BdMO National Higher Secondary 2015#3

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samiul_samin
Posts:1007
Joined:Sat Dec 09, 2017 1:32 pm
BdMO National Higher Secondary 2015#3

Unread post by samiul_samin » Wed Feb 20, 2019 10:34 pm

Let $n$ be a positive integer.Consider the polynomial $p(x)=x^2+x+1$. What is the remainder of $ x^3$ when divided by $x^2+x+1$.For what positive integers values of $n$ is $ x^{2n}+x^n+1$ divisible by $p(x)$?

samiul_samin
Posts:1007
Joined:Sat Dec 09, 2017 1:32 pm

Re: BdMO National Higher Secondary 2015#3

Unread post by samiul_samin » Mon Feb 25, 2019 11:59 pm

I know a different solution but I have got a nice solution from $AoPS$ .
That Solution :
We have ${{x}^{3}}-1\vdots ({{x}^{2}}+x+1)$, so the rest is $1$.
Let $a^2+a+1=0$, so $a^3=1$. We use the Bezout theoreme:
1) If $n=3k$, result ${{a}^{2n}}+{{a}^{n}}+1={{a}^{6k}}+{{a}^{3k}}+1=3\ne 0$, so it is not divisible.
2) If $n=3k+1$, result ${{a}^{2n}}+{{a}^{n}}+1={{a}^{6k+2}}+{{a}^{3k+1}}+1={{a}^{2}}+a+1=0$, so it is divisible.
3) If $n=3k+2$, result ${{a}^{2n}}+{{a}^{n}}+1={{a}^{6k+4}}+{{a}^{3k+2}}+1=a+{{a}^{2}}+1=0$, so it is divisible.

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