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Ok, so maybe this is a basic task but I'm stuck:

$P(x) = (x+1)^{2n+1} + x^{n+2}$, and $Q(x) = x^2 + x + 1$,

I have to show that $P(x)$ is divisible by $Q(x)$ for every $n \in \Bbb N$.

I tried factoring and direct division (got stuck), and tried induction. Ok, I check for $n = 1$, (it's true), but I can't comperhanse how to deal with the next induction step. Maybe I'm totally on the wrong track here, thanks for help in advance.

Jyrki Lahtonen
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Jovan Veljković
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    For divisibility by $x^2+x+1$, the tricks outlined here by Bill Dubuque settle matters often. Here you first need to replace $x+1$ with $-x^2$ (as in lab's answer) to take advantage. BTW, this has nothing whatsoever to do with [tag:division-algebras]. Those are structures from a couple of years into university algebra. – Jyrki Lahtonen Apr 08 '17 at 08:01

2 Answers2

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HINT:

If $x^2+x+1=0, x+1=-x^2$ and $x^3-1=0$

$$(x+1)^{2n+1}+x^{n+2}=(-x^2)^{2n+1}+x^{n+2}=-x^{4n+2}+x^{n+2}=-x^{n+2}(x^{3n}-1)$$

Now $x^{3n}-1=(x^3)^n-1^n$ is divisible by $x^3-1$

lab bhattacharjee
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  • Ok, makes sense, but how do I state that $Q(x) = 0$? Or, why? I'm not intending to find roots of $Q(x)$ :/ edit: I know that if $(x-a)$ divides some $P(x)$ and there is a remainder $R$, then $P(a) = R$ – Jovan Veljković Apr 07 '17 at 10:58
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Induction on $n.$ Let $P_n(x)=(x+1)^{2n+1}+x^{n+2}.$ We have $P_0(x)=Q(x).$ We have $$P_{n+1}(x)=(x+1)^{2n+3}+x^{n+3}=$$ $$=P_n(x)(x+1)^2-x^{n+2}(x+1)^2+x^{n+3}=$$ $$=P_n(x)(x+1)^2-x^{n+2}Q(x).$$

DanielWainfleet
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