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Prove that if $p$ is a prime number, the polynomial $f(x)=1+x+\cdots +x^{p-2}+x^{p-1}$ is irreducible in $\mathbb{Z}[x]$.

I tried using Eisenstein Criterion and the Rational Root Theorem, also, I know that $f$ is irreducible over $\mathbb{Z}$ if and only if it is irreducible over $\mathbb{Q}$ because of the Gauss' Lemma. Maybe writing $x^p-1=(x-1)f(x)$ and proving $x^p-1$ dos not have any other roots may help, but I don't know.

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    Eisenstein criterion has been invented by Eisenstein himself in order to prove what you are trying to prove. You can see it on the Wikipedia page https://en.wikipedia.org/wiki/Eisenstein%27s_criterion#Cyclotomic_polynomials – Crostul Dec 02 '21 at 20:01
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    Not having other roots doesn’t mean it is irreducible. $(x^2+2)(x^2+3)$ doesn’t have rational roots, but it is obviously reducible. – Thomas Andrews Dec 02 '21 at 20:18
  • This is just the $p$ th cyclotomic polynomial for prime $p$ and those are well known to be irreducible. – Peter Dec 03 '21 at 11:17

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You are really close, from $f(x)=\dfrac{x^p-1}{x-1}$ we get using the Binomial Theorem that$$f(x+1)=x^{p-1}+\sum \limits _{k=1}^{p-1}\binom{p}{k}x^{k-1}.$$Using that $p\mid \binom{p}{k}$ for $1\leq k\leq p-1$ and Eisenstein Criterion you get that $f(x+1)$ is irreducible over $\mathbb{Q}$, from Gauss' Lemma follows that $f(x)$ is irreducible over $\mathbb{Z}$.

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    One should also add that $\binom p1 = p$ because for Eisensteins criterion we also need the absolute coefficient $a_0$ (the coefficient of $x^0$) to satisfy not only $p \mid a_0$ but also $p^2 \nmid a_0$. – Lukas Dec 02 '21 at 20:05
  • Yes, you are right. –  Dec 02 '21 at 20:45