Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)?

share|cite|improve this question
@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
up vote 4 down vote accepted

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 $p$)).

share|cite|improve this answer

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?

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.