Irreducibility of some trinomials modulo $p$ - MathOverflow most recent 30 from http://mathoverflow.net2013-05-22T19:44:06Zhttp://mathoverflow.net/feeds/question/56506http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/56506/irreducibility-of-some-trinomials-modulo-pIrreducibility of some trinomials modulo $p$Luis H Gallardo2011-02-24T09:20:32Z2011-05-24T22:59:07Z
<p>Let $n>1$ be an integer. An old result of Selmer,
See Theorem 1, page 289 in</p>
<p><a href="http://www.mscand.dk/article.php?id=1472" rel="nofollow">http://www.mscand.dk/article.php?id=1472</a>, </p>
<p>(If the link does not work try googling: <code>selmer trinomials</code>)</p>
<p>says that</p>
<p>$$
S(n) = x^n-x-1
$$
is irreducible over the the field $k= \mathbb{Q}$ of rational numbers.</p>
<p>Question : What is known about the possible irreducibility (or not) of the sligthly more general trinomial</p>
<p>$$
T(n,m) = x^n - x^m -1
$$</p>
<p>(with $0 < m < n$)</p>
<p>over the prime field</p>
<p>$$
k =GF(p)
$$</p>
<p>such that (say)</p>
<p>(a)
$p>2,$</p>
<p>(since seems there are many known results for binary polynomials)</p>
<p>and</p>
<p>(b)
$n$ goes to infinity when $p$ goes to infinity.</p>
<p>EDIT: Observe that something can be said about the parity of the number
of irreducible factors: Use Stickelberger's parity theorem.</p>
http://mathoverflow.net/questions/56506/irreducibility-of-some-trinomials-modulo-p/56605#56605Answer by Aaron Meyerowitz for Irreducibility of some trinomials modulo $p$Aaron Meyerowitz2011-02-25T06:41:24Z2011-02-25T11:53:43Z<p>Based on a small number of small cases I suspect that the majority of those of those do factor. </p>
<p>Let $r$ be a primitive root $\mod p$ then there is an $n$ such that $r^n=r+1 \mod p$. Then $x-r$ is a factor of $x^n-x-1$ in $\mathbb{Z}_p$. On average $n$ should be about $p/2$. Of course one can use $x^{n+p-1}-x-1$ but that seems like cheating. That is just cases with a linear factor. </p>
<p>For $p=19, $ $x^k-x-1$ factors for $2 \le k \le 18$ except for $k=4$ and $k=15$.</p>