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$x^4+x^3+1$ and $x^2+x+1$ are co-prime in $Z_{2}[x]$, so $[x^4+x^3+1]$ is a unit in $Z_2[x]/(x^2+x+1)$.

Then, $[x^4+x^3+1]$ has an inverse, which I thought I could find by finding $a,b,c,d$ satisfying

$(i)$ $(x^4+x^3+1)(ax+b)+(x^2+x+1)(cx+d)=1$,

where I choose linear factors $ax+b$ and $cx+d$, because any $f(x)$ $\in$ $Z_{2}[x]$ $\Rightarrow$ $f(x)/(x^2+x+1)=(x^2+x+2)g(x)+r(x)$, and $deg$ $r(x)<deg$ $(x^2+x+1)$ $\Rightarrow$ $f(x)$ $\equiv$ $r(x)$ $mod$ $(x^2+x+1)$.

However, when I try to solve $(i)$, I find the system of equations to be inconsistent (unsolvable). Where did I go wrong?

Per Christian Strøm
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  • It might be easier to first reduce the polynomial $x^4 + x^3 + 1$ modulo $x^2 + x + 1$ to a linear polynomial. You can do this by the polynomial division algorithm or by substituting in formulas for $x^4, x^3, x^2$. – CJ Dowd Apr 13 '24 at 01:00
  • @CJDowd Thank you. Euclidean algorithm gave me $x^4+x^3+1=(x^2+x+1)(x^2+1)+x$. Am I understanding you correctly if you mean this reduced form is more suitable to use in the $(i)$-equation above? – Per Christian Strøm Apr 13 '24 at 01:18
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    Use the linked extended Euclidean algorithm, or note $g=x^2+x+1\mid x^3-1$ so it suffices to invert $\bmod x^3-1!:\ f=x^4+x^3+1\equiv x+1+1\equiv x,,$ by $,x^3\equiv 1.,$ But $,x(x^2)\equiv 1\Rightarrow x^{-1}\equiv x^2,,$ so also $!\bmod g!:\ x^{-1}\equiv x^2\equiv -x-1\equiv x+1.\ \ $ – Bill Dubuque Apr 13 '24 at 01:27
  • Note: the above computation of $f^{-1} = x^{-1}\equiv x^2 \pmod{g = x^3-1}$ can be seen as an optimization of the extended Euclidean algorithm when inverting a linear polynomial $f$, where the algorithm terminates in a single step (since then $,g\bmod f,$ is constant, and we can read off the inverse from that (Bezout) equation), see here. – Bill Dubuque Apr 13 '24 at 01:41
  • @BillDubuque Thank you. I'm working on your replies. – Per Christian Strøm Apr 13 '24 at 01:48
  • Where you went wrong: in your undetermined Bezout equation $(i)$, a quintic plus cubic can't sum to $1$. You need to replace $cx+d$ by a cubic to remedy that. But using undetermined coef's is going to be much more work than either of the mentioned methods. – Bill Dubuque Apr 13 '24 at 01:49
  • @BillDubuque Thank you, I appreciate it! (Must practice congruence arithmetic more:) – Per Christian Strøm Apr 13 '24 at 02:07
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    We can also use Gauss's algorithm as below, where $,\color{#c00}{x+1} = (x^2+x+1)\div x$

    $$\bmod, x^2+x+1!:,\ \dfrac{1}x\equiv \dfrac{(\color{#c00}{x+1}),1}{(\color{#c00}{x+1}),x}\equiv \dfrac{{x+1}}{1}\qquad\qquad\qquad$$

     – Bill Dubuque Apr 13 '24 at 02:09

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