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Suppose I have $$x^{(c(p-1))} \equiv y^{(p-1)} \pmod{p}.$$ I would like to take the (p-1) root of both sides to get: $$x^c \equiv y \pmod{p}$$

I really just want to know if this a valid technique and what it would take to make it rigorous (I already know to show that $x,y$ are not congruent to 0)?

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@user10723: Nothing will work. For any $x$, $y$ not congruent to $0$ modulo $p$, each side of your first line is congruent to $1$, by Fermat's (little) Theorem. So we can't recover $(p-1)$-th roots, almost everything is a $(p-1)$-th root of $y$. –  André Nicolas May 10 '11 at 13:30
    
@quanta: I think it should be $2^{4\cdot4}$. –  joriki May 10 '11 at 13:33

2 Answers 2

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This is not true. If $p$ is a prime, both sides of the identity are $1$ (mod $p$), and hence equal, independent of the values of $x$, $y$ and $c$ (as long as $x$ and $y$ are not $0$ (mod $p$)).

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Modulo $\rm\:p\:,\ \ x\not\equiv 0\ \Rightarrow\ x^{p-1}\equiv 1\:.\:$ Thus the map $\rm\:x\to x^{p-1}\:$ is not one-one so not invertible. But we do have $\rm\:x^p\equiv x\:,\:$ therefore $\rm\:x^p \equiv y^p\ \iff\ x\equiv y\:.\:$ Perhaps this is what you intended?

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This is assuming that $p$ is prime (as the letter suggests). –  joriki May 10 '11 at 20:51
    
@Jor That's the OP's hypothesis (see the question title). –  Bill Dubuque May 10 '11 at 21:13

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