For what values of n is the polynomial 1 + x^2 + x^4 + ... + x^(2n -2) divisible by 1 + x + x^2 + ... + x^(n - 1)?

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- Jul 14th 2008, 05:31 AMfardeen_genChallenging question on geometric progression?
For what values of n is the polynomial 1 + x^2 + x^4 + ... + x^(2n -2) divisible by 1 + x + x^2 + ... + x^(n - 1)?

^ - Raised to - Jul 14th 2008, 05:48 AMmr fantastic
$\displaystyle p_1(x) = 1 + x^2 + x^4 + ... + x^{2n -2} = 1 + x^2 + (x^2)^2 + \, .... \, + (x^2)^{n-1} = \frac{1 - x^{2n}}{1 - x^2}$.

$\displaystyle p_2(x) = 1 + x + x^2 + ... + x^{n-1} = \frac{1 - x^{n}}{1 - x}$.

In both cases the formula for the sum of a geometric series has been used: $\displaystyle 1 + r + r^2 + \, ... \, + r^m = \frac{1 - r^{m+1}}{1 - r}$. Note that $\displaystyle r = x^2$ and $\displaystyle m = n-1$ in $\displaystyle p_1(x)$.

$\displaystyle \frac{p_1}{p_2} = \frac{(1 - x^{2n}) (1 - x)}{(1 - x^n) (1 - x^2)} = \frac{1+x^n}{1 + x}$.

And x + 1 is a factor of $\displaystyle p(x) = x^n + 1$ if p(-1) = 0 ......