Showing the irreducibility of $x^8 - 60 x^6 + 1160 x^4 - 7800 x^2 + 8836$ in $\mathbb{Q}[x]$

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?

• 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. – Jyrki Lahtonen Jan 12 '19 at 19:14
• 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. – Jyrki Lahtonen Jan 12 '19 at 19:47
• 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). – Jyrki Lahtonen Jan 12 '19 at 22:37
• 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? – Jyrki Lahtonen Jan 12 '19 at 22:42
• Ok. The depressing trick let's us easily find the zeros of both polynomials. I'm still curious about how you found them. – 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: $$\begin{array}{rl} n & f(n) \\ \pm 1 & 2137 \\ \pm 3 & -4583 \\ \pm 5 & -8039 \\ \pm 7 & 1117657 \\ \pm 13 & 557943577 \\ \pm 15 & 1936431961 \\ \pm 33 & 1330287723097 \\ \pm 37 & 3360699226777 \\ \pm 55 & 82083690591961 \\ \end{array}$$

• Adapted from math.stackexchange.com/a/2942032/589 – lhf Jan 10 '19 at 13:38
• Very interesting idea, didn't know about this test. It seems it requires too much computing, though – Ray Bern Jan 10 '19 at 21:29
• @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]. – 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

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

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

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.

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