How can we prove that $x^n$ isn't divisible by $g(x)=x^4+x+1$ without remainders?
I understand why for all $n<4$ it's working, but how with $n \ge 4$? How we can prove it formally?
Note: $x^n,\:g\left(x\right)\in \mathbb{Z}_2\left[x\right]$
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Since you haven't really showed any attempt, I'll just give a hint:
Suppose there were a polynomial $h(x)$ such that $h(x)\cdot g(x)=x^n$. Now think of the lowest and highest degree terms of the product $h(x)\cdot g(x)$.
paw88789
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$h(x)$ is the result with a remainder? – Mar 17 '21 at 13:03
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$4\le deg\left(h\left(x\right)\cdot g\left(x\right)\right)\le n+4$ ? – Mar 17 '21 at 13:23
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1$h(x)$ is a polynomial. – paw88789 Mar 17 '21 at 13:45
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$x^n$ splits completely in $\mathbb{Z}_2$, so if $x^4+x+1$ were a factor, it would split completely as well. But it has no roots in $\mathbb{Z}_2$.
rogerl
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Write $x^n = q_n(x) (x^4+x+1) + r_n(x)$. Then $$x^{n+4} = x^4 q_n(x) (x^4+x+1) + x ^4 r_n(x) = x^4 q_n(x) (x^4+x+1) + (x+1) r_n(x)$$ and so $r_{n+4} \equiv (x+1) r_n \bmod (x^4+x+1)$.
The result follows by induction because $r_n = x^n \ne 0 $ for $n=0,1,2,3$ and $x+1$ is a unit mod $(x^4+x+1)$.
lhf
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By Euclid: $\,\gcd(g,\color{#c00}x) = \gcd(g\bmod x,x) = \gcd(1,x)=1\,$ thus $\,\gcd(g,\color{#c00}{x^n}) = 1$
Or prime $\,x\nmid g\,$ so $\,g\,$ is coprime to $\,x\,$ so also to prime power $\,x^n$ (by unique factorization)
Bill Dubuque
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