Is $x^{2k+1} - 7x^2 + 1$ irreducible?

Question. Is the polynomial $x^{2k+1} - 7x^2 + 1$ irreducible over $\mathbb{Q}$ for every positive integer $k$?

It is irreducible for all positive integers $k \leq 800$.

• What's the motivation for asking this question? – RP_ Jan 6 '17 at 10:44
• @René motivation comes from various well studied directions: enriching the existing criteria for irreducibility, analogies with formulas for prime numbers (irreducibles in the ring $\mathbb{Z}$), the study of trinomials (and their Galois groups), and so on. Sometimes, polynomials similar to these pop-up in questions related to combinatorics and dynamics. I did not want to elaborate on these too much since the question is, in my opinion, of independent interest. – Pablo Jan 6 '17 at 10:46
• Could it be that the answer is yes if 7 is replaced by any odd prime? – Timothy Chow Jan 6 '17 at 16:25
• @TimothyChow I believe that it could indeed be the case. – Pablo Jan 6 '17 at 16:35

Here is a proof, based on a trick that can be used to prove that $x^n + x + 1$ is irreducible when $n \not\equiv 2 \bmod 3$.

We work with Laurent polynomials in $R = \mathbb Z[x,x^{-1}]$; note that $R$ has unit group $R^\times = \pm x^{\mathbb Z}$. We observe that for $f \in R$, the sum of the squares of the coefficients is given by $$\|f\| = \int_0^1 |f(e^{2 \pi i t})|^2\,dt = \int_0^1 f(e^{2 \pi i t}) f(e^{-2 \pi i t})\,dt = \int_0^1 f(x) f(x^{-1})\big|_{x = e^{2 \pi i t}}\,dt .$$ Now assume that $f(x) = g(x) h(x)$. Then, since $f(x) f(x^{-1}) = \bigl(g(x)h(x^{-1})\bigr)\bigl(g(x^{-1})h(x)\bigr)$, $G(x) := g(x) h(x^{-1})$ satisfies $\|G\| = \|f\|$ and $G(x) G(x^{-1}) = f(x) f(x^{-1})$.

Now we consider $f(x) = x^n - 7 x^2 + 1$; then $\|f\| = 51$. If $f = g h$ as above, then write $G(x) = \pm x^m(1 + a_1 x + a_2 x^2 + \ldots)$ and $G(x^{-1}) = \pm x^l(1 + b_1 x + b_2 x^2 + \ldots)$. The relation $G(x) G(x^{-1}) = f(x) f(x^{-1})$ translates into (equality of signs and) $$(1 + a_1 x + \ldots)(1 + b_1 x + \ldots) = 1 - 7 x^2 + O(x^{n-2}).$$ Assuming that $n > 40$ and considering terms up to $x^{20}$, one can check (see below) that the only solution such that $a_1^2 + a_2^2 + \ldots + a_{20}^2 + b_1^2 + b_2^2 + \ldots + b_{20}^2\le 49$ is, up to the substitution $x \leftarrow x^{-1}$, given by $1 + a_1 x + \ldots = 1 - 7x^2 + O(x^{21})$, $1 + b_1 x + \ldots = 1 + O(x^{21})$. Since the $-7$ (together with the leading and trailing 1) exhausts our allowance for the sum of squares of the coefficients, all other coefficients must be zero, and we obtain that $G(x) = \pm x^a f(x)$ or $G(x) = \pm x^a f(x^{-1})$. Modulo interchanging $g$ and $h$, we can assume that $g(x) h(x^{-1}) = \pm x^a f(x)$, so $h(x^{-1}) = \pm x^a f(x)/g(x) = \pm x^a h(x)$, and $x^{\deg h} h(x^{-1})$ divides $f(x)$. This implies that $h(x)$ divides $x^n f(x^{-1})$, so $h(x)$ must divide $$f(x) - x^n f(x^{-1}) = 7 x^2 (x^{n-4} - 1).$$ So $h$ also divides $$f(x) - x^4 (x^{n-4} - 1) = x^4 - 7 x^2 + 1 = (x^2-3x+1)(x^2+3x+1).$$ Since $h$ also divides $f$, it must divide the difference $x^n - x^4$, which for $n \neq 4$ it clearly doesn't, since the quartic has no roots of absolute value 0 or 1; contradiction.

The argument shows that $x^n - 7 x^2 + 1$ is irreducible for $n > 40$; for smaller $n$, we can ask the Computer Algebra System we trust. This gives:

Theorem. $x^n - 7 x^2 + 1$ is irreducible over $\mathbb Q$ for all positive integers $n$ except $n=4$.

ADDED 2017-01-08: After re-checking the computations, I realized that there was a small mistake that ruled out some partial solutions prematurely. It looks like one needs to consider terms up to $x^{20}$. Here is a file with MAGMA code that verifies that there are no other solutions.

• This looks awesome and potentially quite general! The first part can be expressed without integration in a way that to me is a little more intuitive: The identity $f(x) f(x^{-1}) = G(x) G(x^{-1})$ implies that the sum of squares of coefficients of $f(x)$ equals the sum of squares of coefficients of $G(x)$ because the sum of the squares of coefficients of $p(x)$ is the coefficient of $0$ in $p(x) p(x^{-1})$ by the obvious expansion. – Will Sawin Jan 7 '17 at 9:32
• @Pablo I found it on page 97 of Zannier's "Lecture notes on Diophantine Analysis" (footnote to exercise 3.20): "Selmer proved that actually $f_n$ is irreducible for such $n$. Here is a sketch of an argument due to Ljiungrenn. ..." Alas, there is no "Ljiungrenn" in MathSciNet or ZMATH. This should be "W. Ljunggren", and the original source appears to be this: mscand.dk/article/view/10593/8614. – Michael Stoll Jan 7 '17 at 9:47
• Ljunggren treats the case $x^n \pm x^m \pm 1$ in his paper (Theorem 3). – Michael Stoll Jan 7 '17 at 10:00
• I am slightly confused by the condition $a_1^2 + a_2^2 + \ldots + b_1^2 + b_2^2 + \ldots \le 49$. Why is it not $a_1^2 + a_2^2 + \ldots\le 50$ and $b_1^2 + b_2^2 + \ldots \le 50$? I mean, we only know that $\|G\| = 51$. – GH from MO Jan 8 '17 at 11:30
• @GHfromMO We consider initial segments of the coefficient vector of $G$ from both ends that do not overlap (this leads to the condition $n>40$), so the sum of squares of both together is bounded by the sum of squares of all coefficients of $G$. – Michael Stoll Jan 8 '17 at 11:32

Not an answer, but an observation, which, if proved, will probably give an answer.

For all $k$ that I have computed, your polynomial has two roots smaller than $1$ in absolute value (for large $k$ these converge to $\pm\frac{1}{\sqrt{7}}.$), and the rest of the roots are larger than $1$ in absolute value. Let's assume this is true. Since the constant term of a factor has to have absolute value $1$ each factor should have one of these roots, so there are at most $2$ irreducible factors.

• I think the claim that there are exactly two roots of absolute value less than 1 in $\mathbb C$) should follow from Rouché's theorem, since $|x^{2k+1}| = 1 < 6 \le |-7x^2 + 1|$ on the unit circle. – Michael Stoll Jan 6 '17 at 18:14
• True. Easier is (triangular inequality): $\vert x^{2k+1}+1\vert\leq 2<7=\vert 7x^2\vert$ on the unit circle. – T. Amdeberhan Jan 6 '17 at 18:59
• @T.Amdeberhan I wouldn't necessarily think that $|a+b| \le |a|+|b|$ is so much easier than $|a-b| \ge |a| - |b|$; they are just two equivalent formulations of the same fact. – Michael Stoll Jan 6 '17 at 20:14
• I was just being picky. I agree. :-) – T. Amdeberhan Jan 6 '17 at 20:27
• The statement in the ADDENDUM is wrong (for example for $2k+1 = 19$, there are six factors of degree 3). If $2k+1 = p$ is prime, the degree of the nonlinear factors mod 7 is the order of 7 in the multiplicative group mod $p$, which does not have to be $p-1$. Assuming Artin's primitive root conjecture, we get at least infinitely many $k$ such that the polynomial is irreducible. – Michael Stoll Jan 7 '17 at 8:30

EDIT: I'm afraid that this answer is wrong and I hope you have some way of "unaccepting" it. The reason is that when taking conjugates of $\frac{1}{\sqrt{7}-\beta_1}$ we're allowed to freely change the sign of $\sqrt{7}$ (unless $\sqrt{7}$ is contained in the field generated by $\beta_1$, which seems rather unlikely). And $\frac{1}{-\sqrt{7}-\beta_1}$ is greater than $1$ in absolute value. I found this mistake only when I noticed that my "proof" gives that $x^4-7x^2+1=(x^2+3x+1)(x^2-3x+1)$ is irreducible. It should work with an integer square in place of $7$. Maybe someone can make it work also for $7$? I apologize for raising false hopes!

The question is equivalent (by replacing $x$ with $\frac{1}{x}$) to asking about the irreducibility of $A(x) = x^{2k+1}-7x^{2k-1}+1 = x^{2k-1}(x^2-7)+1$ over $\mathbb{Q}$. An application of Rouché's theorem shows that $A(x)$ has exactly $2k-1$ zeroes (counted with multiplicity) of absolute value less than $1$. There remain two real zeroes $\beta_1 < -1$ and $\beta_2 > 1$. One can deduce from the equation they satisfy that $|\beta_i - (-1)^i\sqrt{7}| \leq \frac{1}{\sqrt{7}+1}$ ($i=1,2$).

Now suppose that $A$ factorizes as $A = BC$ with monic integer polynomials $B, C$. Wlog $B(\beta_1) = 0$. If $B(\beta_2) \neq 0$, then $$\beta_1^{2k-1}(\beta_1+\sqrt{7}) = \frac{1}{\sqrt{7}-\beta_1}$$ is an algebraic integer, all conjugates of which are less than $1$ in absolute value (the right hand side is less than $\frac{1}{2\sqrt{7}-\frac{1}{\sqrt{7}+1}}<1$ and all its conjugates are of absolute value at most $\frac{1}{\sqrt{7}-1}<1$).

So it would have to be $0$, which is absurd. So $B(\beta_2)=0$. But now all roots of $C$ are less than $1$ in absolute value, while its constant term is $1$. It follows that $C=1$ is constant and so $A$ is irreducible.

• I also completely missed the fact that when one considers Galois conjugates, it is possible also to conjugate $\sqrt{7}$ and not only $\beta_1$. – Pablo Jan 6 '17 at 13:45
• Well, you fooled me! :-) I thought I had read your proof very carefully. So it's still instructive, in a way. – Gro-Tsen Jan 7 '17 at 11:31
• @Gro-Tsen: Well, this general method does prove irreducibility in certain cases. AFAIK, it originates in Ankeny, Brauer, Chowla, A Note on the Class-Numbers of Algebraic Number Fields, Amer. J. Math. 78 (1956), 51-61, and I used it to prove irreducibility in Lemma 4.1(b) of my article (impan.pl/en/publishing-house/journals-and-series/…, sorry for the self-promotion, but I thought it might interest you). – Gabriel Dill Jan 8 '17 at 15:53