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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.

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    $\begingroup$ The polynomial is the $12$ th cyclotomic polynomial. $\endgroup$
    – Peter
    Sep 2, 2017 at 18:44
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    $\begingroup$ Or better note that the given polynomial can be written as $x^{4}+2x^{2}+1-3x^{2}$ and thus factorize it as $(x^{2}+\sqrt{3}x+1)(x^{2}-\sqrt{3}x+1)$. $\endgroup$
    – Paramanand Singh
    Sep 2, 2017 at 18:47
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    $\begingroup$ For finding its roots I recommend writing it in the form $(x^6+1)/(x^2+1)$. Assuming you know how to use complex polar form to find all the roots. $\endgroup$ Sep 2, 2017 at 18:49
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    $\begingroup$ @RoddyMacPhee: Since the factorization is unique (via fundamental theorem of algebra) it is clear that the factorization is not possible over $\mathbb{Q} $. $\endgroup$
    – Paramanand Singh
    Sep 2, 2017 at 18:50
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    $\begingroup$ Curiously, this polynomial is not irreducible modulo any prime. That is because its Galois group has no elements of order four (the Galois group is Klein four). Alternatively, the fact $12\mid (p^2-1)$ for all primes $p>3$ implies that $\Phi_{12}$ has a zero in $\Bbb{F}_{p^2}$ and hence a quadratic factor modulo $p$. $\endgroup$ Sep 2, 2017 at 18:55

5 Answers 5

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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 $a^2=3$.

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  • $\begingroup$ The polynomial is $x^4-x^2+1$, not $x^4+x^2+1$. $\endgroup$
    – Dave
    Sep 2, 2017 at 18:59
  • $\begingroup$ Whoops, yes, fixed. Same argument applies, and actually works for $x^4-x^2+1$. :) $\endgroup$ Sep 2, 2017 at 19:01
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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.

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$$(x^4-x^2+1)(x^2+1) = x^6+1$$ 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.

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    $\begingroup$ You were may be the fourth or fifth viewer to make the observation that this is $\Phi_{12}$. So without a proof this is IMO worth a comment only. $\endgroup$ Sep 2, 2017 at 21:08
  • $\begingroup$ @JyrkiLahtonen: a proof of what? The irreducibility of cyclotomic polynomials should be well-known, and the equality between $\Phi_{12}$ and $x^4-x^2+1$ is proved through the shown decomposition, matching the decomposition given by the Moebius inversion formula. $\endgroup$ Sep 2, 2017 at 21:12
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    $\begingroup$ Yes, it is well known, but apparently not to the asker, so it is not useful to just bluntly state that this polynomial is irreducible. Not even a link. And the proof is not as simple as you claim. Of course, by assuming that the Galois group is what it is, the proof does become trivial. $\endgroup$ Sep 2, 2017 at 21:27
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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 $\mathbb{Q}[x]$.

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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.

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