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I would like to show the irreducibility of $x^8 - 60 x^6 + 1160 x^4 - 7800 x^2 + 8836$ and $x^8 - 120 x^6 + 4360 x^4 - 45600 x^2 + 15376$ in $mathbb{Q}[x]$. In both cases Eisenstein criterion fails. I also attempted some linear changes of variables but nothing seems to work. Any help?

Let $f(x)=x^8 - 60 x^6 + 1160 x^4 - 7800 x^2 + 8836$.

$f$ has degree $8$ and assumes prime values at these $18 > 2 cdot 8$ points and so must be irreducible:
$$
begin{array}{rl}
n & f(n) \
pm 1 & 2137 \
pm 3 & -4583 \
pm 5 & -8039 \
pm 7 & 1117657 \
pm 13 & 557943577 \
pm 15 & 1936431961 \
pm 33 & 1330287723097 \
pm 37 & 3360699226777 \
pm 55 & 82083690591961 \
end{array}
$$

Irreducibility of the first polynomial
$$f(x) = x^8 - 60 x^6 + 1160 x^4 - 7800 x^2 + 8836$$
can also be deduced as follows.

Recall the usual business with Gauss's lemma. If $f(x)$ factors in $Bbb{Q}[x]$, it also factors in $Bbb{Z}[x]$. Let's assume contrariwise that
a non-trivial factorization $f(x)=g(x)h(x),g(x),h(x)inBbb{Z}[x]$ exists.
Without loss of generality the leading coefficients of both $g$ and $h$ are equal to one.

A potentially useful feature of $f(x)$ is that modulo five it becomes very sparse. More precisely
$$
f(x)equiv x^8+1pmod 5.
$$
In $Bbb{F}_5[x]$ we have the factorization
$$
x^8+1=x^8-4=(x^4-2)(x^4+2).
$$
These quartic polynomials are actually irreducible in $Bbb{F}_5[x]$. We have
$$x^8+1mid x^{16}-1.$$ Therefore any zero of either factor (in some extension field of $Bbb{F}_5$) must be a root of unity of order sixteen. But $16nmid 5^ell-1$ for $ell=1,2,3$ meaning that the field $Bbb{F}_{5^4}$ is the smallest extension field containing such roots of unity. Therefore their minimal polynomials over $Bbb{F}_5$ have degree four.

At this point we can conclude that the only remaining way $f(x)$ can factor in $Bbb{Z}[x]$ is as a product of two irreducible factors of degree four, and
$$
g(x)equiv x^4+2pmod 5,qquad h(x)equiv x^4-2pmod 5.
$$

Another feature of $f(x)$ is that it has even degree terms only. In other words, $f(x)=f(-x)$. Therefore $f(x)=g(-x)h(-x)$ is another factorization. But, factorization of polynomials is unique, so we can deduce that either $h(x)=g(-x)$ (when also $h(-x)=g(x)$),
or we have both $g(x)=g(-x), h(x)=h(-x)$.

Claim. It is impossible that $h(x)=g(-x)$.

Proof. Assume contrariwise that $h(x)=g(-x)$. If $g(x)=x^4+Ax^3+Bx^2+Cx+D$, then $h(x)=x^4-Ax^3+Bx^2-Cx+D$. Expanding $g(x)h(x)$ we see that the constant term is $D^2=8836=94^2$. Therefore we must have $D=pm94$. But, earlier we saw that the constant terms of $g,h$ must be congruent to $pm2pmod5$. This is a contradiction.

Ok, so we are left with the possibility $g(x)=g(-x)$, $h(x)=h(-x)$. In other words, both $g(x)$ and $h(x)$ share with $f(x)$ the property that they have even degree terms only. Let's define $F(x),G(x),H(x)$ by the formulas
$$f(x)=F(x^2),quad g(x)=G(x^2),quad h(x)=H(x^2).$$
The above considerations can be summarized as follows. If
$$
F(x)=x^4-60x^3+1160x^2-7800x+8836
$$
is irreducible, then so is $f(x)=F(x^2)$. Furthermore, the putative factors must satisfy the congruences
$$G(x)equiv x^2+2pmod5,quad H(x)equiv x^2-2pmod5.$$

A miracle is that depressing $F(x)$ produces a surprise:
$$
R(x)=F(x+15)=x^4-190x^2+961.
$$
The substitution $xmapsto x+15$ does not change anything modulo five, so the only possible factors of $R(x)$ must still be congruent to $x^2pm2pmod5$.

Irreducibility of $R(x)$ follows from this. The constant term of $R(x)$
is
$$
R(0)=961=31^2,
$$
and this has no factors $equivpm2pmod5$.

The other octic surrenders to similar tricks:
$$f(x)=x^8 - 120 x^6 + 4360 x^4 - 45600 x^2 + 15376.$$
Again, $f(x)equiv(x^4-2)(x^4+2)pmod 5$. The constant term
$15376=(2^2cdot31)^2$ is a square of an integer $equivpm1pmod5$,
ruling out a factorization of the form $g(x)g(-x)$. Again, we are reduced
to proving that
$$
F(x)=x^4-120x^3+4360x^2-45600x+15376
$$
is irreducible. Depressing this gives
$$
R(x)=F(x+30)=x^4-1040x^2+141376equiv(x^2-2)(x^2+2)pmod5.
$$
This time the constant term $R(0)=2^6cdot47^2=376^2$ has more factors, so we need a different argument. However, we can repeat the dose! $pm 376equivpm1pmod5$, ruling out the possibility of a factorization of the form $R(x)=G(x)G(-x)$ as above. So the remaining possibility is a factorization of the form
$$
R(x)=(x^2-A)(x^2-B)
$$
with integers $A$ and $B$. But, the equation
$$
x^2-1040x+141376=0
$$
has no integer roots. Irreducibility follows.

Both these polynomials have irreducible remainders in $mathcal{F}_2[x]$, field of integers modulo 2, when we try to divide each by $x^2+x+1$.

The reference is to Artin's Algebra, Proposition 12.4.3

Let $f(x) = a_n x^n + dots + a_0$ be an integer polynomial, and let $p$ be a prime integer that does not divide the leading coefficient $a_n$. If the residue $bar{f}$ of $f$ modulo $p$ is an irreducible element of $mathcal{F}_p[x]$, then $f$ is an irreducible element of $mathcal{Q}[x]$.

This means if we divide our polynomials in field $mathcal{F}_2[x]$, and we are lucky (= this often works) to get a remainder which is an irreducible polynomial in this field, then the original polynomial is irreducible.

In field $mathcal{F}_2[x]$ the following polynomials are irreducible, $x^2+x+1$ and $x+1$.

If we divide the original polynomials by $x^2+x+1$, the remainders are

$$16575 + 8959 x equiv 1+x mod 2$$

$$60855 + 49959 x equiv 1+x mod 2$$

Note

If any of these two polynomials had proper divisors in $mathcal{Q}[x]$, then it had proper divisors in $mathcal{Z}[x]$ as well, Artin 12.3.6.

This might suggest that if any of these polynomials had proper divisors in $mathcal{Z}[x]$, then divisors would be monic polynomials with limited choice of the last term, ie $8836=2^2 times 47^2$. But I do not have a deep insight to pursue it further.