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Polynomial $x^{p^n} + 1$, $p$ is prime and odd, is irreducible in $\mathbb Q[x]$

I can't use Eisenstein's criterion because $1|a_n$ , $a_n=1$.

Since $p$ is odd, $p^n$ is odd too, so $-1$ is zero of $x^{p^n} + 1$ Hence, $$x^{p^n} + 1 = (x+1) ( x^{p^{n-1}} - x^{p^{n-2}} + x^{p^{n-3}} + \dots + x^{p} +1)$$ I don't know how use the fact about $p$ being prime.

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    $\begingroup$ Can you find any zeros? $\endgroup$ – Arthur Oct 22 '13 at 14:05
  • $\begingroup$ Of course not!! $\endgroup$ – user43208 Oct 22 '13 at 17:36
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    $\begingroup$ Hint: What is $(-1)^s$ for odd $s$? $\endgroup$ – achille hui Oct 22 '13 at 17:49
  • $\begingroup$ yes, i have progress with this hint , because -1 is zero if p is odd , how use the fact abou p is prime? $\endgroup$ – andre Oct 22 '13 at 17:57
  • $\begingroup$ The fact that $p$ is prime is hardly necessary, so you don't need to use it at all. The polynomial $x^n+1$ is always divisible by $x+1$ when $n$ is odd. $\endgroup$ – anon Oct 22 '13 at 17:59
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You've shown that $x+1$ is a factor of $x^{p^n}+1$; thus, $x^{p^n}+1$ is reducible.

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  • $\begingroup$ yes, but i dont understand why p need be prime if x+1|$x^{p^n}$+1 when p is odd and p dont need be necessarily prime. $\endgroup$ – andre Oct 22 '13 at 18:13
  • $\begingroup$ @andre: You don't. The only thing you need is that $p$ is odd and greater than $1$ (and $n\gt0$). $\endgroup$ – robjohn Oct 22 '13 at 18:40

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