The number of natural number $n$ in the interval $[1005,2010]$ for which the polynomial
$$1+x+x^2+x^3\dots +x^{(n-1)}$$ divides the polynomial $$1+x^2+x^4\dots+x^{2010}$$ is:
I could realize that the terms in the two polynomials are in GP and I applied the sum to n terms of a GP formula and then divided both the polynomials.
What I get is $(x^{2010}-1)/(x^{n}-1)$.
Now how to proceed?
HINT:
Using this, $$(a^n-1,a^m-1)=a^{(n,m)}-1$$
$$\implies (a^m-1)|(a^n-1)\iff m|n$$
$$\implies (x^n-1)|(x^{2012}-1)\iff n|2012$$
Now, the factors of $2012 $ are $n=1,2,4,503,1006,2012$
Observe that $\displaystyle\frac{x^{2012}-1}{x^n-1}$ must have $(x+1)$ as factor
and we know from (1) of this, $ x^m=1$ does not have a repeated root
If $n$ is even, $(x^{2012}-1,x^n-1)$ will be divisible by $x^2-1,$ hence by $x+1$
$\implies (x+1)$ will not divide $\displaystyle\frac{x^{2012}-1}{x^n-1}$