Let $n$ be a positive integer greater than $1$, and define the polynomial $$p_{n}(x)=\sum_{k=0}^{n}\dfrac{x^k}{k!}$$

Is $p_{n}(x)$ irreducible in $\mathbf{Q}[x]$?

I can show it when $n$ is a prime number, since $$n!p_{n}(x)=x^n+nx^{n-1}+n(n-1)x^{n-2}+\cdots+n!x+n!$$ using Eisenstein's criterion.

  • $\begingroup$ Summation starts from $k=0$, right? $\endgroup$ May 30 '16 at 6:46
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    $\begingroup$ There is a fairly elementary proof at mattbakerblog.wordpress.com/2014/05/02/… together with some more notes on the topic. $\endgroup$ May 30 '16 at 8:50
  • $\begingroup$ @TobiasKildetoft,your link can't open it,Thanks,can you post this paper title? $\endgroup$ May 30 '16 at 8:58
  • $\begingroup$ It works fine for me. It is a blog post, not a paper (though he has some references to papers in it). $\endgroup$ May 30 '16 at 9:00
  • $\begingroup$ Thanks,can you post his blog with title,maybe I can google it $\endgroup$ May 30 '16 at 9:04

That follows from a theorem of Schur saying that any polynomial $\sum_{k=0}^nc_k\frac{x^k}{k!}$ with $c_i\in\mathbf{Z}$, $c_0,c_n\in\{1,-1\}$, $n\ge 1$, is irreducible over $\mathbf{Q}$.

I. Schur, Einige Sätze über Primzahlen mit Anwendungen auf Irreduzibilitätsfragen I, Sitzungsberichte Preuss. Akad. Wiss. Phys.-Math. Klasse 14 (1929), 125–136. Also in Gesammelte Abhandlungen, Band III, 140–151

See also here.

  • $\begingroup$ Thanks,maybe there exist elementary proof? because it's nice result,like the erdos methods? $\endgroup$ May 30 '16 at 8:46

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