# Challenging question on geometric progression?

• July 14th 2008, 06:31 AM
fardeen_gen
Challenging 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
• July 14th 2008, 06:48 AM
mr fantastic
Quote:

Originally Posted by fardeen_gen
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

$p_1(x) = 1 + x^2 + x^4 + ... + x^{2n -2} = 1 + x^2 + (x^2)^2 + \, .... \, + (x^2)^{n-1} = \frac{1 - x^{2n}}{1 - x^2}$.

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

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

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

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