<s>[INST] Remainder when the polynomial $1+x^2+x^4+\cdots +x^{22}$ is divided by $1+x+x^2\cdots+ x^{11}$ 
Question : Find the remainder when the polynomial $1+x^2+x^4+\ldots +x^{22}$ is divided by $1+x+x^2+\cdots+ x^{11}$.

I tried using Euclid's division lemma, I.e.
$$P_1(x)=1+x^2+x^4+\cdots+x^{22}$$
$$P_2(x)=1+x+x^2+\cdots+x^{11}$$
Then for some polynomial $Q(x)$ and $R(x)$; we have 
$$P_1(x)=Q(x)\cdot P_2(x)+R(x)$$
Now, we put the values of $x$ such that $R(x)=0$ and form equations, but this method is way too long and solving the 11 set of equations for 11 variable (Since $R(x)$ a polynomial of at most 10 degree) is impossible to do for a competitive exam where the average time for solving a question is 3 minutes.
Another method is using the original long division method, and following the pattern, we can predict $Q(x)$ and $R(x)$, but it's also very hard and time taking.
I am searching for a simple solution to this problem since last a week and now I doubt even we have a simple solution to this question.
Can you please give me a hint/solution on how to proceed to solve this problem in time?
Thanks!
 [/INST] Let $\,P_n(x)=1+x+\cdots+x^{n-1}=(x^n-1)/(x-1)\,$, then the problem is equivalent to finding the remainder of the division $\,P_{12}(x^2) / P_{12}(x)\,$.
The remainder is $\,2 \,P_6(x^2)\,$, which follows for $\,n=6\,$ from the general identity:
$$
\begin{align}
P_{2n}(x^2) = \frac{x^{4n}-1}{x^2-1} &= \frac{x^{2n}-1}{x-1} \, \frac{x^{2n}+1}{x+1} \\[5px]
 &= \, \frac{x^{2n}-1}{x-1} \, \frac{x^{2n}-1+2}{x+1} \\[5px]
 &= - \, \frac{x^{2n}-1}{x-1} \, \frac{(-x)^{2n}-1}{(-x)-1} + 2 \, \frac{x^{2n}-1}{x^2-1} \\[5px]
 &= - \, P_{2n}(x) P_{2n}(-x) + 2 P_n(x^2)
\end{align}
$$
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