The number of natural number $n$ in the interval of [$1005,2010$ ] for which the polynomial The number of natural number $n$ in the interval of [$1005,2010$ ] for which the polynomial $1$+$x$+$x^2$+ $x^3$ +...+$x^{n-1}$ divides the polynomial   $1$+$x^2$+$x^4$+$x^6$ ....+$x^{2010}$ is

Ans is the number of such numbers is 503. 

My soln:
$$ (x^{503}-1)(x^{503} +1)( x^{1006}+1) $$ to be 
Divided by $ x^n -1$ So n=503.
The polynomial $1+x+x^2+...+x^{n-1}$ divides $1+x^2+x^4+...+x^{2010}$ exactly when
$$
(x^n-1)=(x-1)(1+x+...+x^{n-1})|(x-1)(1+x^2+x^4+...)=\frac{x^2-1}{x+1}(1+x^2+...+x^{2010})=\frac{x^{2012}-1}{x+1}.
$$

If the polynomial $x^n-1$ divides the polynomial $\frac{x^{2012}-1}{x+1},$ then all the (complex) roots of $x^n-1$ must also be roots of $x^{2012}-1,$ but $x=-1$ cannot be a root of $x^n-1.$. Because $-1$ is a root of $x^n-1$ when $n$ is even, the value $n$ must be odd. If $n$ is odd, then it is sufficient that $x^n-1|x^{2012}-1.$ But the complex roots of $x^n-1$ are shared by $x^{2012}-1$ exactly when $n$ divides $2012$. The only value $n$ in the range $[1005, 2010]$ for which this happens is $1006$, but this even, so there are zero solutions for $n$ in the range $[1005,2010].$
If $f(x)=1+x+x^2+\cdots+x^{n-1}$ divides $g(x)=1+x^2+x^4+\cdots+x^{2010}$, then the quotient has integer coefficients. Therefore, $n=f(1)$ divides $1006=g(1)$. There is only one multiple of $1006$ in $[1005,2010]$, which is $1006$ itself. Thus, the only possible solution, if there is a solution, is $n=1006$.