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I need to find algorithm for finding roots of irreducible polynomial $x^4+x^3+x^2+x+1$ over $GF(2)$ in the extension field $GF(2^4)$, which has $x^4+x+1$ as defining polynomial. I tried listing out all elements of $GF(2^4)$ and checking whether it is a root of $f(x)$ and verified that $f(x)$ have roots $t^3$, $t^3+t$, $t^3+t^2$ and $t^3+t^2+t+1$, where $t$ is root of $x^4+x+1$. Is there any other procedure for finding roots of $f(x)$?

I searched online and got this algorithm from IEEE. Can anyone explain this algorithm 1

Akhilesh Ajithan
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    If $t$ is a root of $X^4+X+1$ [you don't say so, but I assume you meant it] then $t$ has order $15$; the roots of $f(X)=(X^5)-1)/(X-1)$ are the elements of order $5$ so for sure $t^3$ is one of them. You now apply the automorphism $a\mapsto a^2$ to see that $t^6$ is also of order $5$, and $t^6=t^3+t^2$. Repeat the squaring and you'll get the other two roots. – ancient mathematician Nov 06 '23 at 07:41
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    For an explanation of the connection between your list of roots and the explanation provided by @ancientmathematician I invite you to also check out an old answer of mine, prepared for referrals like this. Hopefully it is readable after you notice that I denoted by $\gamma$ what you call $t$. – Jyrki Lahtonen Nov 06 '23 at 12:28
  • @ancientmathematician It is not always the case that the root $t$ will have order $p^n-1$. As an example, in $GF(3)[x]$, $x^2+1$ is an irreducible polynomial for constructing $GF(3^2)$ and root of $x^2+1$ has order 4, not 8. If $f(x)$ is not a cyclotomic polynomial, what will be the procedure for finding root. – Akhilesh Ajithan Nov 07 '23 at 07:11
  • @AkhileshAjithan Of course you are right but I didn't make a general statement: I made a specific claim about the root of a specific irreducible over $GF(2)$. If you choose to construct your field with a polynomial $g$ whose roots are not primitive then I'm not sure how to efficiently solve another irreducible $f$. – ancient mathematician Nov 07 '23 at 07:42

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