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Let $p$ be a prime and $b,k\in \mathbb{N}$, if $gcd(b,p)=1$ and $k|p-1$ then $x^k \equiv b \mod{p} $ has exactly $k$ incongruent solution.

Is the above statement correct, I feel like I had got a counterexample for this, but I don't remember now. Can anyone help to show either this is a correct statement or to find a counter-example.

sabeelmsk
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    maybe it's "at most $k$ incongruent solutions" instead of "exactly". For example, consider $x^2\equiv 2\mod 3$ – Alessandro Jan 04 '20 at 14:12
  • Thank you for the comment, Can we say that the number of incongruent solutions modulo p will be either 0 or $k$. In this example, we are getting no solution right? – sabeelmsk Jan 04 '20 at 14:46
  • https://proofwiki.org/wiki/Solution_of_Linear_Congruence/Number_of_Solutions – lab bhattacharjee Jan 04 '20 at 15:14
  • In this link, it is about linear congruence, here we want about a non-linear congruence – sabeelmsk Jan 04 '20 at 15:21
  • @sabeelmsk, exactly – Alessandro Jan 04 '20 at 18:06
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    notice that $x^k=b\mod p$ has a particular solution $\beta$, then every other solution is in the form $x=\beta\delta$, with $\delta^k=1\mod p$, so the problem is "how many solutions does $x^k\equiv 1\mod p$?", this is answered (partially) in https://math.stackexchange.com/questions/320910/number-of-solutions-to-the-congruence-xq-equiv-1-mod-p – Alessandro Jan 04 '20 at 18:21

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