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Is it possible to proove that solution exists for this equation ? $2^x\equiv 30930818124575525495\pmod {3^{41}}$

I know $x < 3^{41}$ if solution exists. I tried brute force approach couldn't find any solution up to $x < 10^{10}$. Any help will be appreciated

safarov
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2 Answers2

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Sticking to brute force methods:

According to wolfram alpha, the order of $2$ $\pmod {3^{41}}$ is $24315330918113857602$ But this is $2\times 3^{40}=\varphi(3^{41})$. Thus $2$ is a primitive root $\pmod {3^{41}}$. As your number is not divisible by $3$, we conclude that there is indeed a solution to your congruence. Granted, this approach doesn't tell us what the solution is.

Note: computing the order of $2$ isn't that hard. After all, it has to be either $2\times 3^n$ or $3^n$ for $n\in \{0,\cdots,40\}$. Even if you just check each case (not necessary), it's only $82$ cases.

lulu
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Hint $ $ Use this theorem: if $g$ is a primitive root modulo an odd prime $p,$ then $g$ is a primitive root modulo all powers $p^k,\,$ unless $g^{p-1}\equiv 1 \pmod{p^2}$; in that case, $\,g+p\,$ is.

Community
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Bill Dubuque
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  • I will look into this, thanks – safarov Feb 07 '17 at 21:51
  • @safarov Note that if you use this result then no large calculations are required, only a few seconds of mental arithmetic.Almost surely this is what was intended by the author of the problem (assuming it was an exercise). – Bill Dubuque Feb 07 '17 at 21:53
  • What about finding actual value of x? Is there any shortcut ? – safarov Feb 08 '17 at 04:46
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    @safarov Your question asks only about existence of a solution - which is solved easily by the above. As for computing a solution $x$, this is a difficult problem in general, know as the discrete logarithm problem. Follow the link in this answer for a good introduction. – Bill Dubuque Feb 08 '17 at 15:50