Is $X^8+3X^4-53$ irreducible over $\mathbb{Z}[X]$? I want to determine whether or not the polynomial $X^8+3X^4-53$ is irreducible over $\mathbb{Z}[X]$. I noticed that it doesn't have integer (or rational) roots but I have no further ideas.
 A: Following criterion of Osada can be applied:

Let $f(x)=x^n+a_1x^{n-1}+\dots+a_{n-1}x\pm p$ be a polynomial with integer coefficients, where $p$ is a prime. If $p>1+|a_1|+\dots+|a_{n-1}|$, then $f$ is irreducible.

Conditions are satisfied by $p=53$ and $p>1+3$.
The criterion can be found for example as Theorem 2.2.7 in Prasolov's book Polynomials.
A: The power of Osada's criterion (see Sil's answer) is apparent. A bit more elementary approach follows.
If $f(x)=x^8+3x^4-53$ factors over $\Bbb{Q}$, it factors over $\Bbb{Z}$, and hence also modulo any prime $p$. Modulo $p=2$ all we get is that the putative factors must be of even degree as $f(x)\equiv(x^2+x+1)^4\pmod 2$ (apply Freshman's Dream twice). Modulo $p=3$ we have
$$
f(x)\equiv x^8+1\pmod3.
$$
The zeros of $x^8+1$ in some extension field $K$ of $\Bbb{F}_3$ clearly have multiplicative order sixteen. The smallest $K$ containing sixteenth roots of unity is $\Bbb{F}_{81}$. This implies that the modulo three factors are of degree four. Indeed,
$$
x^8+1\equiv (x^8+4x^4+4)-4x^2=(x^4+2)^2-(2x)^2\equiv (x^4+x^2+2)(x^4+2x^2+2)\pmod3.
$$
What this tells us that the only possible factorization $g(x)h(x)=f(x)$ has $\deg g=\deg h=4$, and the irreducible factors $g,h$ must be congruent to the above factors modulo three. Assuming those factors exist we get that $g(0)h(0)=-53$. As $g(0)\equiv h(0)\equiv-1\mod 3$, we can deduce (interchanging the roles of $g$ and $h$ if necessary) that $g(0)=-1$ and $h(0)=53$.
But, $f(-x)=f(x)$, so we must also have $g(-x)h(-x)=g(x)h(x)$ implying (by irreducibility) that either $g(-x)=g(x)$ or $g(-x)=h(x)$. In view of what we know about the constant terms, it follows that we must have $g(x)=g(-x)$, and thus also $h(x)=h(-x)$. So both factors are also even, and
$$
g(x)=x^4+Ax^2-1,\qquad h(x)=x^4+Bx^2+53
$$
for some integers $A,B$.
Expanding $g(x)h(x)$ and looking at the terms of degree two and six gives us the constraints
$$
A+B=0\qquad\text{and} -B+53A=0.
$$
The only solution of this system of equations is $A=0=B$. But then the quartic term in $g(x)h(x)$ is wrong.
Therefore no factorization is possible.
A: Let $P=X^8+3X^4-53$. Write $P=QR$, where $Q,R$ are non trivial factors. Comparing leading coefficients, and multiplying by $-1$ if necessary, one may assume that $Q,R$ are monic.
Note that $Q(0)R(0)=-53$, so one may assume that $Q(0)=\pm 1$ for example (since $53$ is prime)
Let's reduce $P$ modulo $53$, so $\bar{P}=\bar{Q}\bar{R}$. Note that $\bar{Q}$ and $\bar{R}$ are monic as well, and has same degree as $Q$ and $R$ respectively. Now $\bar{P}=X^4(X^4+3)$, and $X^4+3$ is irreducible modulo $53$ (you might want to check this as an exercise). Hence $\bar{Q}=X^i(X^4+3)^j$ for some $0\leq i\leq 4$ and $j=0$ or $1$. But $\bar{Q}(0)=\pm 1$, so $i=0$ and $\bar{Q}=X^4+3$ (since $\bar{Q}$ cannot be $1$; otherwise $Q$ would be a trivial factor of $P$), meaning that $\bar{R}=X^4$. Note that $Q$ and $R$ have both degree $4$. In particular, $R=X^4+53S,$ with $\deg(S)\leq 3$, and $Q=X^4+3+53T$ ,  $\deg(T)\leq 3$. Now, $QR=X^8+3X^4+53((X^4+3)S+X^4T)+53^2 ST$. Hence $-53=53\cdot 3S(0) +53^2 S(0)T(0)$, and $-1=S(0)(3+53S(0)T(0))$. This imples $S(0)=\pm 1$ and $3+53S(0)T(0)=\pm 1$. It is then easy to derive a contradiction.
A: $f(0)=-53$, $f(\pm2)=251$, $f(\pm7)=577951$, $f(\pm12)=430043851$, $f(\pm18)=11020275451$, $f(\pm19)=16983953951$, $f(\pm25)=152589062447$, $f(26)=208828435451$, $f(\pm32)=1099514773451$
are $17$ integer places where $f(X)$ is (plus/minus) a prime, hence if $f=gh$,
then at each of these places, one of the factor polynomials is $\pm1$.
Then for some $a\in\{\pm1\}$, there are $9$ integer places where one of $g,h$ is $=a$. But there can be at most $\deg g +\deg h=8$ such places.

With a bit more of thinking we can get along with less computational search for prime values:
If for example $g(7)=g(12)=\pm1$, then $g(X)=(X-7)(X-12)g^*(X)\pm1$ and so $g(-1)=104g^*(-1)\pm1$ and $g(1)=66g^*(1)\pm1$ must be divisors of $f(1)=-49$ and $g(0)=84g^*(0)\pm1$ a divisor of $-53$, which is only possible if also $g(-1)=g(0)=g(1)=g(7)=g(12)$. Similar calculatins ar possible for other cases, reducing the total amount of prime values "needed".
