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I know it's hard to find the $e$th root of a number mod $n=p_1*p_2$, and if it would be possible we could break RSA. But how hard it is to take an $e$th root mod $p$ where $p$ is a prime and $\gcd(e,p-1)=1$?

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    $\begingroup$ Hint: apply the same math as in RSA. $\endgroup$
    – fgrieu
    Dec 15 '15 at 16:51
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It is very easy.

$gcd(e,p-1)=1$ so there exist $k,t$ where $ek+t(p-1)=1$. Let $x$ be the $e$-th root of $y$, so $x^e=y \bmod p$.

$y^k=x^{ek}=x\cdot {(x^{p-1})}^{-t}=x \bmod p$.

Also in decryption of RSA we use this method.

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