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Question: Show that the polynomial $f(x)=(x+1)^{2n} +(x+2)^n - 1$ is divisible by $g(x) = x^2+3x+2$, where $n$ is an integer.

I have tried to use mathematical induction. The basis case wasn't that difficult, but when it comes to the inductive step itself, I got a bit confused.

Is it possible to prove this by mathematical induction, and is binominal expansion required at that step?

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Artem, For some basic information about writing math at this site see e.g. here, here, here and here. – The Chaz 2.0 Nov 3 '12 at 18:58
up vote 10 down vote accepted

Putting the zeros of $x^2+3x+2=0$ i.e., $-1,-2$ one by one, in $f(x)=(x+1)^{2n}+(x+2)^n-1$

we get, $f(-1)=(-1+1)^{2n}+(-1+2)^n-1=0$

and $f(-2)=(-2+1)^{2n}+(-2+2)^n-1=0$

(i)So, using Remainder Theorem,

$(x+1)\mid f(x)$ and $(x+2)\mid f(x)\implies lcm(x+1,x+2)\mid f(x)$

(ii) Alternatively,


Now $x+1(=x+2-1)\mid \{(x+2)^{2n-1}-1\} $ as $(a-b)\mid (a^n-b^n)--->(1)$

So, $(x+1)\mid f(x).$


Using $(1),\{(x+1)^2\}^n-1$ is divisible by $(x+1)^2-1=x^2+2x=x(x+2)$

So, $(x+2)\mid f(x).$

$lcm(x+1,x+2)=(x+1)(x+2)=x^2+3x+2$ (prove this)

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Very nicely put... – SyndicatorBBB Nov 3 '12 at 18:56
@Guy, welcome anytime. – lab bhattacharjee Nov 3 '12 at 18:58
Thank you @lab bhattacharjee! – Artem Nov 3 '12 at 21:38
@Artem, my pleasure. – lab bhattacharjee Nov 4 '12 at 7:49

Hint $\rm\: mod\ g\!:\,\ e_i^2\! = e_i\Rightarrow\: e_1^n + e_2^n =\, e_1\! + e_2.\ $ Google (orthogonal) idempotents to learn more.

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Warning: Non-Induction Proof Put $z = x+1$ and use $z$ rather than $x$ but everything else is the same. So let's define $$f_n(z)=z^{2n} + (z+1)^n - 1.$$ Obviously $z|f_n(z)$ and $z$ is coprime to $z+1$ so we just need to show that $z+1|f_n(z)$ but that is equivalent to $-1$ being a root of the polynomial, which we verify it is by evaluation.

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I don't have an answer for you because I didn't exercise this stuff a long time but I did find a reference that might assist you.

Please try :

Using the remainder theorem ""

Hope It helps a little bit,


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