MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
    
by axccident I found your pos and the following article: on the irrecucibility of the trinomias $x^n \pm x^m \pm 1$, Helge Trevberg, Math. Scand 8 (1960) 121-126 – miracle173 Nov 8 '13 at 2:16
up vote 25 down vote accepted

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

share|cite|improve this answer
    
@Gerry: Thanks! That was quick! – Mark Sapir Feb 25 '11 at 0:13

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.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.