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Prove


For all $n>1$, equation $\sum _{k=1}^n \frac{x^k}{k!}+1=0$ has no rational root.

I'm not sure whether there are two questions,for without brace after Sigma.

My thought is to prove it is not reducible on rational field.

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    $\begingroup$ Do you know the rational roots test? $\endgroup$ – 6005 Jun 11 '13 at 16:54
  • $\begingroup$ @Goos Do you mean the Einsenstain test? or other about r/s rational root and s divides an, r divides a0? $\endgroup$ – HyperGroups Jun 11 '13 at 16:58
  • $\begingroup$ @HyperGroups en.wikipedia.org/wiki/Rational_root_theorem $\endgroup$ – Kaish Jun 11 '13 at 16:58
  • $\begingroup$ @HyperGroups See the above link; it's the second one (the "other about r/s ...") you mention. $\endgroup$ – 6005 Jun 11 '13 at 17:00
  • $\begingroup$ @Kaish xixi, thank you, do you think my formula in the post is interesting? <kbd>tag $\endgroup$ – HyperGroups Jun 11 '13 at 17:00
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Multiplying by $n!$ we can make all coefficients to become integers. By the rational root theorem, the root has to be integer and divides $n!.$ Moreover, since all coefficients except for the first one are divisible by $n,$ $x^n$ is divisible by $n.$ Take any prime divisor $p$ of $n.$ Then $p$ divides $x$ and it is enough to show that $x^k\frac{n!}{k!}$ is divisible by the higher power of $p$ than $n!$ for each $1\le k\le n$ The power of $p$ that $k!$ is divisible is less than $k/p+k/p^2+...=\frac{k}{p-1}\le k$ provided $p\ge 2$ and the result follows.

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    $\begingroup$ +1 I think this is a great solution, and deserves more up votes. $\endgroup$ – Calvin Lin Jun 15 '13 at 0:25

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