I managed to calculate the GCD($x^7 + x^6 + x^3 + x + 1$, $x^8 + x^4 + x^3 + x + 1$) and I did some steps of the xgcd but I don't know exactly how to 'group' the polynomials.
Field: $F2^8 / m(x)$
$m(x) = x^8 + x^4 + x^3 + x + 1$
GCD:
(1) $(x^8 + x^4 + x^3 + x + 1) / (x^7 + x^6 + x^3 + x + 1)$ -> Quotient = $x+1$ Remainder = $x^6 + x^2 + x$
(2) $(x^7 + x^6 + x^3 + x + 1) / (x^6 + x^2 + x)$ -> Quotient = $x+1$ Remainder = $1$
Remainder is 1 so it has inverse.
Now I want to compute the inverse.
(1) $ 1 = (x^7 + x^6 + x^3 + x + 1) + (x^6+x^2+x)*(x+1)$
(2) $ 1 = (x^7 + x^6 + x^3 + x + 1) + [(x^8 + x^4 + x^3 + x + 1) + (x^7 + x^6 + x^3 + x + 1) * (x+1)] * (x+1)$
(3) $ 1 = (x^7 + x^6 + x^3 + x + 1) + (x+1)*(x^8 + x^4 + x^3 + x + 1) + (x^7 + x^6 + x^3 + x + 1) * (x+1) * (x+1)$
(4) ??
I have one part of the Bezout's identity but I don't know how to continue to get the other.
