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# Olympiad Proof Problem - Day 5

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.

Note by Surya Prakash
1 year, 11 months ago

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I 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.

- 1 year, 11 months ago

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

- 1 year, 11 months ago

Well, is cyclotomic polynomials allowed?

- 1 year, 11 months ago

Yeah, sure. Post your solution using cyclotomic polynomials.

- 1 year, 11 months ago

I have written in the proof in the wiki.

- 1 year, 11 months ago

When is Day 6 going to be posted?

- 1 year, 11 months ago