About irreducible trinomials

Question

This question was inspired by this one. For every $n>m>0$ consider the polynomial $p_{m,n}=x^n-x^m-1$.

For which $m,n$ is $p_{m,n}$ irreducible over $\mathbb Q$?

In particular, if $m$ is odd, is it always irreducible?

Answer 1

MR0124313 (23 #A1627) Ljunggren, Wilhelm On the irreducibility of certain trinomials and quadrinomials. Math. Scand. 8 1960 65–70. 12.30

The author considers the irreducibility over the field of rational numbers of the polynomials $f(x)=x^n+ε_1x^m+ε_2x^p+ε_3$, where $ε_1,ε_2,ε_3$ take the values $\pm1$. He proves that if $f(x)$ has no zeros which are roots of unity, then $f(x)$ is irreducible; if $f(x)$ has exactly $q$ such zeros, then $f(x)$ can be factored into two factors with rational coefficients, one of which is of degree $q$ with all these roots of unity as zeros, while the other is irreducible (and possibly merely a constant). He also determines all possible cases where roots of unity can be zeros of $f(x)$. As a corollary he is able to give a complete treatment of the trinomial $g(x)=x^n+ε_1x^m+ε_2$, where $ε_1,ε_2$ take the values $\pm1$. The irreducibility of this trinomial was studied by E. S. Selmer, who gave a partial solution [Math. Scand. 4 (1956), 287--302; MR0085223 (19,7f); see also #A1628]. The methods used are direct and elementary. Reviewed by H. W. Brinkmann

Answer 2

Ljunggren's result on $\pm 1$ trinomials $X^n \pm X^m \pm 1$ amounts to saying that every such trinomial has at most one non-cyclotomic irreducible factor. Since $\{\zeta,\mu\} = \{e^{2\pi i/3},e^{4\pi i/3}\}$ are evidently the only roots of unity solutions to $1 + \zeta + \mu = 0$, the precise factorization is easily derived from this.

Since Ljunggren's proof follows a case by case analysis that may not be very illuminating, I thought I would add an answer outlining a more conceptual proof due to Schinzel. I was reminded of it on rereading Smyth's survey on the Mahler measure, which recounts Schinzel's proof in section 14.1.

Since the logarithmic Mahler measure $m(P) = \int_{S^1} \log{|P(z)|} \, d\theta$ is manifestly additive ($m(PQ) = m(P) + m(Q)$), the proof that the trinomial has not more than a single non-cyclotomic factor is an almost immediate consequence of two general, if not easy, facts about polynomials:

• Arithmetic component: Smyth's theorem that a non-reciprocal integer polynomial $P \in \mathbb{Z}[X]$ with $P(1) \neq 0$ has $m(P) \geq \log{\rho} = 0.281\ldots$, where $\rho^3 = \rho+1$ (the Plastic number).
• Analytic component: Goncalves's inequality. With $z_1,\ldots,z_d$ any ordering of the complex roots of a monic polynomial $P(X) = X^d + a_{1}X^{d-1} + \cdots + a_d \in \mathbb{C}[X]$, this states that $|z_1\cdots z_t|^2 + |z_{t+1} \cdots z_d|^2 \leq \|P\|_2^2 = 1 + \sum_1^d |a_i|^2$ for all $t = 1,\ldots,d$.

In terms of Mahler measure, Goncalves's inequality is expressed as $e^{2m(P)} + e^{-2m(P)} \leq \|P\|_2^2$. Applied to our trinomial $P(X) = X^n \pm X^m \pm 1$, which has $\|P\|_2^2 = 3$, its conclusion is $m(P) \leq \log{\varphi} = 0.4812\ldots$, where $\varphi^2 = \varphi+1$ (the Golden ratio). We had the trivial bound of $\log{2}$, which is not quite enough to conclude with Smyth's theorem. But since $\log{\varphi} < 2\log{\rho}$, while it is readily seen that the only reciprocal factors are cyclotomic (If $\alpha$ and $1/\alpha$ both satisfy the equation then $\ldots$), the desired conclusion follows.