This question is about the number of $k^{th}$ root modulo m. Consider the congruence equation $$x^k\equiv b\pmod m \tag {$\clubsuit$}$$
I want to ask two questions:
(a) Prove that if $\gcd(b, m) = 1$ and $\gcd(k, \varphi(m)) = 1$, there is exactly one $k^{th}$ root modulo m to ($\clubsuit$)
(b) If $\gcd(k, \varphi(m)) > 1$, {$\clubsuit$} has no $k^{th}$ root modulo m or it has at least two $k^{th}$ roots modulo m.
For (a), I can prove the result if the $k^{th}$ root modulo m is relatively prime to m or m is a product of distinct primes with the idea from Finding the $k^{th}$ root modulo m.
For (b), I have no idea what to do.
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Danny Siu
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For part (a) you're already there: if $\gcd(x,m)\neq1$ then you must have $\gcd(b,m)\neq1$; can you see why? – Steven Stadnicki Oct 29 '17 at 16:04
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@StevenStadnicki I see, it is because $b = qm + x^k$ for some integer q and hence $gcd(x, m)$ divides b. Then $gcd(b, m) \ge gcd(x,m)$. So, we must have $gcd(x, m) = 1$. And now I can prove (a), THX! – Danny Siu Oct 29 '17 at 16:22
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If $gcd(k,\phi(n))=d>1$ can we say that, the given congruence has either no solution or $d$ incongruent solutions. – sabeelmsk Jan 04 '20 at 14:05