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Let $F$ be an arbitrary field, I need to prove that $x^2 - x + 1$ divides $x^{10} - x^7 + x^4 + ax + b$ for some $a, b \in F$

The difficulty that I am currently facing is that since $F$ is an arbitrary field, $x^2 - x + 1$ might be irreducible over $F$ and so I cannot solve it by factoring $x^2 - x + 1$. I also try to use the division algorithm to prove it using contradiction by assuming $x^2 - x + 1$ don't divide it, and hopefully something would happen to the remainder but this leads to a dead end as well. So any help would be appreciated.

Bill Dubuque
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    Use long division to divide $x^{10}-x^7+x^4$ by $x^2-x+1$. The remainder has degree $<2$, so _____ – Jyrki Lahtonen Feb 03 '20 at 07:59
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    If you spot that $(x+1)(x^2-x+1)=x^3+1$ then (in any field) $x^3=(x+1)(x^2-x+1)-1$ and, working modulo $x^2-x+1$ (as you can because the remainder is what you are interested in) you get $x^3\equiv -1$. The "trick" involved here requires a little justification, but it does simplify the process of division. – Mark Bennet Feb 03 '20 at 08:13
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    Same as in the linked dupe, except here use the simpler multiple $,x^3+1,$ instead of $,x^5-1\ \ \ $ – Bill Dubuque Feb 03 '20 at 19:49

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The division algorithm produces $$x^{10}-x^7+x^4=(x^2-x+1)q(x)+r(x)$$ in any field (indeed, polynomial ring) $F$, where $r(x)$ is a polynomial with degree less than $2$, i.e. $ax+b$ for some $a,b\in F$. Then $x^{10}-x^7+x^4-r(x)$ is divisible by $x^2-x+1$.

Parcly Taxel
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$$x^{10} - x^7 + x^4 + ax + b\equiv -x-x-x+ax+b=(a-3)x+b. $$ Can you end it now?

Michael Rozenberg
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$$f(x)=x^4(x^6-x^3+1)+(ax+b)$$ Now reduce $f(x)$ by $x^2=x-1$ to have remainder $R(x)$ as $$\implies R(x)=(x-1)^2[(x-1)^3-x(x-1)+1]+(ax+b)$$ $$\implies R(x)=(x-1-2x+1)[(x-1)(x^2-2x+1-x)+1])+ax+b$$ $$\implies R(x)=(x-1)^2[(x-1)(-2x)+1]+ax+b$$ $$\implies R(x)=(x-1)^2[-2(x-1)+2x+1]+ax+b$$ $$\implies R)x)=3(x-1)^2+ax+b \implies R(x)=-3x+ax+b$$ $$\implies R(x)=(a-3)x+b=0\implies a=3,b=0$$

Z Ahmed
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