I proved that if $n$ is a prime, then $p(x)=x^{n-1}+\cdots+x+1$ is irreducible over $\mathbb Z$. But, I don't know how to prove that if $p(x)$ is irreducible over $\mathbb Z$, then $n$ is prime. Can you give me a hints?
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3Prove that if $m$ divides $n$, then $x^{m-1}+\cdots+1$ divides $p$. – Gerry Myerson Oct 02 '15 at 09:27
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What's the mean $x^{m-1}+\cdots+1$ divides p?? – asdagw Oct 02 '15 at 09:29
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It means exactly what it says. What part of it don't you understand? – Gerry Myerson Oct 02 '15 at 09:31
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1Oh... p is p(x)??? – asdagw Oct 02 '15 at 09:32
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4One implication: Irreducibility of $X^{p-1} + \cdots + X+1$ and other posts linked there. The other implication seems rather straightforward: $1+\dots+x^{ab-1}=(1+x+\dots+x^{a-1})(1+x^a+\dots+x^{a(b-1)}$. – Martin Sleziak Oct 02 '15 at 13:20
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Yes, asdagw, $p$ is $p(x)$. – Gerry Myerson Oct 04 '15 at 12:43
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The key is the repeated application of the geometric summation formula $$ \sum_{i=0}^{z-1} q^i = \frac{q^z - 1}{q-1}. $$
Plugging in $z=n, q=x$ and assuming $n = kl$, we get $$p(x) = x^{n-1} + x^{n-2} + \ldots + 1 = \frac{x^n - 1}{x-1} = \frac{x^{kl} - 1}{x-1}.$$
Geometric summation with $z=l$ and $q = x^k$ yields $$ \ldots = \frac{x^k - 1}{x-1} (x^{(l-1)k} + x^{(l-2)k} + \ldots + 1).$$
And finally by geometric summation with $z=k, q=x$ $$ \ldots = (x^{k-1} + x^{k-2} + \ldots + 1)(x^{(l-1)k} + x^{(l-2)k} + \ldots + 1).$$
If $k \geq 2$ and $l \geq 2$, this is a proper factorization of the polynomial $p$.
azimut
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If $n$ is not a prime, say $n=ab$ with $a,b>1$, then $q_a(x)=\frac{x^a-1}{x-1}$ is a divisor of $p(x)=\frac{x^n-1}{x-1}$, since every root of $q_a$ is a root of $p$, and there are no repeated roots.
Jack D'Aurizio
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