Prove $f(x)=x^8-24 x^6+144 x^4-288 x^2+144$ is irreducible over $\mathbb{Q}$ 
How to prove $f(x)=x^8-24 x^6+144 x^4-288 x^2+144$ is irreducible over $\mathbb{Q}$?

I tried Eisenstein criteria on $f(x+n)$ with $n$ ranging from $-10$ to $10$. None can be applied. I tried factoring over mod $p$ for primes up to $1223$. $f(x)$ is always reducible over these. 
$f(x)$ has roots $\pm\sqrt{\left(2\pm\sqrt{2}\right) \left(3\pm\sqrt{3}\right)}$, and according to computation by PARI, should have Galois group isomorphic to the quaternion group. The splitting field of $f(x)$ is $\mathbb{Q}(\sqrt{\left(2+\sqrt{2}\right) \left(3+\sqrt{3}\right)})$, and it contains $\sqrt{2}, \sqrt{3}, \sqrt{6}$, so we know $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is in the splitting field, so $\mathbb{Q}(\sqrt{\left(2+\sqrt{2}\right) \left(3+\sqrt{3}\right)})$ has degree $4$ or $8$.
I tried showing the degree is $8$ by showing that $(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})^2=(2+\sqrt{2})(3+\sqrt{3})$ cannot have a solution with $a, b, c, d \in \mathbb{Q}$, and got these equations:
$$a^2+2b^2+3c^2+6d^2=6$$
$$2ab + 6cd = 3$$
$$ac+2bd = 1$$
$$2ad+2bc = 1$$
which I'm unable to handle.
Addendum: now that I've solved this problem thanks to the answers, I found some additional related information: 
In A Rational Polynomial whose Group is the Quaternions, a very similar polynomial, $$f(x)=x^8 - 72 x^6 + 180 x^4 - 144 x^2 + 36$$ is studied and its Galois group is proven to be the quaternion group. I subjected this polynomial, as well as two related ones: $f(\sqrt{x})$, $f(6\sqrt{x})/36$, to the same battery of tests (Eisenstein; mod p) and these tests also failed to show them to be irreducible. Maybe there's something common about these polynomials.
So I subjected $f(x)$ to the prime numbers test demonstrated by Robert Israel, and found that it is prime at $\pm\{7, 13, 23, 25, 49, 53, 55, 79, 91, 127, 139, 145, 151, 181, 239, 251, 277, 283, 319, 355, 379, 403, 413, 425, 473, 485, 595, 607, 623, 679, 733, 743, 779, 827, 851, 923, 965, ...\}$ and thus $f(x)$ is irreducible.
 A: You just have to show that $f(x)$ is the minimal polynomial of $\eta=\sqrt{(2+\sqrt{2})(3+\sqrt{3})}$ over $\mathbb{Q}$, i.e. to show that $\eta$ is an algebraic number over $\mathbb{Q}$ with degree $8$. It is pretty straightforward to prove that $\eta^2$ is an algebraic number over $\mathbb{Q}$ with degree $4$, hence we just need to rule out $\eta\in\mathbb{Q}(\sqrt{2},\sqrt{3})$. If that were the case, for any large enough prime $p$ such that both $2$ and $3$ are quadratic residues ($p=24k+1$ is a sufficient condition) we would have that $(2+\sqrt{2})(3+\sqrt{3})$ is a quadratic residue too. That contradicts quadratic reciprocity, and for an explicit counterexample, by considering $p=73$ we get that $21^2\equiv 3\pmod{p}$, $41^2\equiv 2\pmod{p}$ but $(2+41)\cdot(3+21)$ is not a quadratic residue $\!\!\pmod{p}$. There are an infinite number of such counterexamples, hence $\eta\not\in\mathbb{Q}(\sqrt{2},\sqrt{3})$ and $f(x)$ is the minimal polynomial of $\eta$ over $\mathbb{Q}$. In particular, $f(x)$ is irreducible over $\mathbb{Q}$.
It is interesting to point out that "standard tricks" do not work here since $f(x)$ factors over any finite field $\mathbb{F}_p$. This kind of polynomial is known as Hilbert polynomial, if I recall it correctly.
A: This polynomial and the field $\Bbb{Q}(\alpha)$, $\alpha=\sqrt{(2+\sqrt2)(3+\sqrt3)}$ recur in exercises (it may be one of the simplest to describe fields with Galois group $Q_8$). I had a reason to look for a more elementary proof that $\alpha\notin\Bbb{Q}(\sqrt2,\sqrt3)$ which is equivalent to the claim. Can't say that my argument would be as elementary as I want it to be. But it is different from those posted already, so I will share.

As a stepping stone I use reduction modulo five. The goal is to show that modulo five we have the factorization
$$\overline{f}(x)=(x^4+2)(x^4+x^2+2),\tag{1}$$
where both quartic factors are irreducible modulo $5$. It is, of course, trivial to verify this factorization. I sidestep a bit and explain a way to find this.

The field $K=\Bbb{F}_{5^2}=\Bbb{F}_5(\sqrt3)$ contains square roots for both $2$ and $3$. Because $3\equiv2^2\cdot2$ we can think of $\sqrt3$ as $2\sqrt2$, but this is really a choice of sign. With that choice made, in the field $K$ we have
$$\alpha^2=(2+\sqrt2)(3+\sqrt3)=6+3\sqrt2+2\sqrt3+\sqrt2\sqrt3=10++7\sqrt2=2\sqrt2.$$
Therefore $\alpha^4=3$ and consequently $\alpha^4+2=0$. This gives the first modular factor and we get the other by polynomial division.
Then I explain why the two quartic factors are irreducible modulo five. Because $\pm2$ are fourth roots of unity modulo five, any root of $x^4+2$ (in some extension field of $K$) must be a root of unity of order sixteen. But $\Bbb{F}_{5^4}$ is the smallest extension of $\Bbb{F}_5$ containing such roots of unity, and hence their minimal polynomial over the prime field is quartic. There are many ways to show that $u(x)=x^4+x^2+2$ is also irreducible over the prime field. The roots of $x^2+x+2$ in $K$ are $2\pm\sqrt2$, and the roots of $u(x)$ are square roots of these. It is easy to verify that neither of those has a square root in $K$, so they, too, reside in $\Bbb{F}_{5^4}\setminus\Bbb{F}_{5^2}$ proving the claim.

With the modulo five factorization clear, the rest is relatively simple. The factorization $(1)$ implies that the only possible way $f(x)$ can factor is as a product of two irreducible quartic polynomials $g(x),h(x)\in\Bbb{Z}[x]$. We can further assume that the factors are monic.
Observe that $f(x)$ is even, so $f(x)=f(-x)=g(-x)h(-x)$. Because the factorization into irreducible polynomials in $\Bbb{Z}[x]$ is unique, we have two possibilities: either $g(-x)=h(x)$ (when also $h(-x)=g(x)$) or $g(-x)=g(x)$ and $h(-x)=h(x)$.

*

*It is impossible to have $g(-x)=h(x)$ because such a relation would persist after reduction modulo five, and $(1)$ shows that this is not the case.

*If $g(x)=g(-x)$ and $h(x)=h(-x)$, then both $g$ and $h$ contain even degree terms only. That is $g(x)=r(x^2)$ and $h(x)=s(x^2)$ for some quadratic polynomials $r(x),s(x)\in\Bbb{Z}[x]$. This would imply that
$$\alpha^2=(2+\sqrt2)(3+\sqrt3)$$
is a root of a quadratic (either $r$ or $s$). But $\alpha^2$ is not an element of any of $\Bbb{Q}(\sqrt n)$, $n=2,3,6$. By well-known Galois theory of $M=\Bbb{Q}(\sqrt2,\sqrt3)$ those are the only quadratic subfields of $M$, so we arrive at a contradiction in this case also.

A: Assume that $f(x) = f_1(x) f_2(x)$.
Note that $f(1) = - 23$, so one of $f_1(1)$, $f_2(1) = \pm 1$, and the other $\mp 23$.
The polynomial factors $\mod 5$ as $(x^4+2)(x^4 + x^2 + 2)$ ( see Jyrki's answer). Say we have $f_1(x)\equiv x^4 + 2$ and $f_2(x) \equiv x^4 + x^2 + 2 \mod 5$.  Now, $f_1(1) \equiv 1^4 + 2 = 3 \mod 5$, and $f_2(1) \equiv 1^4 + 1^2 + 2 \equiv -1 \mod 5$. We conclude that $f_1(1) = 23$, and $f_2(1) = -1$.
Now, $f(x)$ factors $\mod 3$ as $x^8$. We  have $f_1(x) \equiv x^4$, and $f_2(x) \equiv x^4 \mod 3$. This implies $f_1(1), f_2(1) \equiv 1 \mod 3$.  We got a contradiction
$\bf{Added:}$ Another proof:  Assume that $(2+\sqrt{2})(3+ \sqrt{3})$ is  a square in $\mathbb{Q}(\sqrt{2}, \sqrt{3})$. Taking the norm to $\mathbb{Q}(\sqrt{2})$ we get
$(2+\sqrt{2})^2 (3+ \sqrt{3})(3-\sqrt{3})$ is a square in $\mathbb{Q}(\sqrt{2})$, so $6$ is, contradiction.
A: Your polynomial $f(x)$ takes prime (or $-$ prime) values at $x = \pm 1, \pm 7, \pm 11, \pm 13, \pm 23, \pm 67, \pm 85, \pm 109, \pm 145, \pm 197, \pm 205, \pm 209, \pm 241, \pm 373, \pm 397, \pm 403, \pm 421$.  That's $34$ points.  If it factored as $f(x)=g(x) h(x)$, one of $g$ and $h$ must be $\pm 1$ at at least $17$ of these $x$, and either $+1$ at at least $9$ points or $-1$ at at least $9$ points.  But a non-constant polynomial that takes the same value at $9$ points must have degree at least $9$, and $f$ has degree only $8$.
A: It seems that you have proved that $\mathbb{Q}(\sqrt{\left(2+\sqrt{2}\right) \left(3+\sqrt{3}\right)})/\mathbb{Q}$ has degree $8$, and that $\sqrt{\left(2+\sqrt{2}\right) \left(3+\sqrt{3}\right)}$ is a root of $f$,   right ?
If so, then you may apply the following result: 
Lemma. Let $K(\alpha)/K$ be an extension of degree $n$. If $f\in K[X] $ is a monic polynomial satisfies $f(\alpha)=0$ and
$\deg(f)=n,$ then $f=\mu_{\alpha,K}$ (the minimal polynomial of $\alpha$ over $K$). In particular, $f$ is irreducible over $K.$
Proof. We have $[K(\alpha):K]=n=\deg(\mu_{\alpha,K}).$  Since $f(\alpha)=0$,  $\mu_{\alpha,K}\mid f$. But $f$ is monic and has degree $n$. Then $f=\mu_{\alpha,K}.$
