Prove that $f=x^4-4x^2+16\in\mathbb{Q}$ 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?

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The associated quadratic polynomial $t^2-4t+16$ has negative discriminant, so there’s no real root. Then the polynomial can be factorized over the reals as a product of degree two polynomial. You get them by a process similar to completing the square:
These two polynomials have negative discriminant (no need to verify it) and so they’re irreducible in $\mathbb{R}[x]$. If the given polynomial were reducible over the rationals, the two factorizations in $\mathbb{Q}[x]$ and $\mathbb{R}[x]$ would coincide.

Therefore the given polynomial is irreducible over the rationals.

What’s the general rule? Suppose you have $x^4+px^2+q$, with $p,q$ integers and $p^2-4q<0$ (so $q>0$). Write $q=r^2$, with $r>0$ (it need not be integer), and
Note that $2r-p>0$: it’s obvious if $p<0$; if $p\ge0$ it’s the same as $4q>p^2$, which is true by hypothesis. Then
is the decomposition of the polynomial in $\mathbb{R}[x]$. It is in $\mathbb{Q}[x]$ if and only if $\sqrt{q}$ and $\sqrt{2\sqrt{q}-p}$ are integers.

For example, $q=4$ and $p=0$ is a case. For $q=16$ we need $8-p$ to be a square, so $q=16$ and $p=4$ is another case.

The polynomial has no real roots, because it is equal to $(x^2-2)^2+12$. The remaining possibility is thus that it is a product of two quadratic factors. By Gauss’ Lemma these need to have integer coefficients, so we are looking for a possibile factorization like
with some integers $a,b,c,d$. Modulo $3$ we have the factorization
$$p(x)=(x^2-2)^2+12\equiv(x^2+1)^2.$$ Therefore $a$ and $c$ must both be divisible by three, and $b\equiv d\equiv 1\pmod3$. Modulo $5$ we have
p(x)\equiv x^4+x^2+1=(x^2+1)^2-x^2=(x^2+x+1)(x^2-x+1).
This means that $b\equiv d\equiv1\pmod 5$ as well. The Chinese Remainder Theorem (or case-by-case check) then shows that $b\equiv d\equiv 1\pmod{15}$.

Because $bd=16$ the only remaining possibility is that they are $1$ and $16$ in some order. But this is impossible because modulo $2$ we have
$$p(x)\equiv x^4,$$ so all of $a,b,c,d$ must be even.

The conclusion is that $p(x)$ is irreducible.

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 $\mathbb{Q}$.

You’ve seen that $f(x)$ has no roots, so you want to exclude factorizations of the form
$$f(x) = (x^2 + ax + b)(x^2 + cx + d)$$
Since $f(x) = f(-x)$, the above implies
$$f(x) = (x^2 – ax + b)(x^2 – cx + d)$$
Here $a,b,c$, and $d$ are integers by Gauss’s Lemma.

So a given root $r$ of $x^2 – ax + b$ is a root of $x^2 + ax + b$ or $x^2 + cx + d$.

If $r$ is a root of $x^2 + ax + b$, it is the root of the difference $x^2 + ax + b – (x^2 – ax + b) = 2ax$, which implies $a = 0$ since zero is not a root of $f(x)$.

If $r$ is a root of $x^2 + cx + d$ it is similarly a root of the difference $(c + a)x + (d – b)$, and since $f(x)$ has no rational roots $c = -a$ and $d = b$.

So either $a = 0$ or $c = -a$ and $d = b$. Since the argument is entirely symmetric in the two factors, we also either have $c = 0$ or $c = -a$ and $d = b$. Hence we have two possibilities: Either $a = c = 0$ or $c = -a$ and $d = b$.

In the first case we have
$$x^4 – 4x^2 + 16 = (x^2 + b)(x^2 + d)$$
But the roots of $y^2 – 4y + 16$ are irrational (they’re not even real) so this can’t happen.

In the second case we have
$$x^4 – 4x^2 + 16 = (x^2 + ax + b)(x^2 – ax + b) = x^4 + (2b – a^2) x^2 + b^2$$
Hence $b = \pm 4$ and therefore either $8 – a^2 = -4$ or $-8 – a^2 = -4$, neither of which has rational solutions.

Hence $f(x)$ is irreducible.

Suppose that $f$ is reducible, observe that $f$ is monic, hence $f$ can be written as the product of two polynomials $g$ and $h$ of degree at least one and with integer coefficients.
Now observe that $f$ has no integer root, because with rational root theorem an integer root to $f$ must be a divisor of $f(0)=16$ and you can check it’s cases.
So $f$ has no linear factor, hence $g$ and $h$ are two monic polynomials with degree two, with letting
We get
Now if $a\neq0$, then $b=d$,$b+d+ac=-4$, so $b^2=16$, $2b-a^2=\pm8-a^2=-4$, which contradicts.
Now if $a=-c=0$, then $bd=16$, and $b+d=-4$ which again contradicts.

As Jyrki Lahtonen observed in his answer, $p(x)=x^4-4x^2+16=(x^2-2)^2+12$ has no real roots, so the only possible factorization is of the form


which expands to


We conclude that $c=-a$ (from the $x^3$ term), hence* $d=b$ (from the $x$ term), which means $2b-a^2=-4$ and $b^2=16$. Plugging the two possibilities for $b$ into $a^2=2b+4$ gives $a^2=12$ or $-4$, neither of which corresponds to an integer value for $a$.

*Zarrax in comments astutely observes my “hence” is mistaken. It ignores the possibility $a=c=0$. To complete the proof that $p(x)$ is irreducible, we need to note that if $a=c=0$ then, letting $u=x^2$, we would have a factorization


My thanks to Zarrax for pointing out the error.

I think you should try using a method other than Eisenstein’s criterion. By Gauss’ lemma it suffices to show that $f(x)$ does not factor over the integers, so you need only show the following:

  • There is no integer root: ie $f(x) \ne 0 $ for $x$ an integer dividing $16$.
  • There is no quadratic factorization: You cannot write $f(x) = (x^2+ax+b)(x^2+cx+d)$.

The first of these is straightforward to verify, and deriving a contradiction from the second is not unreasonable with this particular $f(x)$.


No rational roots and no factorization into quadratics over the rationals. The polynomial is irreducible over the rationals

edit for those who commented that this is not enough. I factorized over $\mathbb{C}[X]$ and thus proved that there are no rational solution i.e no degree $1$ factors. The only factorization possible is therefore into two quadratics. $\mathbb{C}[X]$ is a UFD and therefore we have


And this is unique. So combining the degree 1 factors in pairs is the only way to factorize in quadratics and there are three different ways to combine and none is rational

f(x) = 0 leads to $x^4$ +16 = 4$x^2$ then for |x| = 0,1, 2 and for |x| ≥ 3 the equality is impossible.Then x can’t be integer. On the other hand we have
($x^2$ – 2)^2 = -12 which gives $x^2$ = 2(1 +sqrt(-3)) and x can’t be rational. What remains is to prove that f has not two quadratic factors of Z[x] wich can be proved with undetermined coefficients. I stop here.