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Prove that $g(\alpha)$=0 if and only if $g'(\alpha)$=0

$g(t)=t^{11}+t^{10}+t^6+t^5+t^4+t^2+1$
$g'(t)=t^{11}+t^9+t^7+t^6+t^5+t+1$

where $\alpha \in F[t]$. We are working in standard binary space.

This is the answer:

$\alpha^{11}+\alpha^{10}+\alpha^6+\alpha^5+\alpha^4+\alpha^2+1=0$
if and only if
$1+\alpha^{-1}+\alpha^{-5}+\alpha^{-6}+\alpha^{11}+\alpha^{-7}+\alpha^{-9}+\alpha^{-11}=0$

I can't seem to get to it??

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    What is $F$? I'm assuming that it is the field of two elements? But you shouldn't keep such things as secrets, when asking a question.2012-05-02

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It seems to me that $g(t)$ and $g'(t)$ are reciprocals of each other: $ t^{11} g(1/t)=t^{11}(t^{-11}+t^{-10}+t^{-6}+t^{-5}+t^{-4}+t^{-2}+1)=1+t+t^5+t^6+t^7+t^9+t^{11}=g'(t). $ This does mean that if one of them has a zero in some field, then so does the other. However, you seem to claim that the zeros coincide (you denote the common zero by $\alpha$). This is false in general. It could be true in some particular case, if the minimal polynomial of $\alpha$ should be a factor of both $g$ and $g'$. That is not the case here, as a quick run of Euclid's algorithm shows that the two polynomials have no common factors.

[Edit] A correct claim is that if $g(\alpha)=0$, then $g'(1/\alpha)=0$. As $g(t)$ is irreducible in $F_2[t]$ so is $g'(t)$. They both have eleven zeros in the field $GF(2048)$, but none of them are common because $\gcd(g(t),g'(t))=1$. [/Edit]

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    argh ! I get completely ruined by typos. Sorry everyone ! But thank you in any case2012-05-02