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How do you find the factors of $x^4+x+1$ in $GF(2^8)$ in terms of polynomials?

Let me explain, We have primitive irreducible polynomial $p(x)=x^2+x+1$ in $GF(2^2)$ which has root $\alpha^2+\alpha$ in $GF(2^4)$.

Example: Using $\alpha^2+\alpha$ in $p\left(\alpha^2+\alpha\right)=0 \mod x^4+x+1$ (where $x^4+x+1$ is primitive irreducible polynomial in $GF(2^4)$.

Then how we can find $x^4+x+1$ roots in $GF(2^8)$ that generate subgroup of degree $15$ which is simple emending of $GF(2^4)$ in $GF(2^8)$?

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  • $\begingroup$ It has no factors in this GF $\endgroup$ – sashas Dec 25 '14 at 7:59
  • $\begingroup$ The polynomial $x^4+x+1$ splits into linear factors already in the subfield $GF(2^4)\subset GF(2^8)$. Are you looking for a way to identify a copy of that subfield? Do you mean a subgroup of order 15? The answer depends on how you have constructed the field $GF(2^8)$. If you don't tell us how you did that, we cannot help. I can guess that you are using either the methof used in AES crypto or QuickResponse codes, but there are others. You need to co-operate a bit. $\endgroup$ – Jyrki Lahtonen Dec 25 '14 at 22:01
  • $\begingroup$ The material linked to in this question and my answer there may help. Emphasis on may, as you need to give the degree 8 polynomial $p(x)$ that you use when you construct $GF(256)$ as $GF(2)[x]/\langle p(x)\rangle$. $\endgroup$ – Jyrki Lahtonen Dec 25 '14 at 22:11
  • $\begingroup$ Dear please let me know how can i find the roots of that irreducible polynomial in GF(256) that gives me unity on 15th power $\endgroup$ – user203139 Jan 10 '15 at 9:15

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