Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
Is that supposed to be $1+x+x^2+\dots+x^{n-1}$? – Brian M. Scott Apr 16 '12 at 18:58
I suggest trying to find a decompostion for this polynomial remembering that a prime number have only the trivial decomposition – Belgi Apr 16 '12 at 19:00
Hint: $k|n \Rightarrow (x^k-1)|(x^n-1)$. – marlu Apr 16 '12 at 19:00
up vote 5 down vote accepted

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).

share|cite|improve this answer

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

share|cite|improve this answer

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 $\dfrac{x^a-1}{x-1}=\;$?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.