt2 + \sqrt3))(x-(\sqrt2 - \sqrt3))(x+(\sqrt2 - \sqrt3))=0$
Expanding this out gives $x^4 -10x^2 +1 =0$ so we have found all the roots and we can confirm that they are the roots right? None of these roots are rational so it must be irreducible?
Please can you advise me on this method that I am using and not direct me to a theorem or any type of shortcut. 
Thanks in advance.
 [/INST] First note that it suffices to show the polynomial is irreducible over $\mathbb{Z}$. We then note that $p(x)=x^4-10x^2-1$ splits either as
$$(x-a)(x^3+bx^2+cx+d)$$
or as
$$(x^2+bx+c)(x^2+dx+e)$$
In the first case $a$ would be a root of $p$, but by the rational root theorem the polynomial has no roots in $\mathbb{Q}$ (and hence not in $\mathbb{Z}$). In the second case
$$(x^2+bx+c)(x^2+dx+e)=x^4+(b+d)x^3+(e+c+bd)x^2+(be+dc)x+ec$$
so $e=c=\pm 1$, $b=-d$, filling this in we find
$$x^4+(b+d)x^3+(e+c+bd)x^2+(be+dc)x+ec=x^4+(2-b^2)+1$$
and $2-b^2=-10$ has no solution in $\mathbb{Z}$.
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