Let $F$ be a finite field. Given an element $a^x$ in $F\setminus\{0\}$, could we find $a$?? I know that finding an integer $x$ is very hard problem (Discrete Logarithm Problem). However, I don't know the hardness of this problem. What's the problem's name? Thanks.
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What is $x$? Is it a natural number? And is it given or not? – Servaes May 18 '16 at 16:53
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sorry, I have many mistakes. x is also given and x is natural number – swain May 18 '16 at 16:55
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1Do you know that $F\backslash{0}$ is a cyclic group? – Servaes May 18 '16 at 16:56
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Related: http://math.stackexchange.com/questions/1177537 – Watson May 18 '16 at 17:05
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@Servaes Yeah, I studied abstract algebra – swain May 18 '16 at 17:05
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@Watson Why this link is related my question? I think they are different. – swain May 18 '16 at 17:10
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@swain : it is related when your finite field is $\Bbb F_p$, and when $x=2$, because the answer links to the Cipolla's and Tonelli-Shanks' algorithms, which are part of the computational number theory (related to your tags). – Watson May 18 '16 at 17:47
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Let $q = |F|$. Since $(F^*,\times)$ is a cyclic group of order $q-1$, you are basically asking when and how we can divide by $x$ in $\Bbb Z/(q-1)\Bbb Z$.
This depends heavily on $x$. If $x$ and $q-1$ are not coprime, then this is impossible (for example if $x=q-1$ well $a^x= 1$ forall $a \in F^*$ so knowing $a^x$ gives you absolutely no information on $a$).
If they are, then $x$ has an inverse $y$ modulo $q-1$, and dividing by $x$ is the same as multiplying by $y$ : in the context of $(F^*,\times)$ this means that $a = (a^x)^y$.
mercio
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Thanks. If x divide q-1 but not q-1, is the case also impossible?? I don't know the reason why this case is impossible – swain May 18 '16 at 17:04
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yes, if $xy = q-1$ then $xy = 0 = x0$ modulo $q-1$ so multiplication by $x$ is not injective and so it's not invertible ( multiplication by $x$ is then $x$-to-$1$) – mercio May 19 '16 at 11:40
