Prove the polynomial $x^2+2x+1$ divides the polynomial $x^{2(2n+1)}+2x^{2n+1}+1$ for every $n\in \mathbb{N}$ I've been trying to prove that the polynomial $x^2+2x+1$ divides the polynomial $x^{2(2n+1)}+2x^{2n+1}+1$ for all $n\in \mathbb{N}$ as follows and reached a "roadblock":
I wrote $x^2+2x+1$ as $(x+1)^2$ and showed that $x=-1$ is a root of the 2nd polynomial, hence by Bezout's Little Theorem the polynomial $x+1$ divides $x^{2(2n+1)}+2x^{2n+1}+1$, but how do I continue from here?
 A: Hint: $x^{2(2n+1)}+2x^{2n+1}+1 = (x^{2n+1}+1)^2$ 
Further hint:

 $x^{2n+1}+1=(x+1)(x^{2m}-x^{2m-1}+\cdots-x+1)$

A: Take the equation
$$(x^{2n}-x^{2n-1}+x^{2n-2}-\ ...\ +x^2-x+1)(x+1) = x^{2n+1}+1$$
This holds true, since
$$(x^{2n}-x^{2n-1}+x^{2n-2}-\ ...\ +x^2-x+1)(x+1) = \\ =x^{2n+1}-x^{2n}+x^{2n-1}-\ ...\ +x^3-x^2+x + \\ +\  x^{2n}-x^{2n-1}+x^{2n-2}-\ ...\ +x^2-x+1 = \\ = x^{2n+1} + 1$$
You can see that everything in the center cancels out, only leaving the first and last terms.
Now squaring both sides of the equation we get:
$$(x^{2n}-x^{2n-1}+x^{2n-2}-\ ...\ +x^2-x+1)^2(x+1)^2 = (x^{2n+1}+1)^2$$
$$(x^{2n}-x^{2n-1}+x^{2n-2}-\ ...\ +x^2-x+1)^2(x^2+2x+1) = x^{2(2n+1)}+2x^{2n+1}+1$$
Which means that $x^2+2x+1$ divides $x^{2(2n+1)}+2x^{2n+1}+1$, and the divisor is:
$$(x^{2n}-x^{2n-1}+x^{2n-2}-\ ...\ +x^2-x+1)^2.$$
A: So you need to show that $x+1$ divides $f(x)=x^{2(2n+1)}+2x^{2n+1}+1$ twice. This is the same as $x=-1$ being a double root of $f(x)$. Simply show that $f'(-1)=0$ and you're done.
To clarify, we are using the following result, which states that for a polynomial $f(x)$ we have:
$f(a)=0\Leftrightarrow x-a$ divides $f(x)$,
$f(a)=f'(a)=0\Leftrightarrow (x-a)^2$ divides $f(x)$,
$f(a)=f'(a)=f''(a)=0\Leftrightarrow (x-a)^3$ divides $f(x)$ etc.
This generalises mutatis mutandis. 
A: Take $x^{2m+1}=a$
Now, $x^{2(2m+1)}+2x^{2m+1}+1=a^2+2a+1$
$\Rightarrow$$(a+1)^2$ $\Rightarrow$$(x^{2m+1}+1)^2$
So, $x+1|x^{2m+1}+1$ $\Rightarrow$$(x+1)^2|x^{2(2m+1)}+2x^{2m+1}+1$
A: Write $t=x+1$. Then you have to show that constatnt and linear part of the polynomial $p(t)=(t-1)^{2(2n+1)}+2(t-1)^{2n+1}+1$ are zero.
To show this we will use the binomial theorem:
\begin{eqnarray*}
  p(t) &=& ...-{4n+2\choose 4n+1}t+{4n+2\choose 4n+2}-2\Big[...+{2n+1\choose 2n}t+{2n+1\choose 2n+1} \Big]+1\\
&=& ... -(4n+2)t+1-2\Big[...(2n+1)t+1\Big]-1\\
&=& ... + 0t+0
\end{eqnarray*}
