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I am working on finding the multiplicative reverse in $GF(2^8)$ using the Euclidean Algorithm but after reading multiple sources, I feel as though I am proceeding incorrectly. Using the irreducible polynomial $m(p)=x^8+x^4+x^3+x+1=0x11B$ I am trying to find the inverse of $x^6+x^4+x+1=0x53$

I know using long division (via http://www.wolframalpha.com/widgets/view.jsp?id=f396eaca9aaccbf858652bccc972324a) I get for the first step $$(x^8+x^4+x^3+x+1)=(x^6+x^4+x+1)*(x^2-1)+(2x^4-x^2+2x+2)$$ but do I keep the negatives and even coefficients? I can't seem to get a reasonable answer and all the examples I have seen use simpler numbers. I know the answer to be $x^7+x^6+x^3+x=0xCA$ I just cannot seem to get there.

Jimmy Dur
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thebigreddude
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    Your coefficients are modulo $2$, so $2 = 0$ and $-1 = 1$. – Andreas Caranti Jan 26 '16 at 21:29
  • To expand on @Andreas' comment: $x^8+x^4+x^3+x+1=(x^6+x^4+x+1)(x^2+1)+x^2$. – TonyK Jan 26 '16 at 21:36
  • Well at least I was on the right track. I am still a little confused then since that means $(x^6+x^4+x+1 )=(x^2)(x^4 )+(x^4+x+1)$ but that would mean $(x^2)=(x^4+x+1)(a)+(r)$ and then I am working with fractional exponents. I must be missing something because I should be getting a remainder of 1 at some point. – thebigreddude Jan 26 '16 at 22:00
  • Recall that the remainder (if non-zero) must have degree smaller that the divisor. – Andreas Caranti Jan 26 '16 at 22:10

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Here are the steps you should obtain.

\begin{align} &x^8+x^4+x^3+x+1 = (x^6+x^4+x_1+1) (x^2 + 1) + x^2\\ &x^6+x^4+x+1 = x^2 (x^4 + x^2) + x + 1\\ &x^2 = (x+1) x + 1. \end{align}

Andreas Caranti
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  • Ah, I have stared at this for a while but now it just slapped me across the face. Long division almost defeated me, of all things. – thebigreddude Jan 26 '16 at 22:23