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Consider the finite field $\mathbb{F}_{\displaystyle{3^{23}}}$ which is constructed from the primitive polynomial ${\bf f}=x^{23}-x^3+1$ over $\mathbb{F}_3$. Let $\alpha$ be a root of $\bf f$. Suppose that $\beta$ is a polynomial based on the $\alpha$ such as $\beta=\alpha^{22}+\alpha^{21}+\alpha^{11}+1 $.

My question: Is there a method (except full search) to find a positive integer number $k$ such that $\beta=\alpha^{k}$?

I know my question is a kind of discrete logarithm problem, but I have no idea except exhaustive search to answer.

Thanks for any suggestions

user0410
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    Index calculus may work. Here is an example run, though in a much smaller field. I'm fairly sure I don't want to try that in your case (without a serious silicon aide) unless the polynomial factors easily. – Jyrki Lahtonen Dec 27 '18 at 06:12
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    $3^{23}-1=2\cdot47\cdot1001523179$, so Chinese remainder theorem is not going to really help. Baby step - giant step will give you square root order reduction. Pollard $\rho$ about the same. – Jyrki Lahtonen Dec 27 '18 at 06:15
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    Index calculus has a wikipedia page. See that for a general description and links to more. I jotted down that example partly to have a semiserious case as an example I can refer to. – Jyrki Lahtonen Dec 27 '18 at 06:28
  • @JyrkiLahtonen At first, I want to thank you professor Lahtonen because of useful link and notations that you mentioned. I have learned from you that finite fields may seem hard, but with some sweet tricks. Because of this, I want to ask my question clearly. Consider $\gamma$ is an element in $\mathbb{F}_{\displaystyle{3^{23}}}$ such that its order is $47$. Now, consider the following polynomial $$ \beta=\gamma^{22}+\gamma^{21}+\gamma^{19}+\gamma^{18}+\gamma^{15}-\gamma^{14}-\gamma^{13}+\gamma^{10}-\gamma^{9}-\gamma^{8}-\gamma^{7}+\gamma^{6}-\gamma^{3}- \gamma+1 $$ – user0410 Dec 27 '18 at 12:18
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    $\gamma$ generates that field, so every element of $\Bbb{F}_{3^{23}}$ can be written as a polynomial in $\gamma$. Also, this expression does not specify $\beta$ uniquely because there are $46$ primitive roots of unity of order $47$, split into two conjugacy classes. This calculation may be one for the computers. I'm afraid I don't see tricks for handling such a $\beta$. – Jyrki Lahtonen Dec 27 '18 at 16:58
  • @JyrkiLahtonen Maybe, it was better to wrote $$ \gamma=\alpha^{\displaystyle{\frac{3^{23}-1}{47}}} $$ One question. The following relation is the decomposition of the mentioned polynomial over $\mathbb{F}_3$. Now, how this spiting is used to find solution of question $$ (\gamma^{3}+\gamma^{2}-\gamma+1) (\gamma^{19}+\gamma^{17}-\gamma^{16}+\gamma^{14}-\gamma^{12}-\gamma^{11}-\gamma^{10}+\gamma^{9}-\gamma^{8}+\gamma^{7}-\gamma^{6}-\gamma^{5}+\gamma^{4}-\gamma^{2}+1) $$ Thanks – user0410 Dec 27 '18 at 17:44

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