Prove that for all for primes \(p\) the polynomial

\[x^{p-1} + x^{p-2} + x^{p-3} + \ldots + 1\]

is irreducible over rational numbers.

Try more proof problems at Olympiad Proof Problems.

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## Comments

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TopNewestI used Eisenstein's Irreducibility Criterion to solve this question.

Let

\[\Phi_p (x) = x^{p-1} + x^{p-2} + \ldots + x^2 + x + 1 = \dfrac {x^p - 1}{x-1}\]

We claim that \(\Phi_p (x)\) is irreducible over the rational numbers. Although \(\Phi_p (x)\) itself does not directly admit application of Eisenstein's criterion, a minor variant of it does. That is, consider

\[f(x) = \Phi_p (x+1) = \dfrac {(x+1)^p - 1}{(x+1)-1} = \dfrac {\displaystyle \sum_{k=0}^{p-1} {p \choose k} x^{p-k}}{x}\]

\[=x^{p-1}+{p \choose 1} x^{p-2} + {p \choose 2} x^{p-3} + \ldots + {p \choose {p-2}} x + {p \choose {p-1}}\]

All the lower coefficients are divisible by \(p\), and the constant coefficient is exactly \(p\), so it is not divisible by \(p^2\). Thus, Eisenstein's criterion applies, and \(f\) is irreducible. Certainly, if \(\Phi_p (x) = g(x) h(x)\) then \(f(x) = \Phi_p (x+1) = g(x+1) h(x+1)\) gives a factorisation of \(f\). Thus, \(\Phi_p\) has no proper factorisation, i.e. it is irreducible.

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Nice solution. But why don't you do it another way or rather prove Eisenstein's criterion?

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Well, is cyclotomic polynomials allowed?

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I

havewritten in the proof in the wiki.Log in to reply

When is Day 6 going to be posted?

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