Show that $x^4-x^2+1$ is irreducible over $\mathbb{Q}$ My attempts: 
I cannot apply the Eisenstein's criteria here, because there is no prime number that divides the constant term i.e. $1$ Taking a translation of the form $x \rightarrow x+a$ does not solve this issue either.  
Next, I tried the mod tests: $\operatorname{mod}2$ doesn't work since $x^4-x^2+1=(x^2+x+1)(x^2-x+1)$, similarly in $\operatorname{mod}3$ $x^4-x^2+1=(x^2+1)^2$. Now I can go on and maybe eventually find a $\operatorname{mod}p$ that works, but that is very time consuming, specially in examinations. 
So, I'll use the rational root test. The possibilities for roots are $\pm 1$ and it is easy to see that neither is a root.
The only possibility left then are quadratic factors, say, $(x^2+ax+b)(x^2+cx+d)=x^4-x^2+1$
This gives me a set of equations $bd=1, a+c=0, b+d+ac=-1$. So either $b=d=1$, in which case $a=\pm \sqrt3 \notin \mathbb{Q}$, or $b=d=-1$, which gives $a=\pm i \notin \mathbb{Q}$. 
So such factorization is not possible and hence the given polynomial is irreducible.
Is this solution correct? Also, is there an easier way to solve this? Thank you. 
 A: Another approach which I saw Robert Israel use here would be to note that $x^4-x^2+1$ takes on prime values for $x=\pm2,\pm3,\pm4,\pm5$ and $\pm9$.  That's ten points, so that one of the quadratic factors would have to take on the value $\pm 1$ at least five times.  Finally one of the quadratic factors would have to take on either $+1$ or $-1$ at least 3 times which is impossible for a quadratic, since a non-constant polynomial that takes the same value three times must have degree at least three.
A: $$(x^4-x^2+1)(x^2+1) = x^6+1$$
implies:
$$ x^4-x^2+1 = \frac{x^6+1}{x^2+1} = \frac{(x^{12}-1)(x^2-1)}{(x^6-1)(x^4-1)} = \Phi_{12}(x) $$
hence the LHS is irreducible over $\mathbb{Q}$ since it is the minimal polynomial of $\exp\left(\frac{2\pi i}{12}\right)$.
The irreducibility of cyclotomic polynomials is a well-known fact, proved here.
A: Let $r(x)$ be the resolvent cubic of your polynomial. Then $r(x)=x^3-2x^2-3x$. The roots of $r(x)$ are $-1$, $0$, and $3$, none of which is the square of a non-zero rational number. Furthermore, your polynomial has no rational root and the coefficient of $x$ in $r(x)$ is not a perfect square in $\mathbb Q$. Therefore your polynomial is irreducible in $\mathbb{Q}[x]$.
A: There are no rational roots, so no linear factors.
If $p(x)$ is a factor of $x^4-x^2+1$ then $p(-x)$ is, too.
If $x^4-x^2+1 = (x^2+a)(x^2+b)$ then $x^2-x+1=(x+a)(x+b)$. Show that $x^2-x+1$ is irreducible.
On the other hand, you'd have to have $x^4-x^2+1=(x^2+ax+b)(x^2-ax+b)$ where $b^2=1$ and $a\neq 0$. This means that $x^4+(2b-a^2)x^2+b^2 = x^4-x^2+1$.
So you need $2b-a^2=-1$. If $b=-1,$ then this means $a^2=-1$, and if $b=1$ then $a^2=3$.
A: You can use wonderful criterion of Murty's (see Theorem 1):

Let $f(x)=a_mx^m+a_{m-1}x^{m-1}+\dots+a_1x+a_0$ be a polynomial of degree $m$ in $\mathbb{Z}[x]$ and set $$H=\max_{0\leq i\leq m-1} |a_i/a_m|.$$
  If $f(n)$ is prime for some integer $n\geq H+2$, then $f(x)$ is irreducible in $\mathbb{Z}[x]$. 

In this case it works because $f(3)=73$ is a prime.
