$(x^3+x+1)^{-1} \mod (x^4+x+1)$ over $\text{GF}(2)$
I understand well how to solve the equation without inverse but don't know how to solve it with inverse.
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1Just find a polynomial $g$ with $g(x)(x^3+x+1)\equiv 1 \bmod x^4+x+1$ using the Euclidean algorithm. – Dietrich Burde Oct 12 '21 at 14:13
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Or use a suitable discrete logarithm table. The table in my answer shows that $x^3+x+1\equiv x^7$. As $x^{15}\equiv1$ we find the inverse to be ______ (you fill in the blank). – Jyrki Lahtonen Oct 13 '21 at 20:00
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For a problem with bigger polynomials, I'd want to try fancier tools. But in this problem, I'd be tempted to compute
\begin{align} A &= (x^3+x+1)(1) \bmod x^4+x+1 \\ B &= (x^3+x+1)(x) \bmod x^4+x+1 \\ C &= (x^3+x+1)(x^2) \bmod x^4+x+1 \\ D &= (x^3+x+1)(x^3) \bmod x^4+x+1 \end{align}
and then see what linear combination of $A, B, C, D$ adds up to $1$. Then the inverse is the same linear combination of $1, x, x^2, x^3$.
(Some of these, like $A$, don't actually take work to compute. We can stop at $D$ because $x^4 \equiv x+1 \pmod{x^4+x+1}$, so a hypothetical $E$ would just be the same as $A+D$.)
Misha Lavrov
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Thanks for answering the question. May I ask where do you get the 1, x, x^2, x^3? – Vale Oct 13 '21 at 06:04
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Because any polynomial is a sum of $1, x, x^2, \dots$ with some coefficients. – Misha Lavrov Oct 13 '21 at 13:46
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Just perform the extended Euclidean algorithm in $\mathbf F_2$ to obtain a Bézout's relation between $X^4+X+1$ and $X^3+X+1$: \begin{array}{r| lll} r(X) & u & v & q \\ \hline X^4+X+1 & 0 & 1 \\ X^3+X+1&1 & 0 & X \\ \hline X^2+1 & X & 1 & X \\ 1 & 1 +X^2 & X \\ \hline \end{array}
Therefore, $(X^2+1)(X^3+X+1)+X(X^4+X+1)=0 $.
Bernard
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