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I try to prove the next statement:

For $n>=0$ show that $5x^n - 1$ is irreducible over Q[x].

I have been doing some research and find einstein's statement but it doesn't fix the problems and the rest of the examples that I found on this page are for a very specific polynomial. I think that probably the fact of not repeated roots can help but I don't know the strategy to prove this.

Thank you in advance.

Nkm20
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  • You can view $x^n-\frac{1}{5}$. – Dietrich Burde Sep 14 '21 at 14:29
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    Do you mean Eisenstein's criterion? – user247327 Sep 14 '21 at 14:49
  • @Infinity_hunter That isn't sufficient. You could have no rational roots, but still split into quadratic/cubic terms. – Calvin Lin Sep 14 '21 at 17:34
  • @Calvin Lin Oh,Yes you are correct – Infinity_hunter Sep 14 '21 at 17:48
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    The reciprocal of the polynomial $f(x)=5x^n-1$ is $$\tilde{f}(x)=x^5f(1/x)=-x^5+5,$$ which is irreducible by Eisenstein. It is easy to see that a polynomial with non-zero constant term is irreducible if and only if its reciprocal is. – Jyrki Lahtonen Sep 15 '21 at 05:13
  • @JyrkiLahtonen Hi, thank u for your help it was very helpful. I already proof the Einsensteion criteria in order to use it in my proof but for the statement "a polynomial with the non-zero constant term is irreducible if and only if its reciprocal is" I don't get it because I think the direction <= isn't true, take for example (taking into a count f*=gh) h(x) be irreducible with nonzero constant term and let f(x)=xh(x). Then (f∗)∗=h(x) and ((f∗)∗)∗=f∗, so f∗ is irreducible even though f is not. Thanks again! – Nkm20 Sep 16 '21 at 04:49
  • You need to have non-zero constant coefficient as mentioned by Jyrki (the $xh(x)$ has zero constant coefficient). – Sil Sep 17 '21 at 06:21

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