<s>[INST] 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?
 [/INST] 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:
$$(x^{2n}-x^{2n-1}+x^{2n-2}-\ ...\ +x^2-x+1)^2.$$
</s>