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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?

Matan Haller
  • 35

5 Answers5

3

Hint: $x^{2(2n+1)}+2x^{2n+1}+1 = (x^{2n+1}+1)^2$

Further hint:

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

lhf
  • 216,483
3

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: $$(x^{2n}-x^{2n-1}+x^{2n-2}-\ ...\ +x^2-x+1)^2.$$

Daniel P
  • 2,710
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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.

Teddy38
  • 3,309
2

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$

nonuser
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Love Invariants
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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*}

nonuser
  • 90,026