Let $p$ be a prime. We learned that the number of solutions to $x^k\equiv 1\pmod{p}$ is $\gcd(k, p-1)$ and I know how to prove it in case $k$ divides $p-1$.
But how do I prove it in the other case?
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Is $p$ considered to be a prime? – Arthur Dec 21 '15 at 14:20
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Note that $x^{jk}=(x^k)^j$ – Ben Grossmann Dec 21 '15 at 14:23
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@Arthur yes, thanks – Nesa Dec 21 '15 at 15:12
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Let $d=\gcd(k,p-1)$ and write $d=ku+(p-1)v$.
Then $x^d \equiv (x^k)^u (x^{p-1})^v \equiv (x^k)^u$. This proves that $x^k\equiv 1$ implies $x^d \equiv 1$.
On the other hand, $x^d\equiv 1$ implies $x^k \equiv 1$ simply because $d$ divides $k$.
Thus, the set of solutions of $x^k \equiv 1$ is the same as the set of solutions of $x^d\equiv 1$, and the problem has been reduced to the case when $k$ divides $p-1$.
lhf
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Using Discrete Logarithm, $k\cdot $ind$(x)\equiv0\pmod{p-1}$
Now use Linear Congruence Theorem
lab bhattacharjee
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