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Can someone show me step by step of "Extended Euclidean Algorithm" Multiplicative inverse of polynomial

gcd of $(x^4+x+1)$ and $(x^2+1)$ over $gf(2^4)$?

what I did : $(x^4+x+1) = (x^2+1)(x^2-1)+(x+2)$

so do I remove the 2 ,since 2 mod 2 = 0?

$(x^2+1) = x(x)+1$?

Please show me the steps using the equation format, not the table that most people use..

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puppylord
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    So you have done the two steps necessary to show that $\gcd(x^4+x+1,x^2+1)=1$ in the ring $GF(2)[x]$. The rest goes just like when computing a modular inverse. – Jyrki Lahtonen Oct 27 '15 at 07:51
  • BTW I don't think it's quite kosher to write "over $GF(2^4)$" in this context. Your polynomials have coefficients in $GF(2)$. The GCD of two polynomials over their common field of coefficients $K$ is the same over any extension field $L$ of $K$. I do realize that this task naturally comes up, when you declare that $GF(2^4)=GF(2)[x]/\langle x^4+x+1\rangle$, and are looking for the inverse of the coset $x^2+1+\langle x^4+x+1\rangle$. – Jyrki Lahtonen Oct 27 '15 at 07:56

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