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Will be glad for a little hint: let x and n be positive integer such that $1+x+x^2+\dots+x^{n-1}$ is a prime number then show that n is prime

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    $\begingroup$ Is that supposed to be $1+x+x^2+\dots+x^{n-1}$? $\endgroup$ – Brian M. Scott Apr 16 '12 at 18:58
  • $\begingroup$ I suggest trying to find a decompostion for this polynomial remembering that a prime number have only the trivial decomposition $\endgroup$ – Belgi Apr 16 '12 at 19:00
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    $\begingroup$ Hint: $k|n \Rightarrow (x^k-1)|(x^n-1)$. $\endgroup$ – marlu Apr 16 '12 at 19:00
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Hint $\ $ The sequence $\rm\:f_n = (x^n-1)/(x-1)\:$ is a divisibility sequence, i.e. $\rm\:m\:|\:n\:$ $\Rightarrow$ $\rm\:f_m\:|\:f_n.\:$

In fact it is a strong divisibility sequence, i.e. $\rm\:(f_m,f_n) = f_{\:\!(m,n)},\:$ which implies an intimate relationship between divisibility properties of the $\rm\:f_n\:$ and integers $\rm\:n\:$ (so, in particular, relation between notions associated with divisibility, such as irreducible = prime).

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Let $n = kl$. Consider $$ (1 + x + ... + x^{k-1})(1 + x^k + x^{2k} + ... + x^{(l-1)k}) = (1 + x + ... + x^{n-1}) $$

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Assuming that you meant that $1+x+x^2+\ldots+x^{n-1}$ is prime, note that this sum is $\frac{x^n-1}{x-1}$. If $n=ab$, we have $$\frac{x^n-1}{x-1}=\frac{(x^a)^b-1}{x-1}=\frac{(x^a-1)(1+x^a+x^{2a}+\ldots+x^{(b-1)a})}{x-1}\;,$$ and $\dfrac{x^a-1}{x-1}=\;$?

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