About irreducible trinomials - MathOverflow most recent 30 from http://mathoverflow.net2013-05-19T19:08:03Zhttp://mathoverflow.net/feeds/question/56579http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/56579/about-irreducible-trinomialsAbout irreducible trinomialsMark Sapir2011-02-24T23:47:46Z2011-02-25T00:08:14Z
<p>This question was inspired by <a href="http://mathoverflow.net/questions/56506/irreducibility-of-some-trinomials-modulo-p" rel="nofollow">this one</a>. For every $n>m>0$ consider the polynomial $p_{m,n}=x^n-x^m-1$. </p>
<p>For which $m,n$ is $p_{m,n}$ irreducible over $\mathbb Q$? </p>
<p>In particular, if $m$ is odd, is it always irreducible?</p>
http://mathoverflow.net/questions/56579/about-irreducible-trinomials/56580#56580Answer by Gerry Myerson for About irreducible trinomialsGerry Myerson2011-02-25T00:08:14Z2011-02-25T00:08:14Z<p>MR0124313 (23 #A1627)
Ljunggren, Wilhelm
On the irreducibility of certain trinomials and quadrinomials.
Math. Scand. 8 1960 65–70.
12.30 </p>
<p>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 $f(x)$. <strong>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 $\pm1$</strong>. 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</p>