A really short task:
Calculate GCD$(x^4+x+1,x^3+x^2)$ and a Bezout Identity in $\mathbb{F_2}.$
I've tried it but my GCD is $1$ and I cannot see where my mistake is.
$x^4+x+1= x \cdot (x^3+x^2) + x^3 +x + 1$
$x^3+x^2 = 1 \cdot (x^3 + x + 1) + x^2 + x + 1$
$x^3+x+1 = x \cdot (x^2 + 1 + x) + x^2 + 1$
$x^2+x+1 = 1 \cdot (x^2+1) + x$
$x^2 + 1 = x \cdot x + 1$
$x = 1 \cdot x + 0$
This is easy when using the augmented-matrix form of the extended Euclidean algorithm.
$\begin{eqnarray} (1)&& &&x^4\!+x+1 \,&=&\, \left<\,\color{#c00}1,\color{#0a0}0\,\right>\ \ \ {\rm i.e.}\,\ \ x^4\!+x+1 = \color{#c00}1\cdot (x^4\!+x+1) + \color{#0a0}0\cdot(x^3\!+x^2)\\ (2)&& && x^3\!+x^2 \,&=&\, \left<\,\color{#c00}0,\color{#0a0}1\,\right>\ \ \ {\rm i.e.}\,\quad\ \ \ x^3\!+x^2 = \color{#c00}0\cdot (x^4\!+x+1) + \color{#0a0}1\cdot(x^3\!+x^2)\\ (3)&=&(1)-x\cdot (2)\quad && x^3\!+x+1 \,&=&\, \left<\,1,\,x\,\right>\\ (4)&=&(2)-(3)\quad && x^2\!+x+1 \,&=&\, \left<\,1,\,x+1\,\right>\\ (5)&=&(2)-x\cdot(4)\quad &&\qquad\quad\ \ \ x \,&=&\, \left<x,\,x^2+x+1\right>\\ (6)&=&(4)-(x\!+\!1)\,(5)\!\!\!\! &&\qquad\quad\ \ \ 1 \,&=&\, \left<\color{#c00}{x^2+x+1},\ \color{#0a0}{x^3+x}\right> \end{eqnarray}$
The Bezout Identity is $\ 1\, =\, (\color{#c00}{x^2\!+x+1})(x^4\!+x+1) + (\color{#0a0}{x^3\!+x})(x^3\!+x^2)$
