<s>[INST] Prove that $f=x^4-4x^2+16\in\mathbb{Q}[x]$ is irreducible 
Prove that $f=x^4-4x^2+16\in\mathbb{Q}[x]$ is irreducible.

I am trying to prove it with Eisenstein's criterion but without success: for p=2, it divides -4 and the constant coefficient 16, don't divide the leading coeficient 1, but its square 4 divides the constant coefficient 16, so doesn't work. Therefore I tried to find $f(x\pm c)$ which is irreducible:

$f(x+1)=x^4+4x^3+2x^2-4x+13$, but 13 has the divisors: 1 and 13, so don't exist a prime number p such that to apply the first condition: $p|a_i, i\ne n$; the same problem for $f(x-1)=x^4+...+13$
For $f(x+2)=x^4+8x^3+20x^2+16x+16$ is the same problem from where we go, if we set p=2, that means $2|8, 2|20, 2|16$, not divide the leading coefficient 1, but its square 4 divide the constant coefficient 16; again, doesn't work.. is same problem for x-2

Now I'll verify for $f(x\pm3)$, but I think it will be fall... I think if I verify all constant $f(x\pm c)$ it doesn't work with this method...  so have any idea how we can prove that $f$ is irreducible?
 [/INST] Below is an explicit proof. Note that $x^4-4x^2+16 = (x^2-2)^2 + 12$, which clearly has no real root. Hence, the only possible way to reduce $x^4-4x^2+16$ over $\mathbb{Q}$ is $(x^2+ax+b)(x^2+cx+d)$. However, the roots of $x^4-4x^2+16$ are $x = \pm \sqrt{2 \pm i\sqrt{12}}$, which are all complex numbers. Since complex roots occur in conjugate pairs, $\sqrt{2 \pm i\sqrt{12}}$ must be the roots of one of the factored quadratic. Hence, the factored quadratic must be
$$x^2-(\sqrt{2+i\sqrt{12}}+\sqrt{2-i\sqrt{12}})x + \sqrt{2+ i\sqrt{12}}\sqrt{2- i\sqrt{12}} = x^2-2\sqrt3 x+4$$
The other factored quadratic must be
$$x^2+(\sqrt{2+i\sqrt{12}}+\sqrt{2-i\sqrt{12}})x + \sqrt{2+ i\sqrt{12}}\sqrt{2- i\sqrt{12}} = x^2+2\sqrt3 x+4$$
Hence, $x^4-4x^2+16$ is irreducible over $\mathbb{Q}$.
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