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 March 30th, 2014, 10:28 PM #1 Newbie   Joined: Mar 2014 From: Matrix Posts: 1 Thanks: 0 Prove divisibility of polynomial Prove that: $Q(x)= x^{2012} + x + 1$ is divisable by $R(x)= x^{2} + x + 1$ I've no idea how to even start.. have spent a couple of hours already Thanks in advance
 March 31st, 2014, 12:51 AM #2 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Hint : Prove by induction (or otherwise) that $x^{3n + 2} + x + 1$ is divisible by $x^2 + x + 1$ for all positive integer $n$. Thanks from mente oscura and duc
 March 31st, 2014, 02:01 AM #3 Senior Member     Joined: May 2013 From: España Posts: 151 Thanks: 4 Hello. And, a bit of brute force, would be worth?: $x^{2012}+x+1=x^{2012}-x^2+x^2+x+1=$ $=x^2(x^{1005}+1)(x^{1005}-1)+x^2+x+1=$ $=x^2(x^{1005}+1)(x-1)(x^{1004}+x^{1003}+ \cdots +1)+x^2+x+1=$ $=x^2(x^{1005}+1)(x-1)[x^{1002}(x^2+x+1)+x^{999}(x^2+x+1)+ \cdots +x^3(x^2+x+1)+x^2+x+1)]+x^2+x+1=$ $=x^2(x^{1005}+1)(x-1)(x^{1002}+x^{999}+ \cdots +x^3+1)(x^2+x+1)+(x^2+x+1)=$ $=[x^2(x^{1005}+1)(x-1)(x^{1002}+x^{999}+ \cdots +x^3+1)+1](x^2+x+1)=end$ Regards
 March 31st, 2014, 03:11 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,746 Thanks: 2133 If x is either root of x² + x + 1 = 0, x³ = x(x²) = -x² - x = 1. Hence x^2012 = x², and so the factor theorem implies that x² + x + 1 divides x^2012 + x + 1. Thanks from mathbalarka
 March 31st, 2014, 05:04 AM #5 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory You're the man, skipper, as usual. (clap)

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### proving divisibility of polynomials

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