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?

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    $\begingroup$ What is the origin of these two polynomials? They are similar in many ways. In addition to those congruences I looked at in my answer and the comments under it, they share the prme factors of the discriminants: $2,3,7,31,47$ (by Mathematica). I would bet against that being a coincidence. $\endgroup$ – Jyrki Lahtonen Jan 12 '19 at 19:14
  • $\begingroup$ Further toying: Both have eight real zeros. Also, both have the curious property that the four local minima are all equal, $-8064$ for the former and $-129024$ for the latter. Speak up, man! It is not given that I can reverse engineer them, even when aided by Mathematica :-) It's just that their origin may also simplify the irreducibility proofs. $\endgroup$ – Jyrki Lahtonen Jan 12 '19 at 19:47
  • $\begingroup$ I'm fairly sure that both the polynomials have Galois group isomorphic to $D_4$ (Chebotarev density analysis). Meaning that they are solvable in radicals (square roots actually). $\endgroup$ – Jyrki Lahtonen Jan 12 '19 at 22:37
  • $\begingroup$ Are these polynomials constructed to have an element $\alpha$ of a field $L$ such that $Gal(L/\Bbb{Q})\simeq D_4$ as a zero? And you want a confirmation of irreducibility to conclude that you have found a primitive element? $\endgroup$ – Jyrki Lahtonen Jan 12 '19 at 22:42
  • $\begingroup$ Ok. The depressing trick let's us easily find the zeros of both polynomials. I'm still curious about how you found them. $\endgroup$ – Jyrki Lahtonen Jan 13 '19 at 16:26

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 $f(x)=g(x)h(x),g(x),h(x)\in\Bbb{Z}[x]$ 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 $$ f(x)\equiv x^8+1\pmod 5. $$ 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 $$x^8+1\mid x^{16}-1.$$ Therefore any zero of either factor (in some extension field of $\Bbb{F}_5$) must be a root of unity of order sixteen. But $16\nmid 5^\ell-1$ 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 $$ g(x)\equiv x^4+2\pmod 5,\qquad h(x)\equiv x^4-2\pmod 5. $$

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 $\pm2\pmod5$. 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 $$G(x)\equiv x^2+2\pmod5,\quad H(x)\equiv x^2-2\pmod5.$$

A miracle is that depressing $F(x)$ produces a surprise: $$ R(x)=F(x+15)=x^4-190x^2+961. $$ The substitution $x\mapsto x+15$ does not change anything modulo five, so the only possible factors of $R(x)$ must still be congruent to $x^2\pm2\pmod5$.

Irreducibility of $R(x)$ follows from this. The constant term of $R(x)$ is $$ R(0)=961=31^2, $$ and this has no factors $\equiv\pm2\pmod5$.

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 $15376=(2^2\cdot31)^2$ is a square of an integer $\equiv\pm1\pmod5$, 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 $$ R(x)=F(x+30)=x^4-1040x^2+141376\equiv(x^2-2)(x^2+2)\pmod5. $$ This time the constant term $R(0)=2^6\cdot47^2=376^2$ has more factors, so we need a different argument. However, we can repeat the dose! $\pm 376\equiv\pm1\pmod5$, 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.


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: $$ \begin{array}{rl} n & f(n) \\ \pm 1 & 2137 \\ \pm 3 & -4583 \\ \pm 5 & -8039 \\ \pm 7 & 1117657 \\ \pm 13 & 557943577 \\ \pm 15 & 1936431961 \\ \pm 33 & 1330287723097 \\ \pm 37 & 3360699226777 \\ \pm 55 & 82083690591961 \\ \end{array} $$

  • $\begingroup$ Adapted from math.stackexchange.com/a/2942032/589 $\endgroup$ – lhf Jan 10 '19 at 13:38
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    $\begingroup$ Very interesting idea, didn't know about this test. It seems it requires too much computing, though $\endgroup$ – Ray Bern Jan 10 '19 at 21:29
  • $\begingroup$ @RayBern, it's one line in Mathematica. Too bad it does not work in WA. Select[Table[x^8-60x^6+1160x^4-7800x^2+8836,{x,0,100}],PrimeQ]. $\endgroup$ – lhf Jan 10 '19 at 23:07

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

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

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


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.

  • $\begingroup$ -1 You seem to have misunderstood the proposition you quote. None of this is logically sound. $\endgroup$ – Servaes Jan 8 at 11:17

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