<s>[INST] Remainder of dividing polynomial of the $n$th degree 
What is the remainder when dividing the polynomial
$$P(x)=x^n+x^{n-1}+\cdots+x+1$$ with the polynomial
$$x^3-x$$ if $n$ is a natural odd number?

So, what I know so far is:
$$P(x)=Q(x)D(x)+R(x)$$
In this case I'll call $Q(x) = x^3-x$
$$Q(x) = 0 \iff x=\pm1$$
So from here:
$\begin{array} {l}P(1):\qquad&1^n+1^{n-1}+\cdots+1^1+1=R(1)\\P(-1):&(-1)^n+(-1)^{n-1}+\cdots+(-1)^1+1=R(-1)\end{array}$
Here is where I assumed that $R(x)=ax^2+b$ since $Q(x)=x^3-x$
And from the equations (assuming $n = \{2k+1 \mid k\in\mathbb{N}\}$):
$\begin{array}{l}P(1):\\&R(1)=n\\P(-1):\\&R(-1)=0\end{array}$
And here I get to:
$$\left\{
\begin{array}{ll}
ax^2+by &=n \\ 
ax^2+by &=0 \\ 
\end{array} 
\right.$$
I would like to know where did I go wrong? Is my assumption for $R(x)$ incorrect, my calculation of the $P(1)$ and $P(-1)$ wrong or is there something else I didn't think about?
 [/INST] Hint:
Experimenting the first few values of $n$: $\:n= 3,5,7,9,11$, you may conjecture the remainder  for the general polynomial $P_n(x)=x^n+x^{n-1}+\dots+x^3+x^2+x+1\:$ is
$$R_n(x)=kx^2+(k+1)x+1,\quad\text{ where } k=\left\lfloor \frac n2 \right\rfloor.$$
and try to prove it by induction on $n$ (odd).
You can prove the inductive step (case $n \implies$ case $n+2$) you can try rewriting
$$ P_{n+2}(x)= x^2 P_n(x)+ (x+1)=x^2\bigl[ Q_n(x)(x^3-x)+R_n(x)\bigr]+(x+1), $$
so the remainder for $P_{n+2}$ will be as well the remainder for
$$x^2 R_n(x)+(x+1)=x^2(kx^2+(k+1)x +1)+(x+1)=kx^4+(k+1)x^3+x^2+x+1.$$
You'll only have to prove that the coefficient of $x^2$ thus obtained is $\left\lfloor\dfrac{n+2}2\right\rfloor$ and the coefficient of $x$ is $1$ more.
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