I'm not absolutely sure if I'm answering this correctly but here is the question.

Let $F=GF(2)$. find polynomials $u(x), v(x) \in F[X]$ satisfying $X^5 + X^2 = (X^3 + X + 1)u(x) + v(x)$

What I think is if $u(x)= x^2$, then $(X^3 + X +1)\cdot X^2 = X^5 + X^3 + X^2$, therefore $v(x)=-X^3$


Just perform division algorithm (long division). $u(x)$ is your quotient and $v(x)$ can be your remainder. I don't think the question is looking for any $u(x)$ and $v(x)$ because if that is the case you can always choose $u(x)=0$ and $v(x)$ as the given polynomial.

  • $\begingroup$ Would I be correct in saying tha u(x)=x^2 -1 and v(x)=(x+1)/(x^3+x+1) then? $\endgroup$ – cele Jun 18 '14 at 19:40
  • 1
    $\begingroup$ Your $u(x)$ is correct but $v(x)=x+1$ not $(x+1)/x^3+x+1$. $\endgroup$ – Anurag A Jun 18 '14 at 22:16

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